47
Op-Amp By By AJAL.A.J ( ASSISTANT PROFESSOR ) AJAL.A.J ( ASSISTANT PROFESSOR ) ECE DEPARTMENT ECE DEPARTMENT MAIL: MAIL: [email protected] MOB: 8907305642 MOB: 8907305642
| Date post: | 01-Dec-2014 |
| Category: |
Engineering |
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Op-Amp
ByBy
AJALAJ ( ASSISTANT PROFESSOR )AJALAJ ( ASSISTANT PROFESSOR )ECE DEPARTMENTECE DEPARTMENTMAIL MAIL ec2reachgmailcomMOB 8907305642MOB 8907305642
WHAT
bull An Operational Amplifier (Op-Amp) is an integrated circuit that uses external voltage to amplify the input through a very high gain
Key WordsKey Words Op Amp Model Ideal Op Amp Op Amp transfer characteristic Feedback Virtual short
History of the Op-Amp
bull The Vacuum Tube Age
bull The First Op-Amp (1930 ndash 1940) Designed by Karl Swartzel for the Bell Labs M9 gun director
bull Uses 3 vacuum tubes only one input and plusmn 350 V to attain a gain of 90 dB
bull Loebe Julie then develops an Op-Amp with two inputs Inverting and Non-inverting
1950rsquos-1960rsquos
The end of Vacuum Tubes was built up during the 1950rsquos-1960rsquos to the advent of solid-state electronics
1The Transistor 2The Integrated Circuit3The Planar Process
1960s
bull 1960s beginning of the Solid State Op-Ampbull Example GAPR P45 (1961 ndash 1971)ndash Runs on plusmn 15 V but costs $118 for 1 ndash 4
bull The GAPR PP65 (1962) makes the Op-Amp into a circuit component as a potted module
1963
bull The solid-state decade saw a proliferation of Op-Ampsndash Model 121 High Speed FET family etc
bull Robert J Widlar develops the μA702 Monolithic IC Op-Amp (1963) and shortly after the μA709
bull Fairchild Semiconductor vs National Semiconductorndash National The LM101 (1967) and then the LM101A (1968) (both by
Widlar)
ndash Fairchild The ldquofamousrdquo μA741 (by Dave Fullager 1968) and then the μA748 (1969)
Mathematics of the Op-Amp
bull The gain of the Op-Amp itself is calculated asG = Vout(V+ ndash V-)
bull The maximum output is the power supply voltage
bull When used in a circuit the gain of the circuit (as opposed to the op-amp component) is
Av = VoutVin
Op-Amp Characteristics
bull Open-loop gain G is typically over 9000bull But closed-loop gain is much smaller
bull Rin is very large (MΩ or larger)bull Rout is small (75Ω or smaller)
bull Effective output impedance in closed loop is very small
Ideal Op-Amp Characteristics
bull Open-loop gain G is infinitebull Rin is infinite
bull Zero input currentbull Rout is zero
Ideal Op-Amp Analysis
To analyze an op-amp feedback circuitbull Assume no current flows into either input terminalbull Assume no current flows out of the output terminalbull Constrain V+ = V-
11
Ideal Vs Practical Op-AmpIdeal Practical
Open Loop gain A 105
Bandwidth BW 10-100Hz
Input Impedance Zin gt1M
Output Impedance Zout 0 10-100
Output Voltage VoutDepends only on Vd = (V+V)
Differential mode signal
Depends slightly on average input Vc = (V++V)2 Common-Mode signal
CMRR 10-100dB
+
~AVin
Vin Vout
Zout=0
Ideal op-amp
+
AVin
Vin Vout
Zout
~Zin
Practical op-amp
Practical opamp3 categories are considering
bull 3 categories are considering
Close-Loop Voltage Gain Input impedance Output impedance
Ref080114HKN Operational Amplifier 13
Ideal Op-Amp ApplicationsAnalysis Method Two ideal Op-Amp Properties(1) The voltage between V+ and V is zero V+ = V
(2) The current into both V+ and V termainals is zero
For ideal Op-Amp circuit(1) Write the kirchhoff node equation at the noninverting
terminal V+ (2) Write the kirchhoff node eqaution at the inverting
terminal V (3) Set V+ = V and solve for the desired closed-loop gain
2 TYPES
1 Inverting Amplifier 2 Non-Inverting Amplifier
1 Inverting Amplifier Analysis
virtual ground
2 Non-Inverting Amplifier Analysis
17
Op-Amp Properties(1) Infinite Open Loop gain
- The gain without feedback- Equal to differential gain- Zero common-mode gain- Pratically Gd = 20000 to 200000
(2) Infinite Input impedance- Input current ii ~0A- T- in high-grade op-amp - m-A input current in low-grade op-
amp
(3) Zero Output Impedance- act as perfect internal voltage source- No internal resistance- Output impedance in series with load- Reducing output voltage to the load- Practically Rout ~ 20-100
+
V1
V2
Vo
+
Vo
i1~0
i2~0
+
Rout
VoRload
outload
loadoload RR
RVV
Op-Amp Buffer
Vout = VinIsolates loading effects
A
High output impedance
B
Low input impedance
Op-Amp Differentiator
20
Op-Amp Differentiator
RCdt
dVv io
+
RC
VoVi
0to t1 t2
0
to t1 t2
Op-Amp Integrator
Op-Amp Summing Amplifier
Op-Amp Differential Amplifier
If R1 = R2 and Rf = Rg
Closed-loop gain
2
21
21
21
R
RRv
RRARAv
v
IN
INo
Closed-loop gain Af=vovin
Suppose A=106 R1=9R R2=R
10INo vv
Gain
INf vA
)1(2
1
R
RAf
Closed-loop gain determined by resistor ratio insensitive to A temperature
Negative feedback
1048774
)1(2
1
R
R
v
vA
i
of
1
2
R
R
v
vA
i
of
bull We can adjust the closed-loop gain by changing the ratio of R2 and R1
bull The closed-loop gain is (ideally) independent of op amp open-loop gain A (if A is large enough) and we can make it arbitrarily large or small and with the desired accuracy depending on the accuracy of the resistors
1048774
The terminal 1 is a virtual ground since terminal 2 is grounded
Inverting configuration
This is a classic example of what negative feedback does It takes an amplifier with very large gain and through negative feedback obtain a gain that is smaller stable and predictable In effect we have traded gain for accuracy This kind of trade off is common in electronic circuit designhellip as we will see more later
Negative feedback
1048774
Inverting configuration
Input ResistanceAssuming an ideal op amp (open-loop gain A = infinity) in the closed-loop inverting configuration the input resistance is R1
111 R
Rv
v
i
vR
i
iiin
Negative feedback
1
2
R
R
v
vA
i
of
Inverting configuration011
oi
in RRi
vR
We can model the closed-loop inverting amplifier (with A = infinite) with the following equivalent circuit using a voltage-controlled voltage sourcehellip
Negative feedback 1048774
29
Frequency-Gain Relationbull Ideally signals are amplified
from DC to the highest AC frequency
bull Practically bandwidth is limitedbull 741 family op-amp have an limit
bandwidth of few KHzbull Unity Gain frequency f1 the gain at
unitybull Cutoff frequency fc the gain drop by
3dB from dc gain Gd
(Voltage Gain)
(frequency)f1
Gd
0707Gd
fc0
1
GB Product f1 = Gd fc
20log(0707)=3dB
30
GB ProductExample Determine the cutoff frequency of an op-amp having a unit gain frequency f1 = 10 MHz and voltage differential gain Gd = 20VmV
Sol
Since f1 = 10 MHz
By using GB production equation
f1 = Gd fc
fc = f1 Gd = 10 MHz 20 VmV
= 10 106 20 103
= 500 Hz
(Voltage Gain)
(frequency)f1
Gd
0707Gd
fc0
1
10MHz
Hz
Op Amp transfer characteristic curve
saturation
)( vvAvAv io
active region
So far we have been looking at the amplification that can be achieved for relatively small (amplitude) signals For a fixed gain as we increase the input signal amplitude there is a limit to how large the output signal can be The output saturates as it approaches the positive andnegative power supply voltages In other words there is limited range across which the gain is linear vv Vvo
vv Vvo
Op Amp transfer characteristic curve
Applications of Op-AmpsFiltersTypesbullLow pass filterbullHigh pass filterbullBand pass filterbullCascading (2 or more filters connected together) R2
+
-
+
V0
__
+ Vcc
- Vcc
-+
R1
C
Low pass filter
Low pass filter Cutoff frequency
Low pass filter transfer function
Applications of Op-Ampsbull Electrocardiogram EKG Amplification
ndash Need to measure difference in voltage from lead 1 and lead 2ndash 60 Hz interference from electrical equipment
Applications of Op-Ampsbull Simple EKG circuit
ndash Uses differential amplifier to cancel common mode signal and amplify differential mode signal
bull Realistic EKG circuitndash Uses two non-inverting
amplifiers to first amplify voltage from each lead followed by differential amplifier
ndash Forms an ldquoinstrumentation amplifierrdquo
Strain Gauge
Use a Wheatstone bridge to determine the strain of an element by measuring the change in resistance of a
strain gauge
(No strain) Balanced Bridge R 1 = R 2
(Strain) Unbalanced Bridge R 1 ne R 2
Strain GaugeHalf-Bridge Arrangement
Using KCL at the inverting and non-inverting terminals of the op amp we find that ε ~ Vo = 2ΔR(Rf R2)
R + ΔR Rf
+
- +
V0
__
+ Vcc
- Vcc
-+
Rf
Vref
R
R - ΔR
R
Op amp used to amplify output from strain gauge
Applications of Op-Ampsbull Piezoelectric Transducerndash Used to measure force pressure accelerationndash Piezoelectric crystal generates an electric charge in
response to deformation
bull Use Charge Amplifierndash Just an integrator op-amp circuit
bullGoal is to have VSET = VOUT
bullRemember that VERROR = VSET ndash VSENSOR
bullOutput Process uses VERROR from the PID controller to adjust Vout such that it is ~VSET
P
I
D
Output Process
Sensor
VERRORVSET VOUT
VSENSOR
PID Controller ndash System Block Diagram
ApplicationsPID Controller ndash System Circuit Diagram
Source httpwwwecircuitcentercomCircuitsop_pidop_pidhtm
Calculates VERROR = -(VSET + VSENSOR)
Signal conditioning allows you to introduce a time delay which could
account for things like inertia
System to control
-VSENSOR
ApplicationsPID Controller ndash PID Controller Circuit Diagram
VERR
Adjust Change
Kp RP1 RP2
Ki RI CI
Kd RD CD
VERR PID
Applications of Op-Ampsbull Example of PI Control
Temperature Control
bull Thermal System we wish to automatically control the temperature of
bull Block Diagram of Control System
Applications of Op-Amps
bull Voltage Error Circuit
bull Proportional-Integral Control Circuit
bull Example of PI Control Temperature Control
44
Multiple Inputs(1) Kirchhoff node equation at V+
yields
(2) Kirchhoff node equation at V yields
(3) Setting V+ = Vndash yields
0V
0_
c
c
b
b
a
a
f
o
R
VV
R
VV
R
VV
R
VV
c
aj j
jf
c
c
b
b
a
afo R
VR
R
V
R
V
R
VRV
+
Rf
Va
VoRb
Ra
RcVb
Vc
Homework
1) Design a circuit to 50i
of V
VA
2) Find the vo=
Review Two fundamental Op Amp
Structure Af Input voltage ( )terminal
Feed back
( )terminal
Inverting Amp
_ _
Non inverting Amp
+ _
WHAT
bull An Operational Amplifier (Op-Amp) is an integrated circuit that uses external voltage to amplify the input through a very high gain
Key WordsKey Words Op Amp Model Ideal Op Amp Op Amp transfer characteristic Feedback Virtual short
History of the Op-Amp
bull The Vacuum Tube Age
bull The First Op-Amp (1930 ndash 1940) Designed by Karl Swartzel for the Bell Labs M9 gun director
bull Uses 3 vacuum tubes only one input and plusmn 350 V to attain a gain of 90 dB
bull Loebe Julie then develops an Op-Amp with two inputs Inverting and Non-inverting
1950rsquos-1960rsquos
The end of Vacuum Tubes was built up during the 1950rsquos-1960rsquos to the advent of solid-state electronics
1The Transistor 2The Integrated Circuit3The Planar Process
1960s
bull 1960s beginning of the Solid State Op-Ampbull Example GAPR P45 (1961 ndash 1971)ndash Runs on plusmn 15 V but costs $118 for 1 ndash 4
bull The GAPR PP65 (1962) makes the Op-Amp into a circuit component as a potted module
1963
bull The solid-state decade saw a proliferation of Op-Ampsndash Model 121 High Speed FET family etc
bull Robert J Widlar develops the μA702 Monolithic IC Op-Amp (1963) and shortly after the μA709
bull Fairchild Semiconductor vs National Semiconductorndash National The LM101 (1967) and then the LM101A (1968) (both by
Widlar)
ndash Fairchild The ldquofamousrdquo μA741 (by Dave Fullager 1968) and then the μA748 (1969)
Mathematics of the Op-Amp
bull The gain of the Op-Amp itself is calculated asG = Vout(V+ ndash V-)
bull The maximum output is the power supply voltage
bull When used in a circuit the gain of the circuit (as opposed to the op-amp component) is
Av = VoutVin
Op-Amp Characteristics
bull Open-loop gain G is typically over 9000bull But closed-loop gain is much smaller
bull Rin is very large (MΩ or larger)bull Rout is small (75Ω or smaller)
bull Effective output impedance in closed loop is very small
Ideal Op-Amp Characteristics
bull Open-loop gain G is infinitebull Rin is infinite
bull Zero input currentbull Rout is zero
Ideal Op-Amp Analysis
To analyze an op-amp feedback circuitbull Assume no current flows into either input terminalbull Assume no current flows out of the output terminalbull Constrain V+ = V-
11
Ideal Vs Practical Op-AmpIdeal Practical
Open Loop gain A 105
Bandwidth BW 10-100Hz
Input Impedance Zin gt1M
Output Impedance Zout 0 10-100
Output Voltage VoutDepends only on Vd = (V+V)
Differential mode signal
Depends slightly on average input Vc = (V++V)2 Common-Mode signal
CMRR 10-100dB
+
~AVin
Vin Vout
Zout=0
Ideal op-amp
+
AVin
Vin Vout
Zout
~Zin
Practical op-amp
Practical opamp3 categories are considering
bull 3 categories are considering
Close-Loop Voltage Gain Input impedance Output impedance
Ref080114HKN Operational Amplifier 13
Ideal Op-Amp ApplicationsAnalysis Method Two ideal Op-Amp Properties(1) The voltage between V+ and V is zero V+ = V
(2) The current into both V+ and V termainals is zero
For ideal Op-Amp circuit(1) Write the kirchhoff node equation at the noninverting
terminal V+ (2) Write the kirchhoff node eqaution at the inverting
terminal V (3) Set V+ = V and solve for the desired closed-loop gain
2 TYPES
1 Inverting Amplifier 2 Non-Inverting Amplifier
1 Inverting Amplifier Analysis
virtual ground
2 Non-Inverting Amplifier Analysis
17
Op-Amp Properties(1) Infinite Open Loop gain
- The gain without feedback- Equal to differential gain- Zero common-mode gain- Pratically Gd = 20000 to 200000
(2) Infinite Input impedance- Input current ii ~0A- T- in high-grade op-amp - m-A input current in low-grade op-
amp
(3) Zero Output Impedance- act as perfect internal voltage source- No internal resistance- Output impedance in series with load- Reducing output voltage to the load- Practically Rout ~ 20-100
+
V1
V2
Vo
+
Vo
i1~0
i2~0
+
Rout
VoRload
outload
loadoload RR
RVV
Op-Amp Buffer
Vout = VinIsolates loading effects
A
High output impedance
B
Low input impedance
Op-Amp Differentiator
20
Op-Amp Differentiator
RCdt
dVv io
+
RC
VoVi
0to t1 t2
0
to t1 t2
Op-Amp Integrator
Op-Amp Summing Amplifier
Op-Amp Differential Amplifier
If R1 = R2 and Rf = Rg
Closed-loop gain
2
21
21
21
R
RRv
RRARAv
v
IN
INo
Closed-loop gain Af=vovin
Suppose A=106 R1=9R R2=R
10INo vv
Gain
INf vA
)1(2
1
R
RAf
Closed-loop gain determined by resistor ratio insensitive to A temperature
Negative feedback
1048774
)1(2
1
R
R
v
vA
i
of
1
2
R
R
v
vA
i
of
bull We can adjust the closed-loop gain by changing the ratio of R2 and R1
bull The closed-loop gain is (ideally) independent of op amp open-loop gain A (if A is large enough) and we can make it arbitrarily large or small and with the desired accuracy depending on the accuracy of the resistors
1048774
The terminal 1 is a virtual ground since terminal 2 is grounded
Inverting configuration
This is a classic example of what negative feedback does It takes an amplifier with very large gain and through negative feedback obtain a gain that is smaller stable and predictable In effect we have traded gain for accuracy This kind of trade off is common in electronic circuit designhellip as we will see more later
Negative feedback
1048774
Inverting configuration
Input ResistanceAssuming an ideal op amp (open-loop gain A = infinity) in the closed-loop inverting configuration the input resistance is R1
111 R
Rv
v
i
vR
i
iiin
Negative feedback
1
2
R
R
v
vA
i
of
Inverting configuration011
oi
in RRi
vR
We can model the closed-loop inverting amplifier (with A = infinite) with the following equivalent circuit using a voltage-controlled voltage sourcehellip
Negative feedback 1048774
29
Frequency-Gain Relationbull Ideally signals are amplified
from DC to the highest AC frequency
bull Practically bandwidth is limitedbull 741 family op-amp have an limit
bandwidth of few KHzbull Unity Gain frequency f1 the gain at
unitybull Cutoff frequency fc the gain drop by
3dB from dc gain Gd
(Voltage Gain)
(frequency)f1
Gd
0707Gd
fc0
1
GB Product f1 = Gd fc
20log(0707)=3dB
30
GB ProductExample Determine the cutoff frequency of an op-amp having a unit gain frequency f1 = 10 MHz and voltage differential gain Gd = 20VmV
Sol
Since f1 = 10 MHz
By using GB production equation
f1 = Gd fc
fc = f1 Gd = 10 MHz 20 VmV
= 10 106 20 103
= 500 Hz
(Voltage Gain)
(frequency)f1
Gd
0707Gd
fc0
1
10MHz
Hz
Op Amp transfer characteristic curve
saturation
)( vvAvAv io
active region
So far we have been looking at the amplification that can be achieved for relatively small (amplitude) signals For a fixed gain as we increase the input signal amplitude there is a limit to how large the output signal can be The output saturates as it approaches the positive andnegative power supply voltages In other words there is limited range across which the gain is linear vv Vvo
vv Vvo
Op Amp transfer characteristic curve
Applications of Op-AmpsFiltersTypesbullLow pass filterbullHigh pass filterbullBand pass filterbullCascading (2 or more filters connected together) R2
+
-
+
V0
__
+ Vcc
- Vcc
-+
R1
C
Low pass filter
Low pass filter Cutoff frequency
Low pass filter transfer function
Applications of Op-Ampsbull Electrocardiogram EKG Amplification
ndash Need to measure difference in voltage from lead 1 and lead 2ndash 60 Hz interference from electrical equipment
Applications of Op-Ampsbull Simple EKG circuit
ndash Uses differential amplifier to cancel common mode signal and amplify differential mode signal
bull Realistic EKG circuitndash Uses two non-inverting
amplifiers to first amplify voltage from each lead followed by differential amplifier
ndash Forms an ldquoinstrumentation amplifierrdquo
Strain Gauge
Use a Wheatstone bridge to determine the strain of an element by measuring the change in resistance of a
strain gauge
(No strain) Balanced Bridge R 1 = R 2
(Strain) Unbalanced Bridge R 1 ne R 2
Strain GaugeHalf-Bridge Arrangement
Using KCL at the inverting and non-inverting terminals of the op amp we find that ε ~ Vo = 2ΔR(Rf R2)
R + ΔR Rf
+
- +
V0
__
+ Vcc
- Vcc
-+
Rf
Vref
R
R - ΔR
R
Op amp used to amplify output from strain gauge
Applications of Op-Ampsbull Piezoelectric Transducerndash Used to measure force pressure accelerationndash Piezoelectric crystal generates an electric charge in
response to deformation
bull Use Charge Amplifierndash Just an integrator op-amp circuit
bullGoal is to have VSET = VOUT
bullRemember that VERROR = VSET ndash VSENSOR
bullOutput Process uses VERROR from the PID controller to adjust Vout such that it is ~VSET
P
I
D
Output Process
Sensor
VERRORVSET VOUT
VSENSOR
PID Controller ndash System Block Diagram
ApplicationsPID Controller ndash System Circuit Diagram
Source httpwwwecircuitcentercomCircuitsop_pidop_pidhtm
Calculates VERROR = -(VSET + VSENSOR)
Signal conditioning allows you to introduce a time delay which could
account for things like inertia
System to control
-VSENSOR
ApplicationsPID Controller ndash PID Controller Circuit Diagram
VERR
Adjust Change
Kp RP1 RP2
Ki RI CI
Kd RD CD
VERR PID
Applications of Op-Ampsbull Example of PI Control
Temperature Control
bull Thermal System we wish to automatically control the temperature of
bull Block Diagram of Control System
Applications of Op-Amps
bull Voltage Error Circuit
bull Proportional-Integral Control Circuit
bull Example of PI Control Temperature Control
44
Multiple Inputs(1) Kirchhoff node equation at V+
yields
(2) Kirchhoff node equation at V yields
(3) Setting V+ = Vndash yields
0V
0_
c
c
b
b
a
a
f
o
R
VV
R
VV
R
VV
R
VV
c
aj j
jf
c
c
b
b
a
afo R
VR
R
V
R
V
R
VRV
+
Rf
Va
VoRb
Ra
RcVb
Vc
Homework
1) Design a circuit to 50i
of V
VA
2) Find the vo=
Review Two fundamental Op Amp
Structure Af Input voltage ( )terminal
Feed back
( )terminal
Inverting Amp
_ _
Non inverting Amp
+ _
History of the Op-Amp
bull The Vacuum Tube Age
bull The First Op-Amp (1930 ndash 1940) Designed by Karl Swartzel for the Bell Labs M9 gun director
bull Uses 3 vacuum tubes only one input and plusmn 350 V to attain a gain of 90 dB
bull Loebe Julie then develops an Op-Amp with two inputs Inverting and Non-inverting
1950rsquos-1960rsquos
The end of Vacuum Tubes was built up during the 1950rsquos-1960rsquos to the advent of solid-state electronics
1The Transistor 2The Integrated Circuit3The Planar Process
1960s
bull 1960s beginning of the Solid State Op-Ampbull Example GAPR P45 (1961 ndash 1971)ndash Runs on plusmn 15 V but costs $118 for 1 ndash 4
bull The GAPR PP65 (1962) makes the Op-Amp into a circuit component as a potted module
1963
bull The solid-state decade saw a proliferation of Op-Ampsndash Model 121 High Speed FET family etc
bull Robert J Widlar develops the μA702 Monolithic IC Op-Amp (1963) and shortly after the μA709
bull Fairchild Semiconductor vs National Semiconductorndash National The LM101 (1967) and then the LM101A (1968) (both by
Widlar)
ndash Fairchild The ldquofamousrdquo μA741 (by Dave Fullager 1968) and then the μA748 (1969)
Mathematics of the Op-Amp
bull The gain of the Op-Amp itself is calculated asG = Vout(V+ ndash V-)
bull The maximum output is the power supply voltage
bull When used in a circuit the gain of the circuit (as opposed to the op-amp component) is
Av = VoutVin
Op-Amp Characteristics
bull Open-loop gain G is typically over 9000bull But closed-loop gain is much smaller
bull Rin is very large (MΩ or larger)bull Rout is small (75Ω or smaller)
bull Effective output impedance in closed loop is very small
Ideal Op-Amp Characteristics
bull Open-loop gain G is infinitebull Rin is infinite
bull Zero input currentbull Rout is zero
Ideal Op-Amp Analysis
To analyze an op-amp feedback circuitbull Assume no current flows into either input terminalbull Assume no current flows out of the output terminalbull Constrain V+ = V-
11
Ideal Vs Practical Op-AmpIdeal Practical
Open Loop gain A 105
Bandwidth BW 10-100Hz
Input Impedance Zin gt1M
Output Impedance Zout 0 10-100
Output Voltage VoutDepends only on Vd = (V+V)
Differential mode signal
Depends slightly on average input Vc = (V++V)2 Common-Mode signal
CMRR 10-100dB
+
~AVin
Vin Vout
Zout=0
Ideal op-amp
+
AVin
Vin Vout
Zout
~Zin
Practical op-amp
Practical opamp3 categories are considering
bull 3 categories are considering
Close-Loop Voltage Gain Input impedance Output impedance
Ref080114HKN Operational Amplifier 13
Ideal Op-Amp ApplicationsAnalysis Method Two ideal Op-Amp Properties(1) The voltage between V+ and V is zero V+ = V
(2) The current into both V+ and V termainals is zero
For ideal Op-Amp circuit(1) Write the kirchhoff node equation at the noninverting
terminal V+ (2) Write the kirchhoff node eqaution at the inverting
terminal V (3) Set V+ = V and solve for the desired closed-loop gain
2 TYPES
1 Inverting Amplifier 2 Non-Inverting Amplifier
1 Inverting Amplifier Analysis
virtual ground
2 Non-Inverting Amplifier Analysis
17
Op-Amp Properties(1) Infinite Open Loop gain
- The gain without feedback- Equal to differential gain- Zero common-mode gain- Pratically Gd = 20000 to 200000
(2) Infinite Input impedance- Input current ii ~0A- T- in high-grade op-amp - m-A input current in low-grade op-
amp
(3) Zero Output Impedance- act as perfect internal voltage source- No internal resistance- Output impedance in series with load- Reducing output voltage to the load- Practically Rout ~ 20-100
+
V1
V2
Vo
+
Vo
i1~0
i2~0
+
Rout
VoRload
outload
loadoload RR
RVV
Op-Amp Buffer
Vout = VinIsolates loading effects
A
High output impedance
B
Low input impedance
Op-Amp Differentiator
20
Op-Amp Differentiator
RCdt
dVv io
+
RC
VoVi
0to t1 t2
0
to t1 t2
Op-Amp Integrator
Op-Amp Summing Amplifier
Op-Amp Differential Amplifier
If R1 = R2 and Rf = Rg
Closed-loop gain
2
21
21
21
R
RRv
RRARAv
v
IN
INo
Closed-loop gain Af=vovin
Suppose A=106 R1=9R R2=R
10INo vv
Gain
INf vA
)1(2
1
R
RAf
Closed-loop gain determined by resistor ratio insensitive to A temperature
Negative feedback
1048774
)1(2
1
R
R
v
vA
i
of
1
2
R
R
v
vA
i
of
bull We can adjust the closed-loop gain by changing the ratio of R2 and R1
bull The closed-loop gain is (ideally) independent of op amp open-loop gain A (if A is large enough) and we can make it arbitrarily large or small and with the desired accuracy depending on the accuracy of the resistors
1048774
The terminal 1 is a virtual ground since terminal 2 is grounded
Inverting configuration
This is a classic example of what negative feedback does It takes an amplifier with very large gain and through negative feedback obtain a gain that is smaller stable and predictable In effect we have traded gain for accuracy This kind of trade off is common in electronic circuit designhellip as we will see more later
Negative feedback
1048774
Inverting configuration
Input ResistanceAssuming an ideal op amp (open-loop gain A = infinity) in the closed-loop inverting configuration the input resistance is R1
111 R
Rv
v
i
vR
i
iiin
Negative feedback
1
2
R
R
v
vA
i
of
Inverting configuration011
oi
in RRi
vR
We can model the closed-loop inverting amplifier (with A = infinite) with the following equivalent circuit using a voltage-controlled voltage sourcehellip
Negative feedback 1048774
29
Frequency-Gain Relationbull Ideally signals are amplified
from DC to the highest AC frequency
bull Practically bandwidth is limitedbull 741 family op-amp have an limit
bandwidth of few KHzbull Unity Gain frequency f1 the gain at
unitybull Cutoff frequency fc the gain drop by
3dB from dc gain Gd
(Voltage Gain)
(frequency)f1
Gd
0707Gd
fc0
1
GB Product f1 = Gd fc
20log(0707)=3dB
30
GB ProductExample Determine the cutoff frequency of an op-amp having a unit gain frequency f1 = 10 MHz and voltage differential gain Gd = 20VmV
Sol
Since f1 = 10 MHz
By using GB production equation
f1 = Gd fc
fc = f1 Gd = 10 MHz 20 VmV
= 10 106 20 103
= 500 Hz
(Voltage Gain)
(frequency)f1
Gd
0707Gd
fc0
1
10MHz
Hz
Op Amp transfer characteristic curve
saturation
)( vvAvAv io
active region
So far we have been looking at the amplification that can be achieved for relatively small (amplitude) signals For a fixed gain as we increase the input signal amplitude there is a limit to how large the output signal can be The output saturates as it approaches the positive andnegative power supply voltages In other words there is limited range across which the gain is linear vv Vvo
vv Vvo
Op Amp transfer characteristic curve
Applications of Op-AmpsFiltersTypesbullLow pass filterbullHigh pass filterbullBand pass filterbullCascading (2 or more filters connected together) R2
+
-
+
V0
__
+ Vcc
- Vcc
-+
R1
C
Low pass filter
Low pass filter Cutoff frequency
Low pass filter transfer function
Applications of Op-Ampsbull Electrocardiogram EKG Amplification
ndash Need to measure difference in voltage from lead 1 and lead 2ndash 60 Hz interference from electrical equipment
Applications of Op-Ampsbull Simple EKG circuit
ndash Uses differential amplifier to cancel common mode signal and amplify differential mode signal
bull Realistic EKG circuitndash Uses two non-inverting
amplifiers to first amplify voltage from each lead followed by differential amplifier
ndash Forms an ldquoinstrumentation amplifierrdquo
Strain Gauge
Use a Wheatstone bridge to determine the strain of an element by measuring the change in resistance of a
strain gauge
(No strain) Balanced Bridge R 1 = R 2
(Strain) Unbalanced Bridge R 1 ne R 2
Strain GaugeHalf-Bridge Arrangement
Using KCL at the inverting and non-inverting terminals of the op amp we find that ε ~ Vo = 2ΔR(Rf R2)
R + ΔR Rf
+
- +
V0
__
+ Vcc
- Vcc
-+
Rf
Vref
R
R - ΔR
R
Op amp used to amplify output from strain gauge
Applications of Op-Ampsbull Piezoelectric Transducerndash Used to measure force pressure accelerationndash Piezoelectric crystal generates an electric charge in
response to deformation
bull Use Charge Amplifierndash Just an integrator op-amp circuit
bullGoal is to have VSET = VOUT
bullRemember that VERROR = VSET ndash VSENSOR
bullOutput Process uses VERROR from the PID controller to adjust Vout such that it is ~VSET
P
I
D
Output Process
Sensor
VERRORVSET VOUT
VSENSOR
PID Controller ndash System Block Diagram
ApplicationsPID Controller ndash System Circuit Diagram
Source httpwwwecircuitcentercomCircuitsop_pidop_pidhtm
Calculates VERROR = -(VSET + VSENSOR)
Signal conditioning allows you to introduce a time delay which could
account for things like inertia
System to control
-VSENSOR
ApplicationsPID Controller ndash PID Controller Circuit Diagram
VERR
Adjust Change
Kp RP1 RP2
Ki RI CI
Kd RD CD
VERR PID
Applications of Op-Ampsbull Example of PI Control
Temperature Control
bull Thermal System we wish to automatically control the temperature of
bull Block Diagram of Control System
Applications of Op-Amps
bull Voltage Error Circuit
bull Proportional-Integral Control Circuit
bull Example of PI Control Temperature Control
44
Multiple Inputs(1) Kirchhoff node equation at V+
yields
(2) Kirchhoff node equation at V yields
(3) Setting V+ = Vndash yields
0V
0_
c
c
b
b
a
a
f
o
R
VV
R
VV
R
VV
R
VV
c
aj j
jf
c
c
b
b
a
afo R
VR
R
V
R
V
R
VRV
+
Rf
Va
VoRb
Ra
RcVb
Vc
Homework
1) Design a circuit to 50i
of V
VA
2) Find the vo=
Review Two fundamental Op Amp
Structure Af Input voltage ( )terminal
Feed back
( )terminal
Inverting Amp
_ _
Non inverting Amp
+ _
1950rsquos-1960rsquos
The end of Vacuum Tubes was built up during the 1950rsquos-1960rsquos to the advent of solid-state electronics
1The Transistor 2The Integrated Circuit3The Planar Process
1960s
bull 1960s beginning of the Solid State Op-Ampbull Example GAPR P45 (1961 ndash 1971)ndash Runs on plusmn 15 V but costs $118 for 1 ndash 4
bull The GAPR PP65 (1962) makes the Op-Amp into a circuit component as a potted module
1963
bull The solid-state decade saw a proliferation of Op-Ampsndash Model 121 High Speed FET family etc
bull Robert J Widlar develops the μA702 Monolithic IC Op-Amp (1963) and shortly after the μA709
bull Fairchild Semiconductor vs National Semiconductorndash National The LM101 (1967) and then the LM101A (1968) (both by
Widlar)
ndash Fairchild The ldquofamousrdquo μA741 (by Dave Fullager 1968) and then the μA748 (1969)
Mathematics of the Op-Amp
bull The gain of the Op-Amp itself is calculated asG = Vout(V+ ndash V-)
bull The maximum output is the power supply voltage
bull When used in a circuit the gain of the circuit (as opposed to the op-amp component) is
Av = VoutVin
Op-Amp Characteristics
bull Open-loop gain G is typically over 9000bull But closed-loop gain is much smaller
bull Rin is very large (MΩ or larger)bull Rout is small (75Ω or smaller)
bull Effective output impedance in closed loop is very small
Ideal Op-Amp Characteristics
bull Open-loop gain G is infinitebull Rin is infinite
bull Zero input currentbull Rout is zero
Ideal Op-Amp Analysis
To analyze an op-amp feedback circuitbull Assume no current flows into either input terminalbull Assume no current flows out of the output terminalbull Constrain V+ = V-
11
Ideal Vs Practical Op-AmpIdeal Practical
Open Loop gain A 105
Bandwidth BW 10-100Hz
Input Impedance Zin gt1M
Output Impedance Zout 0 10-100
Output Voltage VoutDepends only on Vd = (V+V)
Differential mode signal
Depends slightly on average input Vc = (V++V)2 Common-Mode signal
CMRR 10-100dB
+
~AVin
Vin Vout
Zout=0
Ideal op-amp
+
AVin
Vin Vout
Zout
~Zin
Practical op-amp
Practical opamp3 categories are considering
bull 3 categories are considering
Close-Loop Voltage Gain Input impedance Output impedance
Ref080114HKN Operational Amplifier 13
Ideal Op-Amp ApplicationsAnalysis Method Two ideal Op-Amp Properties(1) The voltage between V+ and V is zero V+ = V
(2) The current into both V+ and V termainals is zero
For ideal Op-Amp circuit(1) Write the kirchhoff node equation at the noninverting
terminal V+ (2) Write the kirchhoff node eqaution at the inverting
terminal V (3) Set V+ = V and solve for the desired closed-loop gain
2 TYPES
1 Inverting Amplifier 2 Non-Inverting Amplifier
1 Inverting Amplifier Analysis
virtual ground
2 Non-Inverting Amplifier Analysis
17
Op-Amp Properties(1) Infinite Open Loop gain
- The gain without feedback- Equal to differential gain- Zero common-mode gain- Pratically Gd = 20000 to 200000
(2) Infinite Input impedance- Input current ii ~0A- T- in high-grade op-amp - m-A input current in low-grade op-
amp
(3) Zero Output Impedance- act as perfect internal voltage source- No internal resistance- Output impedance in series with load- Reducing output voltage to the load- Practically Rout ~ 20-100
+
V1
V2
Vo
+
Vo
i1~0
i2~0
+
Rout
VoRload
outload
loadoload RR
RVV
Op-Amp Buffer
Vout = VinIsolates loading effects
A
High output impedance
B
Low input impedance
Op-Amp Differentiator
20
Op-Amp Differentiator
RCdt
dVv io
+
RC
VoVi
0to t1 t2
0
to t1 t2
Op-Amp Integrator
Op-Amp Summing Amplifier
Op-Amp Differential Amplifier
If R1 = R2 and Rf = Rg
Closed-loop gain
2
21
21
21
R
RRv
RRARAv
v
IN
INo
Closed-loop gain Af=vovin
Suppose A=106 R1=9R R2=R
10INo vv
Gain
INf vA
)1(2
1
R
RAf
Closed-loop gain determined by resistor ratio insensitive to A temperature
Negative feedback
1048774
)1(2
1
R
R
v
vA
i
of
1
2
R
R
v
vA
i
of
bull We can adjust the closed-loop gain by changing the ratio of R2 and R1
bull The closed-loop gain is (ideally) independent of op amp open-loop gain A (if A is large enough) and we can make it arbitrarily large or small and with the desired accuracy depending on the accuracy of the resistors
1048774
The terminal 1 is a virtual ground since terminal 2 is grounded
Inverting configuration
This is a classic example of what negative feedback does It takes an amplifier with very large gain and through negative feedback obtain a gain that is smaller stable and predictable In effect we have traded gain for accuracy This kind of trade off is common in electronic circuit designhellip as we will see more later
Negative feedback
1048774
Inverting configuration
Input ResistanceAssuming an ideal op amp (open-loop gain A = infinity) in the closed-loop inverting configuration the input resistance is R1
111 R
Rv
v
i
vR
i
iiin
Negative feedback
1
2
R
R
v
vA
i
of
Inverting configuration011
oi
in RRi
vR
We can model the closed-loop inverting amplifier (with A = infinite) with the following equivalent circuit using a voltage-controlled voltage sourcehellip
Negative feedback 1048774
29
Frequency-Gain Relationbull Ideally signals are amplified
from DC to the highest AC frequency
bull Practically bandwidth is limitedbull 741 family op-amp have an limit
bandwidth of few KHzbull Unity Gain frequency f1 the gain at
unitybull Cutoff frequency fc the gain drop by
3dB from dc gain Gd
(Voltage Gain)
(frequency)f1
Gd
0707Gd
fc0
1
GB Product f1 = Gd fc
20log(0707)=3dB
30
GB ProductExample Determine the cutoff frequency of an op-amp having a unit gain frequency f1 = 10 MHz and voltage differential gain Gd = 20VmV
Sol
Since f1 = 10 MHz
By using GB production equation
f1 = Gd fc
fc = f1 Gd = 10 MHz 20 VmV
= 10 106 20 103
= 500 Hz
(Voltage Gain)
(frequency)f1
Gd
0707Gd
fc0
1
10MHz
Hz
Op Amp transfer characteristic curve
saturation
)( vvAvAv io
active region
So far we have been looking at the amplification that can be achieved for relatively small (amplitude) signals For a fixed gain as we increase the input signal amplitude there is a limit to how large the output signal can be The output saturates as it approaches the positive andnegative power supply voltages In other words there is limited range across which the gain is linear vv Vvo
vv Vvo
Op Amp transfer characteristic curve
Applications of Op-AmpsFiltersTypesbullLow pass filterbullHigh pass filterbullBand pass filterbullCascading (2 or more filters connected together) R2
+
-
+
V0
__
+ Vcc
- Vcc
-+
R1
C
Low pass filter
Low pass filter Cutoff frequency
Low pass filter transfer function
Applications of Op-Ampsbull Electrocardiogram EKG Amplification
ndash Need to measure difference in voltage from lead 1 and lead 2ndash 60 Hz interference from electrical equipment
Applications of Op-Ampsbull Simple EKG circuit
ndash Uses differential amplifier to cancel common mode signal and amplify differential mode signal
bull Realistic EKG circuitndash Uses two non-inverting
amplifiers to first amplify voltage from each lead followed by differential amplifier
ndash Forms an ldquoinstrumentation amplifierrdquo
Strain Gauge
Use a Wheatstone bridge to determine the strain of an element by measuring the change in resistance of a
strain gauge
(No strain) Balanced Bridge R 1 = R 2
(Strain) Unbalanced Bridge R 1 ne R 2
Strain GaugeHalf-Bridge Arrangement
Using KCL at the inverting and non-inverting terminals of the op amp we find that ε ~ Vo = 2ΔR(Rf R2)
R + ΔR Rf
+
- +
V0
__
+ Vcc
- Vcc
-+
Rf
Vref
R
R - ΔR
R
Op amp used to amplify output from strain gauge
Applications of Op-Ampsbull Piezoelectric Transducerndash Used to measure force pressure accelerationndash Piezoelectric crystal generates an electric charge in
response to deformation
bull Use Charge Amplifierndash Just an integrator op-amp circuit
bullGoal is to have VSET = VOUT
bullRemember that VERROR = VSET ndash VSENSOR
bullOutput Process uses VERROR from the PID controller to adjust Vout such that it is ~VSET
P
I
D
Output Process
Sensor
VERRORVSET VOUT
VSENSOR
PID Controller ndash System Block Diagram
ApplicationsPID Controller ndash System Circuit Diagram
Source httpwwwecircuitcentercomCircuitsop_pidop_pidhtm
Calculates VERROR = -(VSET + VSENSOR)
Signal conditioning allows you to introduce a time delay which could
account for things like inertia
System to control
-VSENSOR
ApplicationsPID Controller ndash PID Controller Circuit Diagram
VERR
Adjust Change
Kp RP1 RP2
Ki RI CI
Kd RD CD
VERR PID
Applications of Op-Ampsbull Example of PI Control
Temperature Control
bull Thermal System we wish to automatically control the temperature of
bull Block Diagram of Control System
Applications of Op-Amps
bull Voltage Error Circuit
bull Proportional-Integral Control Circuit
bull Example of PI Control Temperature Control
44
Multiple Inputs(1) Kirchhoff node equation at V+
yields
(2) Kirchhoff node equation at V yields
(3) Setting V+ = Vndash yields
0V
0_
c
c
b
b
a
a
f
o
R
VV
R
VV
R
VV
R
VV
c
aj j
jf
c
c
b
b
a
afo R
VR
R
V
R
V
R
VRV
+
Rf
Va
VoRb
Ra
RcVb
Vc
Homework
1) Design a circuit to 50i
of V
VA
2) Find the vo=
Review Two fundamental Op Amp
Structure Af Input voltage ( )terminal
Feed back
( )terminal
Inverting Amp
_ _
Non inverting Amp
+ _
1960s
bull 1960s beginning of the Solid State Op-Ampbull Example GAPR P45 (1961 ndash 1971)ndash Runs on plusmn 15 V but costs $118 for 1 ndash 4
bull The GAPR PP65 (1962) makes the Op-Amp into a circuit component as a potted module
1963
bull The solid-state decade saw a proliferation of Op-Ampsndash Model 121 High Speed FET family etc
bull Robert J Widlar develops the μA702 Monolithic IC Op-Amp (1963) and shortly after the μA709
bull Fairchild Semiconductor vs National Semiconductorndash National The LM101 (1967) and then the LM101A (1968) (both by
Widlar)
ndash Fairchild The ldquofamousrdquo μA741 (by Dave Fullager 1968) and then the μA748 (1969)
Mathematics of the Op-Amp
bull The gain of the Op-Amp itself is calculated asG = Vout(V+ ndash V-)
bull The maximum output is the power supply voltage
bull When used in a circuit the gain of the circuit (as opposed to the op-amp component) is
Av = VoutVin
Op-Amp Characteristics
bull Open-loop gain G is typically over 9000bull But closed-loop gain is much smaller
bull Rin is very large (MΩ or larger)bull Rout is small (75Ω or smaller)
bull Effective output impedance in closed loop is very small
Ideal Op-Amp Characteristics
bull Open-loop gain G is infinitebull Rin is infinite
bull Zero input currentbull Rout is zero
Ideal Op-Amp Analysis
To analyze an op-amp feedback circuitbull Assume no current flows into either input terminalbull Assume no current flows out of the output terminalbull Constrain V+ = V-
11
Ideal Vs Practical Op-AmpIdeal Practical
Open Loop gain A 105
Bandwidth BW 10-100Hz
Input Impedance Zin gt1M
Output Impedance Zout 0 10-100
Output Voltage VoutDepends only on Vd = (V+V)
Differential mode signal
Depends slightly on average input Vc = (V++V)2 Common-Mode signal
CMRR 10-100dB
+
~AVin
Vin Vout
Zout=0
Ideal op-amp
+
AVin
Vin Vout
Zout
~Zin
Practical op-amp
Practical opamp3 categories are considering
bull 3 categories are considering
Close-Loop Voltage Gain Input impedance Output impedance
Ref080114HKN Operational Amplifier 13
Ideal Op-Amp ApplicationsAnalysis Method Two ideal Op-Amp Properties(1) The voltage between V+ and V is zero V+ = V
(2) The current into both V+ and V termainals is zero
For ideal Op-Amp circuit(1) Write the kirchhoff node equation at the noninverting
terminal V+ (2) Write the kirchhoff node eqaution at the inverting
terminal V (3) Set V+ = V and solve for the desired closed-loop gain
2 TYPES
1 Inverting Amplifier 2 Non-Inverting Amplifier
1 Inverting Amplifier Analysis
virtual ground
2 Non-Inverting Amplifier Analysis
17
Op-Amp Properties(1) Infinite Open Loop gain
- The gain without feedback- Equal to differential gain- Zero common-mode gain- Pratically Gd = 20000 to 200000
(2) Infinite Input impedance- Input current ii ~0A- T- in high-grade op-amp - m-A input current in low-grade op-
amp
(3) Zero Output Impedance- act as perfect internal voltage source- No internal resistance- Output impedance in series with load- Reducing output voltage to the load- Practically Rout ~ 20-100
+
V1
V2
Vo
+
Vo
i1~0
i2~0
+
Rout
VoRload
outload
loadoload RR
RVV
Op-Amp Buffer
Vout = VinIsolates loading effects
A
High output impedance
B
Low input impedance
Op-Amp Differentiator
20
Op-Amp Differentiator
RCdt
dVv io
+
RC
VoVi
0to t1 t2
0
to t1 t2
Op-Amp Integrator
Op-Amp Summing Amplifier
Op-Amp Differential Amplifier
If R1 = R2 and Rf = Rg
Closed-loop gain
2
21
21
21
R
RRv
RRARAv
v
IN
INo
Closed-loop gain Af=vovin
Suppose A=106 R1=9R R2=R
10INo vv
Gain
INf vA
)1(2
1
R
RAf
Closed-loop gain determined by resistor ratio insensitive to A temperature
Negative feedback
1048774
)1(2
1
R
R
v
vA
i
of
1
2
R
R
v
vA
i
of
bull We can adjust the closed-loop gain by changing the ratio of R2 and R1
bull The closed-loop gain is (ideally) independent of op amp open-loop gain A (if A is large enough) and we can make it arbitrarily large or small and with the desired accuracy depending on the accuracy of the resistors
1048774
The terminal 1 is a virtual ground since terminal 2 is grounded
Inverting configuration
This is a classic example of what negative feedback does It takes an amplifier with very large gain and through negative feedback obtain a gain that is smaller stable and predictable In effect we have traded gain for accuracy This kind of trade off is common in electronic circuit designhellip as we will see more later
Negative feedback
1048774
Inverting configuration
Input ResistanceAssuming an ideal op amp (open-loop gain A = infinity) in the closed-loop inverting configuration the input resistance is R1
111 R
Rv
v
i
vR
i
iiin
Negative feedback
1
2
R
R
v
vA
i
of
Inverting configuration011
oi
in RRi
vR
We can model the closed-loop inverting amplifier (with A = infinite) with the following equivalent circuit using a voltage-controlled voltage sourcehellip
Negative feedback 1048774
29
Frequency-Gain Relationbull Ideally signals are amplified
from DC to the highest AC frequency
bull Practically bandwidth is limitedbull 741 family op-amp have an limit
bandwidth of few KHzbull Unity Gain frequency f1 the gain at
unitybull Cutoff frequency fc the gain drop by
3dB from dc gain Gd
(Voltage Gain)
(frequency)f1
Gd
0707Gd
fc0
1
GB Product f1 = Gd fc
20log(0707)=3dB
30
GB ProductExample Determine the cutoff frequency of an op-amp having a unit gain frequency f1 = 10 MHz and voltage differential gain Gd = 20VmV
Sol
Since f1 = 10 MHz
By using GB production equation
f1 = Gd fc
fc = f1 Gd = 10 MHz 20 VmV
= 10 106 20 103
= 500 Hz
(Voltage Gain)
(frequency)f1
Gd
0707Gd
fc0
1
10MHz
Hz
Op Amp transfer characteristic curve
saturation
)( vvAvAv io
active region
So far we have been looking at the amplification that can be achieved for relatively small (amplitude) signals For a fixed gain as we increase the input signal amplitude there is a limit to how large the output signal can be The output saturates as it approaches the positive andnegative power supply voltages In other words there is limited range across which the gain is linear vv Vvo
vv Vvo
Op Amp transfer characteristic curve
Applications of Op-AmpsFiltersTypesbullLow pass filterbullHigh pass filterbullBand pass filterbullCascading (2 or more filters connected together) R2
+
-
+
V0
__
+ Vcc
- Vcc
-+
R1
C
Low pass filter
Low pass filter Cutoff frequency
Low pass filter transfer function
Applications of Op-Ampsbull Electrocardiogram EKG Amplification
ndash Need to measure difference in voltage from lead 1 and lead 2ndash 60 Hz interference from electrical equipment
Applications of Op-Ampsbull Simple EKG circuit
ndash Uses differential amplifier to cancel common mode signal and amplify differential mode signal
bull Realistic EKG circuitndash Uses two non-inverting
amplifiers to first amplify voltage from each lead followed by differential amplifier
ndash Forms an ldquoinstrumentation amplifierrdquo
Strain Gauge
Use a Wheatstone bridge to determine the strain of an element by measuring the change in resistance of a
strain gauge
(No strain) Balanced Bridge R 1 = R 2
(Strain) Unbalanced Bridge R 1 ne R 2
Strain GaugeHalf-Bridge Arrangement
Using KCL at the inverting and non-inverting terminals of the op amp we find that ε ~ Vo = 2ΔR(Rf R2)
R + ΔR Rf
+
- +
V0
__
+ Vcc
- Vcc
-+
Rf
Vref
R
R - ΔR
R
Op amp used to amplify output from strain gauge
Applications of Op-Ampsbull Piezoelectric Transducerndash Used to measure force pressure accelerationndash Piezoelectric crystal generates an electric charge in
response to deformation
bull Use Charge Amplifierndash Just an integrator op-amp circuit
bullGoal is to have VSET = VOUT
bullRemember that VERROR = VSET ndash VSENSOR
bullOutput Process uses VERROR from the PID controller to adjust Vout such that it is ~VSET
P
I
D
Output Process
Sensor
VERRORVSET VOUT
VSENSOR
PID Controller ndash System Block Diagram
ApplicationsPID Controller ndash System Circuit Diagram
Source httpwwwecircuitcentercomCircuitsop_pidop_pidhtm
Calculates VERROR = -(VSET + VSENSOR)
Signal conditioning allows you to introduce a time delay which could
account for things like inertia
System to control
-VSENSOR
ApplicationsPID Controller ndash PID Controller Circuit Diagram
VERR
Adjust Change
Kp RP1 RP2
Ki RI CI
Kd RD CD
VERR PID
Applications of Op-Ampsbull Example of PI Control
Temperature Control
bull Thermal System we wish to automatically control the temperature of
bull Block Diagram of Control System
Applications of Op-Amps
bull Voltage Error Circuit
bull Proportional-Integral Control Circuit
bull Example of PI Control Temperature Control
44
Multiple Inputs(1) Kirchhoff node equation at V+
yields
(2) Kirchhoff node equation at V yields
(3) Setting V+ = Vndash yields
0V
0_
c
c
b
b
a
a
f
o
R
VV
R
VV
R
VV
R
VV
c
aj j
jf
c
c
b
b
a
afo R
VR
R
V
R
V
R
VRV
+
Rf
Va
VoRb
Ra
RcVb
Vc
Homework
1) Design a circuit to 50i
of V
VA
2) Find the vo=
Review Two fundamental Op Amp
Structure Af Input voltage ( )terminal
Feed back
( )terminal
Inverting Amp
_ _
Non inverting Amp
+ _
1963
bull The solid-state decade saw a proliferation of Op-Ampsndash Model 121 High Speed FET family etc
bull Robert J Widlar develops the μA702 Monolithic IC Op-Amp (1963) and shortly after the μA709
bull Fairchild Semiconductor vs National Semiconductorndash National The LM101 (1967) and then the LM101A (1968) (both by
Widlar)
ndash Fairchild The ldquofamousrdquo μA741 (by Dave Fullager 1968) and then the μA748 (1969)
Mathematics of the Op-Amp
bull The gain of the Op-Amp itself is calculated asG = Vout(V+ ndash V-)
bull The maximum output is the power supply voltage
bull When used in a circuit the gain of the circuit (as opposed to the op-amp component) is
Av = VoutVin
Op-Amp Characteristics
bull Open-loop gain G is typically over 9000bull But closed-loop gain is much smaller
bull Rin is very large (MΩ or larger)bull Rout is small (75Ω or smaller)
bull Effective output impedance in closed loop is very small
Ideal Op-Amp Characteristics
bull Open-loop gain G is infinitebull Rin is infinite
bull Zero input currentbull Rout is zero
Ideal Op-Amp Analysis
To analyze an op-amp feedback circuitbull Assume no current flows into either input terminalbull Assume no current flows out of the output terminalbull Constrain V+ = V-
11
Ideal Vs Practical Op-AmpIdeal Practical
Open Loop gain A 105
Bandwidth BW 10-100Hz
Input Impedance Zin gt1M
Output Impedance Zout 0 10-100
Output Voltage VoutDepends only on Vd = (V+V)
Differential mode signal
Depends slightly on average input Vc = (V++V)2 Common-Mode signal
CMRR 10-100dB
+
~AVin
Vin Vout
Zout=0
Ideal op-amp
+
AVin
Vin Vout
Zout
~Zin
Practical op-amp
Practical opamp3 categories are considering
bull 3 categories are considering
Close-Loop Voltage Gain Input impedance Output impedance
Ref080114HKN Operational Amplifier 13
Ideal Op-Amp ApplicationsAnalysis Method Two ideal Op-Amp Properties(1) The voltage between V+ and V is zero V+ = V
(2) The current into both V+ and V termainals is zero
For ideal Op-Amp circuit(1) Write the kirchhoff node equation at the noninverting
terminal V+ (2) Write the kirchhoff node eqaution at the inverting
terminal V (3) Set V+ = V and solve for the desired closed-loop gain
2 TYPES
1 Inverting Amplifier 2 Non-Inverting Amplifier
1 Inverting Amplifier Analysis
virtual ground
2 Non-Inverting Amplifier Analysis
17
Op-Amp Properties(1) Infinite Open Loop gain
- The gain without feedback- Equal to differential gain- Zero common-mode gain- Pratically Gd = 20000 to 200000
(2) Infinite Input impedance- Input current ii ~0A- T- in high-grade op-amp - m-A input current in low-grade op-
amp
(3) Zero Output Impedance- act as perfect internal voltage source- No internal resistance- Output impedance in series with load- Reducing output voltage to the load- Practically Rout ~ 20-100
+
V1
V2
Vo
+
Vo
i1~0
i2~0
+
Rout
VoRload
outload
loadoload RR
RVV
Op-Amp Buffer
Vout = VinIsolates loading effects
A
High output impedance
B
Low input impedance
Op-Amp Differentiator
20
Op-Amp Differentiator
RCdt
dVv io
+
RC
VoVi
0to t1 t2
0
to t1 t2
Op-Amp Integrator
Op-Amp Summing Amplifier
Op-Amp Differential Amplifier
If R1 = R2 and Rf = Rg
Closed-loop gain
2
21
21
21
R
RRv
RRARAv
v
IN
INo
Closed-loop gain Af=vovin
Suppose A=106 R1=9R R2=R
10INo vv
Gain
INf vA
)1(2
1
R
RAf
Closed-loop gain determined by resistor ratio insensitive to A temperature
Negative feedback
1048774
)1(2
1
R
R
v
vA
i
of
1
2
R
R
v
vA
i
of
bull We can adjust the closed-loop gain by changing the ratio of R2 and R1
bull The closed-loop gain is (ideally) independent of op amp open-loop gain A (if A is large enough) and we can make it arbitrarily large or small and with the desired accuracy depending on the accuracy of the resistors
1048774
The terminal 1 is a virtual ground since terminal 2 is grounded
Inverting configuration
This is a classic example of what negative feedback does It takes an amplifier with very large gain and through negative feedback obtain a gain that is smaller stable and predictable In effect we have traded gain for accuracy This kind of trade off is common in electronic circuit designhellip as we will see more later
Negative feedback
1048774
Inverting configuration
Input ResistanceAssuming an ideal op amp (open-loop gain A = infinity) in the closed-loop inverting configuration the input resistance is R1
111 R
Rv
v
i
vR
i
iiin
Negative feedback
1
2
R
R
v
vA
i
of
Inverting configuration011
oi
in RRi
vR
We can model the closed-loop inverting amplifier (with A = infinite) with the following equivalent circuit using a voltage-controlled voltage sourcehellip
Negative feedback 1048774
29
Frequency-Gain Relationbull Ideally signals are amplified
from DC to the highest AC frequency
bull Practically bandwidth is limitedbull 741 family op-amp have an limit
bandwidth of few KHzbull Unity Gain frequency f1 the gain at
unitybull Cutoff frequency fc the gain drop by
3dB from dc gain Gd
(Voltage Gain)
(frequency)f1
Gd
0707Gd
fc0
1
GB Product f1 = Gd fc
20log(0707)=3dB
30
GB ProductExample Determine the cutoff frequency of an op-amp having a unit gain frequency f1 = 10 MHz and voltage differential gain Gd = 20VmV
Sol
Since f1 = 10 MHz
By using GB production equation
f1 = Gd fc
fc = f1 Gd = 10 MHz 20 VmV
= 10 106 20 103
= 500 Hz
(Voltage Gain)
(frequency)f1
Gd
0707Gd
fc0
1
10MHz
Hz
Op Amp transfer characteristic curve
saturation
)( vvAvAv io
active region
So far we have been looking at the amplification that can be achieved for relatively small (amplitude) signals For a fixed gain as we increase the input signal amplitude there is a limit to how large the output signal can be The output saturates as it approaches the positive andnegative power supply voltages In other words there is limited range across which the gain is linear vv Vvo
vv Vvo
Op Amp transfer characteristic curve
Applications of Op-AmpsFiltersTypesbullLow pass filterbullHigh pass filterbullBand pass filterbullCascading (2 or more filters connected together) R2
+
-
+
V0
__
+ Vcc
- Vcc
-+
R1
C
Low pass filter
Low pass filter Cutoff frequency
Low pass filter transfer function
Applications of Op-Ampsbull Electrocardiogram EKG Amplification
ndash Need to measure difference in voltage from lead 1 and lead 2ndash 60 Hz interference from electrical equipment
Applications of Op-Ampsbull Simple EKG circuit
ndash Uses differential amplifier to cancel common mode signal and amplify differential mode signal
bull Realistic EKG circuitndash Uses two non-inverting
amplifiers to first amplify voltage from each lead followed by differential amplifier
ndash Forms an ldquoinstrumentation amplifierrdquo
Strain Gauge
Use a Wheatstone bridge to determine the strain of an element by measuring the change in resistance of a
strain gauge
(No strain) Balanced Bridge R 1 = R 2
(Strain) Unbalanced Bridge R 1 ne R 2
Strain GaugeHalf-Bridge Arrangement
Using KCL at the inverting and non-inverting terminals of the op amp we find that ε ~ Vo = 2ΔR(Rf R2)
R + ΔR Rf
+
- +
V0
__
+ Vcc
- Vcc
-+
Rf
Vref
R
R - ΔR
R
Op amp used to amplify output from strain gauge
Applications of Op-Ampsbull Piezoelectric Transducerndash Used to measure force pressure accelerationndash Piezoelectric crystal generates an electric charge in
response to deformation
bull Use Charge Amplifierndash Just an integrator op-amp circuit
bullGoal is to have VSET = VOUT
bullRemember that VERROR = VSET ndash VSENSOR
bullOutput Process uses VERROR from the PID controller to adjust Vout such that it is ~VSET
P
I
D
Output Process
Sensor
VERRORVSET VOUT
VSENSOR
PID Controller ndash System Block Diagram
ApplicationsPID Controller ndash System Circuit Diagram
Source httpwwwecircuitcentercomCircuitsop_pidop_pidhtm
Calculates VERROR = -(VSET + VSENSOR)
Signal conditioning allows you to introduce a time delay which could
account for things like inertia
System to control
-VSENSOR
ApplicationsPID Controller ndash PID Controller Circuit Diagram
VERR
Adjust Change
Kp RP1 RP2
Ki RI CI
Kd RD CD
VERR PID
Applications of Op-Ampsbull Example of PI Control
Temperature Control
bull Thermal System we wish to automatically control the temperature of
bull Block Diagram of Control System
Applications of Op-Amps
bull Voltage Error Circuit
bull Proportional-Integral Control Circuit
bull Example of PI Control Temperature Control
44
Multiple Inputs(1) Kirchhoff node equation at V+
yields
(2) Kirchhoff node equation at V yields
(3) Setting V+ = Vndash yields
0V
0_
c
c
b
b
a
a
f
o
R
VV
R
VV
R
VV
R
VV
c
aj j
jf
c
c
b
b
a
afo R
VR
R
V
R
V
R
VRV
+
Rf
Va
VoRb
Ra
RcVb
Vc
Homework
1) Design a circuit to 50i
of V
VA
2) Find the vo=
Review Two fundamental Op Amp
Structure Af Input voltage ( )terminal
Feed back
( )terminal
Inverting Amp
_ _
Non inverting Amp
+ _
Mathematics of the Op-Amp
bull The gain of the Op-Amp itself is calculated asG = Vout(V+ ndash V-)
bull The maximum output is the power supply voltage
bull When used in a circuit the gain of the circuit (as opposed to the op-amp component) is
Av = VoutVin
Op-Amp Characteristics
bull Open-loop gain G is typically over 9000bull But closed-loop gain is much smaller
bull Rin is very large (MΩ or larger)bull Rout is small (75Ω or smaller)
bull Effective output impedance in closed loop is very small
Ideal Op-Amp Characteristics
bull Open-loop gain G is infinitebull Rin is infinite
bull Zero input currentbull Rout is zero
Ideal Op-Amp Analysis
To analyze an op-amp feedback circuitbull Assume no current flows into either input terminalbull Assume no current flows out of the output terminalbull Constrain V+ = V-
11
Ideal Vs Practical Op-AmpIdeal Practical
Open Loop gain A 105
Bandwidth BW 10-100Hz
Input Impedance Zin gt1M
Output Impedance Zout 0 10-100
Output Voltage VoutDepends only on Vd = (V+V)
Differential mode signal
Depends slightly on average input Vc = (V++V)2 Common-Mode signal
CMRR 10-100dB
+
~AVin
Vin Vout
Zout=0
Ideal op-amp
+
AVin
Vin Vout
Zout
~Zin
Practical op-amp
Practical opamp3 categories are considering
bull 3 categories are considering
Close-Loop Voltage Gain Input impedance Output impedance
Ref080114HKN Operational Amplifier 13
Ideal Op-Amp ApplicationsAnalysis Method Two ideal Op-Amp Properties(1) The voltage between V+ and V is zero V+ = V
(2) The current into both V+ and V termainals is zero
For ideal Op-Amp circuit(1) Write the kirchhoff node equation at the noninverting
terminal V+ (2) Write the kirchhoff node eqaution at the inverting
terminal V (3) Set V+ = V and solve for the desired closed-loop gain
2 TYPES
1 Inverting Amplifier 2 Non-Inverting Amplifier
1 Inverting Amplifier Analysis
virtual ground
2 Non-Inverting Amplifier Analysis
17
Op-Amp Properties(1) Infinite Open Loop gain
- The gain without feedback- Equal to differential gain- Zero common-mode gain- Pratically Gd = 20000 to 200000
(2) Infinite Input impedance- Input current ii ~0A- T- in high-grade op-amp - m-A input current in low-grade op-
amp
(3) Zero Output Impedance- act as perfect internal voltage source- No internal resistance- Output impedance in series with load- Reducing output voltage to the load- Practically Rout ~ 20-100
+
V1
V2
Vo
+
Vo
i1~0
i2~0
+
Rout
VoRload
outload
loadoload RR
RVV
Op-Amp Buffer
Vout = VinIsolates loading effects
A
High output impedance
B
Low input impedance
Op-Amp Differentiator
20
Op-Amp Differentiator
RCdt
dVv io
+
RC
VoVi
0to t1 t2
0
to t1 t2
Op-Amp Integrator
Op-Amp Summing Amplifier
Op-Amp Differential Amplifier
If R1 = R2 and Rf = Rg
Closed-loop gain
2
21
21
21
R
RRv
RRARAv
v
IN
INo
Closed-loop gain Af=vovin
Suppose A=106 R1=9R R2=R
10INo vv
Gain
INf vA
)1(2
1
R
RAf
Closed-loop gain determined by resistor ratio insensitive to A temperature
Negative feedback
1048774
)1(2
1
R
R
v
vA
i
of
1
2
R
R
v
vA
i
of
bull We can adjust the closed-loop gain by changing the ratio of R2 and R1
bull The closed-loop gain is (ideally) independent of op amp open-loop gain A (if A is large enough) and we can make it arbitrarily large or small and with the desired accuracy depending on the accuracy of the resistors
1048774
The terminal 1 is a virtual ground since terminal 2 is grounded
Inverting configuration
This is a classic example of what negative feedback does It takes an amplifier with very large gain and through negative feedback obtain a gain that is smaller stable and predictable In effect we have traded gain for accuracy This kind of trade off is common in electronic circuit designhellip as we will see more later
Negative feedback
1048774
Inverting configuration
Input ResistanceAssuming an ideal op amp (open-loop gain A = infinity) in the closed-loop inverting configuration the input resistance is R1
111 R
Rv
v
i
vR
i
iiin
Negative feedback
1
2
R
R
v
vA
i
of
Inverting configuration011
oi
in RRi
vR
We can model the closed-loop inverting amplifier (with A = infinite) with the following equivalent circuit using a voltage-controlled voltage sourcehellip
Negative feedback 1048774
29
Frequency-Gain Relationbull Ideally signals are amplified
from DC to the highest AC frequency
bull Practically bandwidth is limitedbull 741 family op-amp have an limit
bandwidth of few KHzbull Unity Gain frequency f1 the gain at
unitybull Cutoff frequency fc the gain drop by
3dB from dc gain Gd
(Voltage Gain)
(frequency)f1
Gd
0707Gd
fc0
1
GB Product f1 = Gd fc
20log(0707)=3dB
30
GB ProductExample Determine the cutoff frequency of an op-amp having a unit gain frequency f1 = 10 MHz and voltage differential gain Gd = 20VmV
Sol
Since f1 = 10 MHz
By using GB production equation
f1 = Gd fc
fc = f1 Gd = 10 MHz 20 VmV
= 10 106 20 103
= 500 Hz
(Voltage Gain)
(frequency)f1
Gd
0707Gd
fc0
1
10MHz
Hz
Op Amp transfer characteristic curve
saturation
)( vvAvAv io
active region
So far we have been looking at the amplification that can be achieved for relatively small (amplitude) signals For a fixed gain as we increase the input signal amplitude there is a limit to how large the output signal can be The output saturates as it approaches the positive andnegative power supply voltages In other words there is limited range across which the gain is linear vv Vvo
vv Vvo
Op Amp transfer characteristic curve
Applications of Op-AmpsFiltersTypesbullLow pass filterbullHigh pass filterbullBand pass filterbullCascading (2 or more filters connected together) R2
+
-
+
V0
__
+ Vcc
- Vcc
-+
R1
C
Low pass filter
Low pass filter Cutoff frequency
Low pass filter transfer function
Applications of Op-Ampsbull Electrocardiogram EKG Amplification
ndash Need to measure difference in voltage from lead 1 and lead 2ndash 60 Hz interference from electrical equipment
Applications of Op-Ampsbull Simple EKG circuit
ndash Uses differential amplifier to cancel common mode signal and amplify differential mode signal
bull Realistic EKG circuitndash Uses two non-inverting
amplifiers to first amplify voltage from each lead followed by differential amplifier
ndash Forms an ldquoinstrumentation amplifierrdquo
Strain Gauge
Use a Wheatstone bridge to determine the strain of an element by measuring the change in resistance of a
strain gauge
(No strain) Balanced Bridge R 1 = R 2
(Strain) Unbalanced Bridge R 1 ne R 2
Strain GaugeHalf-Bridge Arrangement
Using KCL at the inverting and non-inverting terminals of the op amp we find that ε ~ Vo = 2ΔR(Rf R2)
R + ΔR Rf
+
- +
V0
__
+ Vcc
- Vcc
-+
Rf
Vref
R
R - ΔR
R
Op amp used to amplify output from strain gauge
Applications of Op-Ampsbull Piezoelectric Transducerndash Used to measure force pressure accelerationndash Piezoelectric crystal generates an electric charge in
response to deformation
bull Use Charge Amplifierndash Just an integrator op-amp circuit
bullGoal is to have VSET = VOUT
bullRemember that VERROR = VSET ndash VSENSOR
bullOutput Process uses VERROR from the PID controller to adjust Vout such that it is ~VSET
P
I
D
Output Process
Sensor
VERRORVSET VOUT
VSENSOR
PID Controller ndash System Block Diagram
ApplicationsPID Controller ndash System Circuit Diagram
Source httpwwwecircuitcentercomCircuitsop_pidop_pidhtm
Calculates VERROR = -(VSET + VSENSOR)
Signal conditioning allows you to introduce a time delay which could
account for things like inertia
System to control
-VSENSOR
ApplicationsPID Controller ndash PID Controller Circuit Diagram
VERR
Adjust Change
Kp RP1 RP2
Ki RI CI
Kd RD CD
VERR PID
Applications of Op-Ampsbull Example of PI Control
Temperature Control
bull Thermal System we wish to automatically control the temperature of
bull Block Diagram of Control System
Applications of Op-Amps
bull Voltage Error Circuit
bull Proportional-Integral Control Circuit
bull Example of PI Control Temperature Control
44
Multiple Inputs(1) Kirchhoff node equation at V+
yields
(2) Kirchhoff node equation at V yields
(3) Setting V+ = Vndash yields
0V
0_
c
c
b
b
a
a
f
o
R
VV
R
VV
R
VV
R
VV
c
aj j
jf
c
c
b
b
a
afo R
VR
R
V
R
V
R
VRV
+
Rf
Va
VoRb
Ra
RcVb
Vc
Homework
1) Design a circuit to 50i
of V
VA
2) Find the vo=
Review Two fundamental Op Amp
Structure Af Input voltage ( )terminal
Feed back
( )terminal
Inverting Amp
_ _
Non inverting Amp
+ _
Op-Amp Characteristics
bull Open-loop gain G is typically over 9000bull But closed-loop gain is much smaller
bull Rin is very large (MΩ or larger)bull Rout is small (75Ω or smaller)
bull Effective output impedance in closed loop is very small
Ideal Op-Amp Characteristics
bull Open-loop gain G is infinitebull Rin is infinite
bull Zero input currentbull Rout is zero
Ideal Op-Amp Analysis
To analyze an op-amp feedback circuitbull Assume no current flows into either input terminalbull Assume no current flows out of the output terminalbull Constrain V+ = V-
11
Ideal Vs Practical Op-AmpIdeal Practical
Open Loop gain A 105
Bandwidth BW 10-100Hz
Input Impedance Zin gt1M
Output Impedance Zout 0 10-100
Output Voltage VoutDepends only on Vd = (V+V)
Differential mode signal
Depends slightly on average input Vc = (V++V)2 Common-Mode signal
CMRR 10-100dB
+
~AVin
Vin Vout
Zout=0
Ideal op-amp
+
AVin
Vin Vout
Zout
~Zin
Practical op-amp
Practical opamp3 categories are considering
bull 3 categories are considering
Close-Loop Voltage Gain Input impedance Output impedance
Ref080114HKN Operational Amplifier 13
Ideal Op-Amp ApplicationsAnalysis Method Two ideal Op-Amp Properties(1) The voltage between V+ and V is zero V+ = V
(2) The current into both V+ and V termainals is zero
For ideal Op-Amp circuit(1) Write the kirchhoff node equation at the noninverting
terminal V+ (2) Write the kirchhoff node eqaution at the inverting
terminal V (3) Set V+ = V and solve for the desired closed-loop gain
2 TYPES
1 Inverting Amplifier 2 Non-Inverting Amplifier
1 Inverting Amplifier Analysis
virtual ground
2 Non-Inverting Amplifier Analysis
17
Op-Amp Properties(1) Infinite Open Loop gain
- The gain without feedback- Equal to differential gain- Zero common-mode gain- Pratically Gd = 20000 to 200000
(2) Infinite Input impedance- Input current ii ~0A- T- in high-grade op-amp - m-A input current in low-grade op-
amp
(3) Zero Output Impedance- act as perfect internal voltage source- No internal resistance- Output impedance in series with load- Reducing output voltage to the load- Practically Rout ~ 20-100
+
V1
V2
Vo
+
Vo
i1~0
i2~0
+
Rout
VoRload
outload
loadoload RR
RVV
Op-Amp Buffer
Vout = VinIsolates loading effects
A
High output impedance
B
Low input impedance
Op-Amp Differentiator
20
Op-Amp Differentiator
RCdt
dVv io
+
RC
VoVi
0to t1 t2
0
to t1 t2
Op-Amp Integrator
Op-Amp Summing Amplifier
Op-Amp Differential Amplifier
If R1 = R2 and Rf = Rg
Closed-loop gain
2
21
21
21
R
RRv
RRARAv
v
IN
INo
Closed-loop gain Af=vovin
Suppose A=106 R1=9R R2=R
10INo vv
Gain
INf vA
)1(2
1
R
RAf
Closed-loop gain determined by resistor ratio insensitive to A temperature
Negative feedback
1048774
)1(2
1
R
R
v
vA
i
of
1
2
R
R
v
vA
i
of
bull We can adjust the closed-loop gain by changing the ratio of R2 and R1
bull The closed-loop gain is (ideally) independent of op amp open-loop gain A (if A is large enough) and we can make it arbitrarily large or small and with the desired accuracy depending on the accuracy of the resistors
1048774
The terminal 1 is a virtual ground since terminal 2 is grounded
Inverting configuration
This is a classic example of what negative feedback does It takes an amplifier with very large gain and through negative feedback obtain a gain that is smaller stable and predictable In effect we have traded gain for accuracy This kind of trade off is common in electronic circuit designhellip as we will see more later
Negative feedback
1048774
Inverting configuration
Input ResistanceAssuming an ideal op amp (open-loop gain A = infinity) in the closed-loop inverting configuration the input resistance is R1
111 R
Rv
v
i
vR
i
iiin
Negative feedback
1
2
R
R
v
vA
i
of
Inverting configuration011
oi
in RRi
vR
We can model the closed-loop inverting amplifier (with A = infinite) with the following equivalent circuit using a voltage-controlled voltage sourcehellip
Negative feedback 1048774
29
Frequency-Gain Relationbull Ideally signals are amplified
from DC to the highest AC frequency
bull Practically bandwidth is limitedbull 741 family op-amp have an limit
bandwidth of few KHzbull Unity Gain frequency f1 the gain at
unitybull Cutoff frequency fc the gain drop by
3dB from dc gain Gd
(Voltage Gain)
(frequency)f1
Gd
0707Gd
fc0
1
GB Product f1 = Gd fc
20log(0707)=3dB
30
GB ProductExample Determine the cutoff frequency of an op-amp having a unit gain frequency f1 = 10 MHz and voltage differential gain Gd = 20VmV
Sol
Since f1 = 10 MHz
By using GB production equation
f1 = Gd fc
fc = f1 Gd = 10 MHz 20 VmV
= 10 106 20 103
= 500 Hz
(Voltage Gain)
(frequency)f1
Gd
0707Gd
fc0
1
10MHz
Hz
Op Amp transfer characteristic curve
saturation
)( vvAvAv io
active region
So far we have been looking at the amplification that can be achieved for relatively small (amplitude) signals For a fixed gain as we increase the input signal amplitude there is a limit to how large the output signal can be The output saturates as it approaches the positive andnegative power supply voltages In other words there is limited range across which the gain is linear vv Vvo
vv Vvo
Op Amp transfer characteristic curve
Applications of Op-AmpsFiltersTypesbullLow pass filterbullHigh pass filterbullBand pass filterbullCascading (2 or more filters connected together) R2
+
-
+
V0
__
+ Vcc
- Vcc
-+
R1
C
Low pass filter
Low pass filter Cutoff frequency
Low pass filter transfer function
Applications of Op-Ampsbull Electrocardiogram EKG Amplification
ndash Need to measure difference in voltage from lead 1 and lead 2ndash 60 Hz interference from electrical equipment
Applications of Op-Ampsbull Simple EKG circuit
ndash Uses differential amplifier to cancel common mode signal and amplify differential mode signal
bull Realistic EKG circuitndash Uses two non-inverting
amplifiers to first amplify voltage from each lead followed by differential amplifier
ndash Forms an ldquoinstrumentation amplifierrdquo
Strain Gauge
Use a Wheatstone bridge to determine the strain of an element by measuring the change in resistance of a
strain gauge
(No strain) Balanced Bridge R 1 = R 2
(Strain) Unbalanced Bridge R 1 ne R 2
Strain GaugeHalf-Bridge Arrangement
Using KCL at the inverting and non-inverting terminals of the op amp we find that ε ~ Vo = 2ΔR(Rf R2)
R + ΔR Rf
+
- +
V0
__
+ Vcc
- Vcc
-+
Rf
Vref
R
R - ΔR
R
Op amp used to amplify output from strain gauge
Applications of Op-Ampsbull Piezoelectric Transducerndash Used to measure force pressure accelerationndash Piezoelectric crystal generates an electric charge in
response to deformation
bull Use Charge Amplifierndash Just an integrator op-amp circuit
bullGoal is to have VSET = VOUT
bullRemember that VERROR = VSET ndash VSENSOR
bullOutput Process uses VERROR from the PID controller to adjust Vout such that it is ~VSET
P
I
D
Output Process
Sensor
VERRORVSET VOUT
VSENSOR
PID Controller ndash System Block Diagram
ApplicationsPID Controller ndash System Circuit Diagram
Source httpwwwecircuitcentercomCircuitsop_pidop_pidhtm
Calculates VERROR = -(VSET + VSENSOR)
Signal conditioning allows you to introduce a time delay which could
account for things like inertia
System to control
-VSENSOR
ApplicationsPID Controller ndash PID Controller Circuit Diagram
VERR
Adjust Change
Kp RP1 RP2
Ki RI CI
Kd RD CD
VERR PID
Applications of Op-Ampsbull Example of PI Control
Temperature Control
bull Thermal System we wish to automatically control the temperature of
bull Block Diagram of Control System
Applications of Op-Amps
bull Voltage Error Circuit
bull Proportional-Integral Control Circuit
bull Example of PI Control Temperature Control
44
Multiple Inputs(1) Kirchhoff node equation at V+
yields
(2) Kirchhoff node equation at V yields
(3) Setting V+ = Vndash yields
0V
0_
c
c
b
b
a
a
f
o
R
VV
R
VV
R
VV
R
VV
c
aj j
jf
c
c
b
b
a
afo R
VR
R
V
R
V
R
VRV
+
Rf
Va
VoRb
Ra
RcVb
Vc
Homework
1) Design a circuit to 50i
of V
VA
2) Find the vo=
Review Two fundamental Op Amp
Structure Af Input voltage ( )terminal
Feed back
( )terminal
Inverting Amp
_ _
Non inverting Amp
+ _
Ideal Op-Amp Characteristics
bull Open-loop gain G is infinitebull Rin is infinite
bull Zero input currentbull Rout is zero
Ideal Op-Amp Analysis
To analyze an op-amp feedback circuitbull Assume no current flows into either input terminalbull Assume no current flows out of the output terminalbull Constrain V+ = V-
11
Ideal Vs Practical Op-AmpIdeal Practical
Open Loop gain A 105
Bandwidth BW 10-100Hz
Input Impedance Zin gt1M
Output Impedance Zout 0 10-100
Output Voltage VoutDepends only on Vd = (V+V)
Differential mode signal
Depends slightly on average input Vc = (V++V)2 Common-Mode signal
CMRR 10-100dB
+
~AVin
Vin Vout
Zout=0
Ideal op-amp
+
AVin
Vin Vout
Zout
~Zin
Practical op-amp
Practical opamp3 categories are considering
bull 3 categories are considering
Close-Loop Voltage Gain Input impedance Output impedance
Ref080114HKN Operational Amplifier 13
Ideal Op-Amp ApplicationsAnalysis Method Two ideal Op-Amp Properties(1) The voltage between V+ and V is zero V+ = V
(2) The current into both V+ and V termainals is zero
For ideal Op-Amp circuit(1) Write the kirchhoff node equation at the noninverting
terminal V+ (2) Write the kirchhoff node eqaution at the inverting
terminal V (3) Set V+ = V and solve for the desired closed-loop gain
2 TYPES
1 Inverting Amplifier 2 Non-Inverting Amplifier
1 Inverting Amplifier Analysis
virtual ground
2 Non-Inverting Amplifier Analysis
17
Op-Amp Properties(1) Infinite Open Loop gain
- The gain without feedback- Equal to differential gain- Zero common-mode gain- Pratically Gd = 20000 to 200000
(2) Infinite Input impedance- Input current ii ~0A- T- in high-grade op-amp - m-A input current in low-grade op-
amp
(3) Zero Output Impedance- act as perfect internal voltage source- No internal resistance- Output impedance in series with load- Reducing output voltage to the load- Practically Rout ~ 20-100
+
V1
V2
Vo
+
Vo
i1~0
i2~0
+
Rout
VoRload
outload
loadoload RR
RVV
Op-Amp Buffer
Vout = VinIsolates loading effects
A
High output impedance
B
Low input impedance
Op-Amp Differentiator
20
Op-Amp Differentiator
RCdt
dVv io
+
RC
VoVi
0to t1 t2
0
to t1 t2
Op-Amp Integrator
Op-Amp Summing Amplifier
Op-Amp Differential Amplifier
If R1 = R2 and Rf = Rg
Closed-loop gain
2
21
21
21
R
RRv
RRARAv
v
IN
INo
Closed-loop gain Af=vovin
Suppose A=106 R1=9R R2=R
10INo vv
Gain
INf vA
)1(2
1
R
RAf
Closed-loop gain determined by resistor ratio insensitive to A temperature
Negative feedback
1048774
)1(2
1
R
R
v
vA
i
of
1
2
R
R
v
vA
i
of
bull We can adjust the closed-loop gain by changing the ratio of R2 and R1
bull The closed-loop gain is (ideally) independent of op amp open-loop gain A (if A is large enough) and we can make it arbitrarily large or small and with the desired accuracy depending on the accuracy of the resistors
1048774
The terminal 1 is a virtual ground since terminal 2 is grounded
Inverting configuration
This is a classic example of what negative feedback does It takes an amplifier with very large gain and through negative feedback obtain a gain that is smaller stable and predictable In effect we have traded gain for accuracy This kind of trade off is common in electronic circuit designhellip as we will see more later
Negative feedback
1048774
Inverting configuration
Input ResistanceAssuming an ideal op amp (open-loop gain A = infinity) in the closed-loop inverting configuration the input resistance is R1
111 R
Rv
v
i
vR
i
iiin
Negative feedback
1
2
R
R
v
vA
i
of
Inverting configuration011
oi
in RRi
vR
We can model the closed-loop inverting amplifier (with A = infinite) with the following equivalent circuit using a voltage-controlled voltage sourcehellip
Negative feedback 1048774
29
Frequency-Gain Relationbull Ideally signals are amplified
from DC to the highest AC frequency
bull Practically bandwidth is limitedbull 741 family op-amp have an limit
bandwidth of few KHzbull Unity Gain frequency f1 the gain at
unitybull Cutoff frequency fc the gain drop by
3dB from dc gain Gd
(Voltage Gain)
(frequency)f1
Gd
0707Gd
fc0
1
GB Product f1 = Gd fc
20log(0707)=3dB
30
GB ProductExample Determine the cutoff frequency of an op-amp having a unit gain frequency f1 = 10 MHz and voltage differential gain Gd = 20VmV
Sol
Since f1 = 10 MHz
By using GB production equation
f1 = Gd fc
fc = f1 Gd = 10 MHz 20 VmV
= 10 106 20 103
= 500 Hz
(Voltage Gain)
(frequency)f1
Gd
0707Gd
fc0
1
10MHz
Hz
Op Amp transfer characteristic curve
saturation
)( vvAvAv io
active region
So far we have been looking at the amplification that can be achieved for relatively small (amplitude) signals For a fixed gain as we increase the input signal amplitude there is a limit to how large the output signal can be The output saturates as it approaches the positive andnegative power supply voltages In other words there is limited range across which the gain is linear vv Vvo
vv Vvo
Op Amp transfer characteristic curve
Applications of Op-AmpsFiltersTypesbullLow pass filterbullHigh pass filterbullBand pass filterbullCascading (2 or more filters connected together) R2
+
-
+
V0
__
+ Vcc
- Vcc
-+
R1
C
Low pass filter
Low pass filter Cutoff frequency
Low pass filter transfer function
Applications of Op-Ampsbull Electrocardiogram EKG Amplification
ndash Need to measure difference in voltage from lead 1 and lead 2ndash 60 Hz interference from electrical equipment
Applications of Op-Ampsbull Simple EKG circuit
ndash Uses differential amplifier to cancel common mode signal and amplify differential mode signal
bull Realistic EKG circuitndash Uses two non-inverting
amplifiers to first amplify voltage from each lead followed by differential amplifier
ndash Forms an ldquoinstrumentation amplifierrdquo
Strain Gauge
Use a Wheatstone bridge to determine the strain of an element by measuring the change in resistance of a
strain gauge
(No strain) Balanced Bridge R 1 = R 2
(Strain) Unbalanced Bridge R 1 ne R 2
Strain GaugeHalf-Bridge Arrangement
Using KCL at the inverting and non-inverting terminals of the op amp we find that ε ~ Vo = 2ΔR(Rf R2)
R + ΔR Rf
+
- +
V0
__
+ Vcc
- Vcc
-+
Rf
Vref
R
R - ΔR
R
Op amp used to amplify output from strain gauge
Applications of Op-Ampsbull Piezoelectric Transducerndash Used to measure force pressure accelerationndash Piezoelectric crystal generates an electric charge in
response to deformation
bull Use Charge Amplifierndash Just an integrator op-amp circuit
bullGoal is to have VSET = VOUT
bullRemember that VERROR = VSET ndash VSENSOR
bullOutput Process uses VERROR from the PID controller to adjust Vout such that it is ~VSET
P
I
D
Output Process
Sensor
VERRORVSET VOUT
VSENSOR
PID Controller ndash System Block Diagram
ApplicationsPID Controller ndash System Circuit Diagram
Source httpwwwecircuitcentercomCircuitsop_pidop_pidhtm
Calculates VERROR = -(VSET + VSENSOR)
Signal conditioning allows you to introduce a time delay which could
account for things like inertia
System to control
-VSENSOR
ApplicationsPID Controller ndash PID Controller Circuit Diagram
VERR
Adjust Change
Kp RP1 RP2
Ki RI CI
Kd RD CD
VERR PID
Applications of Op-Ampsbull Example of PI Control
Temperature Control
bull Thermal System we wish to automatically control the temperature of
bull Block Diagram of Control System
Applications of Op-Amps
bull Voltage Error Circuit
bull Proportional-Integral Control Circuit
bull Example of PI Control Temperature Control
44
Multiple Inputs(1) Kirchhoff node equation at V+
yields
(2) Kirchhoff node equation at V yields
(3) Setting V+ = Vndash yields
0V
0_
c
c
b
b
a
a
f
o
R
VV
R
VV
R
VV
R
VV
c
aj j
jf
c
c
b
b
a
afo R
VR
R
V
R
V
R
VRV
+
Rf
Va
VoRb
Ra
RcVb
Vc
Homework
1) Design a circuit to 50i
of V
VA
2) Find the vo=
Review Two fundamental Op Amp
Structure Af Input voltage ( )terminal
Feed back
( )terminal
Inverting Amp
_ _
Non inverting Amp
+ _
Ideal Op-Amp Analysis
To analyze an op-amp feedback circuitbull Assume no current flows into either input terminalbull Assume no current flows out of the output terminalbull Constrain V+ = V-
11
Ideal Vs Practical Op-AmpIdeal Practical
Open Loop gain A 105
Bandwidth BW 10-100Hz
Input Impedance Zin gt1M
Output Impedance Zout 0 10-100
Output Voltage VoutDepends only on Vd = (V+V)
Differential mode signal
Depends slightly on average input Vc = (V++V)2 Common-Mode signal
CMRR 10-100dB
+
~AVin
Vin Vout
Zout=0
Ideal op-amp
+
AVin
Vin Vout
Zout
~Zin
Practical op-amp
Practical opamp3 categories are considering
bull 3 categories are considering
Close-Loop Voltage Gain Input impedance Output impedance
Ref080114HKN Operational Amplifier 13
Ideal Op-Amp ApplicationsAnalysis Method Two ideal Op-Amp Properties(1) The voltage between V+ and V is zero V+ = V
(2) The current into both V+ and V termainals is zero
For ideal Op-Amp circuit(1) Write the kirchhoff node equation at the noninverting
terminal V+ (2) Write the kirchhoff node eqaution at the inverting
terminal V (3) Set V+ = V and solve for the desired closed-loop gain
2 TYPES
1 Inverting Amplifier 2 Non-Inverting Amplifier
1 Inverting Amplifier Analysis
virtual ground
2 Non-Inverting Amplifier Analysis
17
Op-Amp Properties(1) Infinite Open Loop gain
- The gain without feedback- Equal to differential gain- Zero common-mode gain- Pratically Gd = 20000 to 200000
(2) Infinite Input impedance- Input current ii ~0A- T- in high-grade op-amp - m-A input current in low-grade op-
amp
(3) Zero Output Impedance- act as perfect internal voltage source- No internal resistance- Output impedance in series with load- Reducing output voltage to the load- Practically Rout ~ 20-100
+
V1
V2
Vo
+
Vo
i1~0
i2~0
+
Rout
VoRload
outload
loadoload RR
RVV
Op-Amp Buffer
Vout = VinIsolates loading effects
A
High output impedance
B
Low input impedance
Op-Amp Differentiator
20
Op-Amp Differentiator
RCdt
dVv io
+
RC
VoVi
0to t1 t2
0
to t1 t2
Op-Amp Integrator
Op-Amp Summing Amplifier
Op-Amp Differential Amplifier
If R1 = R2 and Rf = Rg
Closed-loop gain
2
21
21
21
R
RRv
RRARAv
v
IN
INo
Closed-loop gain Af=vovin
Suppose A=106 R1=9R R2=R
10INo vv
Gain
INf vA
)1(2
1
R
RAf
Closed-loop gain determined by resistor ratio insensitive to A temperature
Negative feedback
1048774
)1(2
1
R
R
v
vA
i
of
1
2
R
R
v
vA
i
of
bull We can adjust the closed-loop gain by changing the ratio of R2 and R1
bull The closed-loop gain is (ideally) independent of op amp open-loop gain A (if A is large enough) and we can make it arbitrarily large or small and with the desired accuracy depending on the accuracy of the resistors
1048774
The terminal 1 is a virtual ground since terminal 2 is grounded
Inverting configuration
This is a classic example of what negative feedback does It takes an amplifier with very large gain and through negative feedback obtain a gain that is smaller stable and predictable In effect we have traded gain for accuracy This kind of trade off is common in electronic circuit designhellip as we will see more later
Negative feedback
1048774
Inverting configuration
Input ResistanceAssuming an ideal op amp (open-loop gain A = infinity) in the closed-loop inverting configuration the input resistance is R1
111 R
Rv
v
i
vR
i
iiin
Negative feedback
1
2
R
R
v
vA
i
of
Inverting configuration011
oi
in RRi
vR
We can model the closed-loop inverting amplifier (with A = infinite) with the following equivalent circuit using a voltage-controlled voltage sourcehellip
Negative feedback 1048774
29
Frequency-Gain Relationbull Ideally signals are amplified
from DC to the highest AC frequency
bull Practically bandwidth is limitedbull 741 family op-amp have an limit
bandwidth of few KHzbull Unity Gain frequency f1 the gain at
unitybull Cutoff frequency fc the gain drop by
3dB from dc gain Gd
(Voltage Gain)
(frequency)f1
Gd
0707Gd
fc0
1
GB Product f1 = Gd fc
20log(0707)=3dB
30
GB ProductExample Determine the cutoff frequency of an op-amp having a unit gain frequency f1 = 10 MHz and voltage differential gain Gd = 20VmV
Sol
Since f1 = 10 MHz
By using GB production equation
f1 = Gd fc
fc = f1 Gd = 10 MHz 20 VmV
= 10 106 20 103
= 500 Hz
(Voltage Gain)
(frequency)f1
Gd
0707Gd
fc0
1
10MHz
Hz
Op Amp transfer characteristic curve
saturation
)( vvAvAv io
active region
So far we have been looking at the amplification that can be achieved for relatively small (amplitude) signals For a fixed gain as we increase the input signal amplitude there is a limit to how large the output signal can be The output saturates as it approaches the positive andnegative power supply voltages In other words there is limited range across which the gain is linear vv Vvo
vv Vvo
Op Amp transfer characteristic curve
Applications of Op-AmpsFiltersTypesbullLow pass filterbullHigh pass filterbullBand pass filterbullCascading (2 or more filters connected together) R2
+
-
+
V0
__
+ Vcc
- Vcc
-+
R1
C
Low pass filter
Low pass filter Cutoff frequency
Low pass filter transfer function
Applications of Op-Ampsbull Electrocardiogram EKG Amplification
ndash Need to measure difference in voltage from lead 1 and lead 2ndash 60 Hz interference from electrical equipment
Applications of Op-Ampsbull Simple EKG circuit
ndash Uses differential amplifier to cancel common mode signal and amplify differential mode signal
bull Realistic EKG circuitndash Uses two non-inverting
amplifiers to first amplify voltage from each lead followed by differential amplifier
ndash Forms an ldquoinstrumentation amplifierrdquo
Strain Gauge
Use a Wheatstone bridge to determine the strain of an element by measuring the change in resistance of a
strain gauge
(No strain) Balanced Bridge R 1 = R 2
(Strain) Unbalanced Bridge R 1 ne R 2
Strain GaugeHalf-Bridge Arrangement
Using KCL at the inverting and non-inverting terminals of the op amp we find that ε ~ Vo = 2ΔR(Rf R2)
R + ΔR Rf
+
- +
V0
__
+ Vcc
- Vcc
-+
Rf
Vref
R
R - ΔR
R
Op amp used to amplify output from strain gauge
Applications of Op-Ampsbull Piezoelectric Transducerndash Used to measure force pressure accelerationndash Piezoelectric crystal generates an electric charge in
response to deformation
bull Use Charge Amplifierndash Just an integrator op-amp circuit
bullGoal is to have VSET = VOUT
bullRemember that VERROR = VSET ndash VSENSOR
bullOutput Process uses VERROR from the PID controller to adjust Vout such that it is ~VSET
P
I
D
Output Process
Sensor
VERRORVSET VOUT
VSENSOR
PID Controller ndash System Block Diagram
ApplicationsPID Controller ndash System Circuit Diagram
Source httpwwwecircuitcentercomCircuitsop_pidop_pidhtm
Calculates VERROR = -(VSET + VSENSOR)
Signal conditioning allows you to introduce a time delay which could
account for things like inertia
System to control
-VSENSOR
ApplicationsPID Controller ndash PID Controller Circuit Diagram
VERR
Adjust Change
Kp RP1 RP2
Ki RI CI
Kd RD CD
VERR PID
Applications of Op-Ampsbull Example of PI Control
Temperature Control
bull Thermal System we wish to automatically control the temperature of
bull Block Diagram of Control System
Applications of Op-Amps
bull Voltage Error Circuit
bull Proportional-Integral Control Circuit
bull Example of PI Control Temperature Control
44
Multiple Inputs(1) Kirchhoff node equation at V+
yields
(2) Kirchhoff node equation at V yields
(3) Setting V+ = Vndash yields
0V
0_
c
c
b
b
a
a
f
o
R
VV
R
VV
R
VV
R
VV
c
aj j
jf
c
c
b
b
a
afo R
VR
R
V
R
V
R
VRV
+
Rf
Va
VoRb
Ra
RcVb
Vc
Homework
1) Design a circuit to 50i
of V
VA
2) Find the vo=
Review Two fundamental Op Amp
Structure Af Input voltage ( )terminal
Feed back
( )terminal
Inverting Amp
_ _
Non inverting Amp
+ _
11
Ideal Vs Practical Op-AmpIdeal Practical
Open Loop gain A 105
Bandwidth BW 10-100Hz
Input Impedance Zin gt1M
Output Impedance Zout 0 10-100
Output Voltage VoutDepends only on Vd = (V+V)
Differential mode signal
Depends slightly on average input Vc = (V++V)2 Common-Mode signal
CMRR 10-100dB
+
~AVin
Vin Vout
Zout=0
Ideal op-amp
+
AVin
Vin Vout
Zout
~Zin
Practical op-amp
Practical opamp3 categories are considering
bull 3 categories are considering
Close-Loop Voltage Gain Input impedance Output impedance
Ref080114HKN Operational Amplifier 13
Ideal Op-Amp ApplicationsAnalysis Method Two ideal Op-Amp Properties(1) The voltage between V+ and V is zero V+ = V
(2) The current into both V+ and V termainals is zero
For ideal Op-Amp circuit(1) Write the kirchhoff node equation at the noninverting
terminal V+ (2) Write the kirchhoff node eqaution at the inverting
terminal V (3) Set V+ = V and solve for the desired closed-loop gain
2 TYPES
1 Inverting Amplifier 2 Non-Inverting Amplifier
1 Inverting Amplifier Analysis
virtual ground
2 Non-Inverting Amplifier Analysis
17
Op-Amp Properties(1) Infinite Open Loop gain
- The gain without feedback- Equal to differential gain- Zero common-mode gain- Pratically Gd = 20000 to 200000
(2) Infinite Input impedance- Input current ii ~0A- T- in high-grade op-amp - m-A input current in low-grade op-
amp
(3) Zero Output Impedance- act as perfect internal voltage source- No internal resistance- Output impedance in series with load- Reducing output voltage to the load- Practically Rout ~ 20-100
+
V1
V2
Vo
+
Vo
i1~0
i2~0
+
Rout
VoRload
outload
loadoload RR
RVV
Op-Amp Buffer
Vout = VinIsolates loading effects
A
High output impedance
B
Low input impedance
Op-Amp Differentiator
20
Op-Amp Differentiator
RCdt
dVv io
+
RC
VoVi
0to t1 t2
0
to t1 t2
Op-Amp Integrator
Op-Amp Summing Amplifier
Op-Amp Differential Amplifier
If R1 = R2 and Rf = Rg
Closed-loop gain
2
21
21
21
R
RRv
RRARAv
v
IN
INo
Closed-loop gain Af=vovin
Suppose A=106 R1=9R R2=R
10INo vv
Gain
INf vA
)1(2
1
R
RAf
Closed-loop gain determined by resistor ratio insensitive to A temperature
Negative feedback
1048774
)1(2
1
R
R
v
vA
i
of
1
2
R
R
v
vA
i
of
bull We can adjust the closed-loop gain by changing the ratio of R2 and R1
bull The closed-loop gain is (ideally) independent of op amp open-loop gain A (if A is large enough) and we can make it arbitrarily large or small and with the desired accuracy depending on the accuracy of the resistors
1048774
The terminal 1 is a virtual ground since terminal 2 is grounded
Inverting configuration
This is a classic example of what negative feedback does It takes an amplifier with very large gain and through negative feedback obtain a gain that is smaller stable and predictable In effect we have traded gain for accuracy This kind of trade off is common in electronic circuit designhellip as we will see more later
Negative feedback
1048774
Inverting configuration
Input ResistanceAssuming an ideal op amp (open-loop gain A = infinity) in the closed-loop inverting configuration the input resistance is R1
111 R
Rv
v
i
vR
i
iiin
Negative feedback
1
2
R
R
v
vA
i
of
Inverting configuration011
oi
in RRi
vR
We can model the closed-loop inverting amplifier (with A = infinite) with the following equivalent circuit using a voltage-controlled voltage sourcehellip
Negative feedback 1048774
29
Frequency-Gain Relationbull Ideally signals are amplified
from DC to the highest AC frequency
bull Practically bandwidth is limitedbull 741 family op-amp have an limit
bandwidth of few KHzbull Unity Gain frequency f1 the gain at
unitybull Cutoff frequency fc the gain drop by
3dB from dc gain Gd
(Voltage Gain)
(frequency)f1
Gd
0707Gd
fc0
1
GB Product f1 = Gd fc
20log(0707)=3dB
30
GB ProductExample Determine the cutoff frequency of an op-amp having a unit gain frequency f1 = 10 MHz and voltage differential gain Gd = 20VmV
Sol
Since f1 = 10 MHz
By using GB production equation
f1 = Gd fc
fc = f1 Gd = 10 MHz 20 VmV
= 10 106 20 103
= 500 Hz
(Voltage Gain)
(frequency)f1
Gd
0707Gd
fc0
1
10MHz
Hz
Op Amp transfer characteristic curve
saturation
)( vvAvAv io
active region
So far we have been looking at the amplification that can be achieved for relatively small (amplitude) signals For a fixed gain as we increase the input signal amplitude there is a limit to how large the output signal can be The output saturates as it approaches the positive andnegative power supply voltages In other words there is limited range across which the gain is linear vv Vvo
vv Vvo
Op Amp transfer characteristic curve
Applications of Op-AmpsFiltersTypesbullLow pass filterbullHigh pass filterbullBand pass filterbullCascading (2 or more filters connected together) R2
+
-
+
V0
__
+ Vcc
- Vcc
-+
R1
C
Low pass filter
Low pass filter Cutoff frequency
Low pass filter transfer function
Applications of Op-Ampsbull Electrocardiogram EKG Amplification
ndash Need to measure difference in voltage from lead 1 and lead 2ndash 60 Hz interference from electrical equipment
Applications of Op-Ampsbull Simple EKG circuit
ndash Uses differential amplifier to cancel common mode signal and amplify differential mode signal
bull Realistic EKG circuitndash Uses two non-inverting
amplifiers to first amplify voltage from each lead followed by differential amplifier
ndash Forms an ldquoinstrumentation amplifierrdquo
Strain Gauge
Use a Wheatstone bridge to determine the strain of an element by measuring the change in resistance of a
strain gauge
(No strain) Balanced Bridge R 1 = R 2
(Strain) Unbalanced Bridge R 1 ne R 2
Strain GaugeHalf-Bridge Arrangement
Using KCL at the inverting and non-inverting terminals of the op amp we find that ε ~ Vo = 2ΔR(Rf R2)
R + ΔR Rf
+
- +
V0
__
+ Vcc
- Vcc
-+
Rf
Vref
R
R - ΔR
R
Op amp used to amplify output from strain gauge
Applications of Op-Ampsbull Piezoelectric Transducerndash Used to measure force pressure accelerationndash Piezoelectric crystal generates an electric charge in
response to deformation
bull Use Charge Amplifierndash Just an integrator op-amp circuit
bullGoal is to have VSET = VOUT
bullRemember that VERROR = VSET ndash VSENSOR
bullOutput Process uses VERROR from the PID controller to adjust Vout such that it is ~VSET
P
I
D
Output Process
Sensor
VERRORVSET VOUT
VSENSOR
PID Controller ndash System Block Diagram
ApplicationsPID Controller ndash System Circuit Diagram
Source httpwwwecircuitcentercomCircuitsop_pidop_pidhtm
Calculates VERROR = -(VSET + VSENSOR)
Signal conditioning allows you to introduce a time delay which could
account for things like inertia
System to control
-VSENSOR
ApplicationsPID Controller ndash PID Controller Circuit Diagram
VERR
Adjust Change
Kp RP1 RP2
Ki RI CI
Kd RD CD
VERR PID
Applications of Op-Ampsbull Example of PI Control
Temperature Control
bull Thermal System we wish to automatically control the temperature of
bull Block Diagram of Control System
Applications of Op-Amps
bull Voltage Error Circuit
bull Proportional-Integral Control Circuit
bull Example of PI Control Temperature Control
44
Multiple Inputs(1) Kirchhoff node equation at V+
yields
(2) Kirchhoff node equation at V yields
(3) Setting V+ = Vndash yields
0V
0_
c
c
b
b
a
a
f
o
R
VV
R
VV
R
VV
R
VV
c
aj j
jf
c
c
b
b
a
afo R
VR
R
V
R
V
R
VRV
+
Rf
Va
VoRb
Ra
RcVb
Vc
Homework
1) Design a circuit to 50i
of V
VA
2) Find the vo=
Review Two fundamental Op Amp
Structure Af Input voltage ( )terminal
Feed back
( )terminal
Inverting Amp
_ _
Non inverting Amp
+ _
Practical opamp3 categories are considering
bull 3 categories are considering
Close-Loop Voltage Gain Input impedance Output impedance
Ref080114HKN Operational Amplifier 13
Ideal Op-Amp ApplicationsAnalysis Method Two ideal Op-Amp Properties(1) The voltage between V+ and V is zero V+ = V
(2) The current into both V+ and V termainals is zero
For ideal Op-Amp circuit(1) Write the kirchhoff node equation at the noninverting
terminal V+ (2) Write the kirchhoff node eqaution at the inverting
terminal V (3) Set V+ = V and solve for the desired closed-loop gain
2 TYPES
1 Inverting Amplifier 2 Non-Inverting Amplifier
1 Inverting Amplifier Analysis
virtual ground
2 Non-Inverting Amplifier Analysis
17
Op-Amp Properties(1) Infinite Open Loop gain
- The gain without feedback- Equal to differential gain- Zero common-mode gain- Pratically Gd = 20000 to 200000
(2) Infinite Input impedance- Input current ii ~0A- T- in high-grade op-amp - m-A input current in low-grade op-
amp
(3) Zero Output Impedance- act as perfect internal voltage source- No internal resistance- Output impedance in series with load- Reducing output voltage to the load- Practically Rout ~ 20-100
+
V1
V2
Vo
+
Vo
i1~0
i2~0
+
Rout
VoRload
outload
loadoload RR
RVV
Op-Amp Buffer
Vout = VinIsolates loading effects
A
High output impedance
B
Low input impedance
Op-Amp Differentiator
20
Op-Amp Differentiator
RCdt
dVv io
+
RC
VoVi
0to t1 t2
0
to t1 t2
Op-Amp Integrator
Op-Amp Summing Amplifier
Op-Amp Differential Amplifier
If R1 = R2 and Rf = Rg
Closed-loop gain
2
21
21
21
R
RRv
RRARAv
v
IN
INo
Closed-loop gain Af=vovin
Suppose A=106 R1=9R R2=R
10INo vv
Gain
INf vA
)1(2
1
R
RAf
Closed-loop gain determined by resistor ratio insensitive to A temperature
Negative feedback
1048774
)1(2
1
R
R
v
vA
i
of
1
2
R
R
v
vA
i
of
bull We can adjust the closed-loop gain by changing the ratio of R2 and R1
bull The closed-loop gain is (ideally) independent of op amp open-loop gain A (if A is large enough) and we can make it arbitrarily large or small and with the desired accuracy depending on the accuracy of the resistors
1048774
The terminal 1 is a virtual ground since terminal 2 is grounded
Inverting configuration
This is a classic example of what negative feedback does It takes an amplifier with very large gain and through negative feedback obtain a gain that is smaller stable and predictable In effect we have traded gain for accuracy This kind of trade off is common in electronic circuit designhellip as we will see more later
Negative feedback
1048774
Inverting configuration
Input ResistanceAssuming an ideal op amp (open-loop gain A = infinity) in the closed-loop inverting configuration the input resistance is R1
111 R
Rv
v
i
vR
i
iiin
Negative feedback
1
2
R
R
v
vA
i
of
Inverting configuration011
oi
in RRi
vR
We can model the closed-loop inverting amplifier (with A = infinite) with the following equivalent circuit using a voltage-controlled voltage sourcehellip
Negative feedback 1048774
29
Frequency-Gain Relationbull Ideally signals are amplified
from DC to the highest AC frequency
bull Practically bandwidth is limitedbull 741 family op-amp have an limit
bandwidth of few KHzbull Unity Gain frequency f1 the gain at
unitybull Cutoff frequency fc the gain drop by
3dB from dc gain Gd
(Voltage Gain)
(frequency)f1
Gd
0707Gd
fc0
1
GB Product f1 = Gd fc
20log(0707)=3dB
30
GB ProductExample Determine the cutoff frequency of an op-amp having a unit gain frequency f1 = 10 MHz and voltage differential gain Gd = 20VmV
Sol
Since f1 = 10 MHz
By using GB production equation
f1 = Gd fc
fc = f1 Gd = 10 MHz 20 VmV
= 10 106 20 103
= 500 Hz
(Voltage Gain)
(frequency)f1
Gd
0707Gd
fc0
1
10MHz
Hz
Op Amp transfer characteristic curve
saturation
)( vvAvAv io
active region
So far we have been looking at the amplification that can be achieved for relatively small (amplitude) signals For a fixed gain as we increase the input signal amplitude there is a limit to how large the output signal can be The output saturates as it approaches the positive andnegative power supply voltages In other words there is limited range across which the gain is linear vv Vvo
vv Vvo
Op Amp transfer characteristic curve
Applications of Op-AmpsFiltersTypesbullLow pass filterbullHigh pass filterbullBand pass filterbullCascading (2 or more filters connected together) R2
+
-
+
V0
__
+ Vcc
- Vcc
-+
R1
C
Low pass filter
Low pass filter Cutoff frequency
Low pass filter transfer function
Applications of Op-Ampsbull Electrocardiogram EKG Amplification
ndash Need to measure difference in voltage from lead 1 and lead 2ndash 60 Hz interference from electrical equipment
Applications of Op-Ampsbull Simple EKG circuit
ndash Uses differential amplifier to cancel common mode signal and amplify differential mode signal
bull Realistic EKG circuitndash Uses two non-inverting
amplifiers to first amplify voltage from each lead followed by differential amplifier
ndash Forms an ldquoinstrumentation amplifierrdquo
Strain Gauge
Use a Wheatstone bridge to determine the strain of an element by measuring the change in resistance of a
strain gauge
(No strain) Balanced Bridge R 1 = R 2
(Strain) Unbalanced Bridge R 1 ne R 2
Strain GaugeHalf-Bridge Arrangement
Using KCL at the inverting and non-inverting terminals of the op amp we find that ε ~ Vo = 2ΔR(Rf R2)
R + ΔR Rf
+
- +
V0
__
+ Vcc
- Vcc
-+
Rf
Vref
R
R - ΔR
R
Op amp used to amplify output from strain gauge
Applications of Op-Ampsbull Piezoelectric Transducerndash Used to measure force pressure accelerationndash Piezoelectric crystal generates an electric charge in
response to deformation
bull Use Charge Amplifierndash Just an integrator op-amp circuit
bullGoal is to have VSET = VOUT
bullRemember that VERROR = VSET ndash VSENSOR
bullOutput Process uses VERROR from the PID controller to adjust Vout such that it is ~VSET
P
I
D
Output Process
Sensor
VERRORVSET VOUT
VSENSOR
PID Controller ndash System Block Diagram
ApplicationsPID Controller ndash System Circuit Diagram
Source httpwwwecircuitcentercomCircuitsop_pidop_pidhtm
Calculates VERROR = -(VSET + VSENSOR)
Signal conditioning allows you to introduce a time delay which could
account for things like inertia
System to control
-VSENSOR
ApplicationsPID Controller ndash PID Controller Circuit Diagram
VERR
Adjust Change
Kp RP1 RP2
Ki RI CI
Kd RD CD
VERR PID
Applications of Op-Ampsbull Example of PI Control
Temperature Control
bull Thermal System we wish to automatically control the temperature of
bull Block Diagram of Control System
Applications of Op-Amps
bull Voltage Error Circuit
bull Proportional-Integral Control Circuit
bull Example of PI Control Temperature Control
44
Multiple Inputs(1) Kirchhoff node equation at V+
yields
(2) Kirchhoff node equation at V yields
(3) Setting V+ = Vndash yields
0V
0_
c
c
b
b
a
a
f
o
R
VV
R
VV
R
VV
R
VV
c
aj j
jf
c
c
b
b
a
afo R
VR
R
V
R
V
R
VRV
+
Rf
Va
VoRb
Ra
RcVb
Vc
Homework
1) Design a circuit to 50i
of V
VA
2) Find the vo=
Review Two fundamental Op Amp
Structure Af Input voltage ( )terminal
Feed back
( )terminal
Inverting Amp
_ _
Non inverting Amp
+ _
Ref080114HKN Operational Amplifier 13
Ideal Op-Amp ApplicationsAnalysis Method Two ideal Op-Amp Properties(1) The voltage between V+ and V is zero V+ = V
(2) The current into both V+ and V termainals is zero
For ideal Op-Amp circuit(1) Write the kirchhoff node equation at the noninverting
terminal V+ (2) Write the kirchhoff node eqaution at the inverting
terminal V (3) Set V+ = V and solve for the desired closed-loop gain
2 TYPES
1 Inverting Amplifier 2 Non-Inverting Amplifier
1 Inverting Amplifier Analysis
virtual ground
2 Non-Inverting Amplifier Analysis
17
Op-Amp Properties(1) Infinite Open Loop gain
- The gain without feedback- Equal to differential gain- Zero common-mode gain- Pratically Gd = 20000 to 200000
(2) Infinite Input impedance- Input current ii ~0A- T- in high-grade op-amp - m-A input current in low-grade op-
amp
(3) Zero Output Impedance- act as perfect internal voltage source- No internal resistance- Output impedance in series with load- Reducing output voltage to the load- Practically Rout ~ 20-100
+
V1
V2
Vo
+
Vo
i1~0
i2~0
+
Rout
VoRload
outload
loadoload RR
RVV
Op-Amp Buffer
Vout = VinIsolates loading effects
A
High output impedance
B
Low input impedance
Op-Amp Differentiator
20
Op-Amp Differentiator
RCdt
dVv io
+
RC
VoVi
0to t1 t2
0
to t1 t2
Op-Amp Integrator
Op-Amp Summing Amplifier
Op-Amp Differential Amplifier
If R1 = R2 and Rf = Rg
Closed-loop gain
2
21
21
21
R
RRv
RRARAv
v
IN
INo
Closed-loop gain Af=vovin
Suppose A=106 R1=9R R2=R
10INo vv
Gain
INf vA
)1(2
1
R
RAf
Closed-loop gain determined by resistor ratio insensitive to A temperature
Negative feedback
1048774
)1(2
1
R
R
v
vA
i
of
1
2
R
R
v
vA
i
of
bull We can adjust the closed-loop gain by changing the ratio of R2 and R1
bull The closed-loop gain is (ideally) independent of op amp open-loop gain A (if A is large enough) and we can make it arbitrarily large or small and with the desired accuracy depending on the accuracy of the resistors
1048774
The terminal 1 is a virtual ground since terminal 2 is grounded
Inverting configuration
This is a classic example of what negative feedback does It takes an amplifier with very large gain and through negative feedback obtain a gain that is smaller stable and predictable In effect we have traded gain for accuracy This kind of trade off is common in electronic circuit designhellip as we will see more later
Negative feedback
1048774
Inverting configuration
Input ResistanceAssuming an ideal op amp (open-loop gain A = infinity) in the closed-loop inverting configuration the input resistance is R1
111 R
Rv
v
i
vR
i
iiin
Negative feedback
1
2
R
R
v
vA
i
of
Inverting configuration011
oi
in RRi
vR
We can model the closed-loop inverting amplifier (with A = infinite) with the following equivalent circuit using a voltage-controlled voltage sourcehellip
Negative feedback 1048774
29
Frequency-Gain Relationbull Ideally signals are amplified
from DC to the highest AC frequency
bull Practically bandwidth is limitedbull 741 family op-amp have an limit
bandwidth of few KHzbull Unity Gain frequency f1 the gain at
unitybull Cutoff frequency fc the gain drop by
3dB from dc gain Gd
(Voltage Gain)
(frequency)f1
Gd
0707Gd
fc0
1
GB Product f1 = Gd fc
20log(0707)=3dB
30
GB ProductExample Determine the cutoff frequency of an op-amp having a unit gain frequency f1 = 10 MHz and voltage differential gain Gd = 20VmV
Sol
Since f1 = 10 MHz
By using GB production equation
f1 = Gd fc
fc = f1 Gd = 10 MHz 20 VmV
= 10 106 20 103
= 500 Hz
(Voltage Gain)
(frequency)f1
Gd
0707Gd
fc0
1
10MHz
Hz
Op Amp transfer characteristic curve
saturation
)( vvAvAv io
active region
So far we have been looking at the amplification that can be achieved for relatively small (amplitude) signals For a fixed gain as we increase the input signal amplitude there is a limit to how large the output signal can be The output saturates as it approaches the positive andnegative power supply voltages In other words there is limited range across which the gain is linear vv Vvo
vv Vvo
Op Amp transfer characteristic curve
Applications of Op-AmpsFiltersTypesbullLow pass filterbullHigh pass filterbullBand pass filterbullCascading (2 or more filters connected together) R2
+
-
+
V0
__
+ Vcc
- Vcc
-+
R1
C
Low pass filter
Low pass filter Cutoff frequency
Low pass filter transfer function
Applications of Op-Ampsbull Electrocardiogram EKG Amplification
ndash Need to measure difference in voltage from lead 1 and lead 2ndash 60 Hz interference from electrical equipment
Applications of Op-Ampsbull Simple EKG circuit
ndash Uses differential amplifier to cancel common mode signal and amplify differential mode signal
bull Realistic EKG circuitndash Uses two non-inverting
amplifiers to first amplify voltage from each lead followed by differential amplifier
ndash Forms an ldquoinstrumentation amplifierrdquo
Strain Gauge
Use a Wheatstone bridge to determine the strain of an element by measuring the change in resistance of a
strain gauge
(No strain) Balanced Bridge R 1 = R 2
(Strain) Unbalanced Bridge R 1 ne R 2
Strain GaugeHalf-Bridge Arrangement
Using KCL at the inverting and non-inverting terminals of the op amp we find that ε ~ Vo = 2ΔR(Rf R2)
R + ΔR Rf
+
- +
V0
__
+ Vcc
- Vcc
-+
Rf
Vref
R
R - ΔR
R
Op amp used to amplify output from strain gauge
Applications of Op-Ampsbull Piezoelectric Transducerndash Used to measure force pressure accelerationndash Piezoelectric crystal generates an electric charge in
response to deformation
bull Use Charge Amplifierndash Just an integrator op-amp circuit
bullGoal is to have VSET = VOUT
bullRemember that VERROR = VSET ndash VSENSOR
bullOutput Process uses VERROR from the PID controller to adjust Vout such that it is ~VSET
P
I
D
Output Process
Sensor
VERRORVSET VOUT
VSENSOR
PID Controller ndash System Block Diagram
ApplicationsPID Controller ndash System Circuit Diagram
Source httpwwwecircuitcentercomCircuitsop_pidop_pidhtm
Calculates VERROR = -(VSET + VSENSOR)
Signal conditioning allows you to introduce a time delay which could
account for things like inertia
System to control
-VSENSOR
ApplicationsPID Controller ndash PID Controller Circuit Diagram
VERR
Adjust Change
Kp RP1 RP2
Ki RI CI
Kd RD CD
VERR PID
Applications of Op-Ampsbull Example of PI Control
Temperature Control
bull Thermal System we wish to automatically control the temperature of
bull Block Diagram of Control System
Applications of Op-Amps
bull Voltage Error Circuit
bull Proportional-Integral Control Circuit
bull Example of PI Control Temperature Control
44
Multiple Inputs(1) Kirchhoff node equation at V+
yields
(2) Kirchhoff node equation at V yields
(3) Setting V+ = Vndash yields
0V
0_
c
c
b
b
a
a
f
o
R
VV
R
VV
R
VV
R
VV
c
aj j
jf
c
c
b
b
a
afo R
VR
R
V
R
V
R
VRV
+
Rf
Va
VoRb
Ra
RcVb
Vc
Homework
1) Design a circuit to 50i
of V
VA
2) Find the vo=
Review Two fundamental Op Amp
Structure Af Input voltage ( )terminal
Feed back
( )terminal
Inverting Amp
_ _
Non inverting Amp
+ _
2 TYPES
1 Inverting Amplifier 2 Non-Inverting Amplifier
1 Inverting Amplifier Analysis
virtual ground
2 Non-Inverting Amplifier Analysis
17
Op-Amp Properties(1) Infinite Open Loop gain
- The gain without feedback- Equal to differential gain- Zero common-mode gain- Pratically Gd = 20000 to 200000
(2) Infinite Input impedance- Input current ii ~0A- T- in high-grade op-amp - m-A input current in low-grade op-
amp
(3) Zero Output Impedance- act as perfect internal voltage source- No internal resistance- Output impedance in series with load- Reducing output voltage to the load- Practically Rout ~ 20-100
+
V1
V2
Vo
+
Vo
i1~0
i2~0
+
Rout
VoRload
outload
loadoload RR
RVV
Op-Amp Buffer
Vout = VinIsolates loading effects
A
High output impedance
B
Low input impedance
Op-Amp Differentiator
20
Op-Amp Differentiator
RCdt
dVv io
+
RC
VoVi
0to t1 t2
0
to t1 t2
Op-Amp Integrator
Op-Amp Summing Amplifier
Op-Amp Differential Amplifier
If R1 = R2 and Rf = Rg
Closed-loop gain
2
21
21
21
R
RRv
RRARAv
v
IN
INo
Closed-loop gain Af=vovin
Suppose A=106 R1=9R R2=R
10INo vv
Gain
INf vA
)1(2
1
R
RAf
Closed-loop gain determined by resistor ratio insensitive to A temperature
Negative feedback
1048774
)1(2
1
R
R
v
vA
i
of
1
2
R
R
v
vA
i
of
bull We can adjust the closed-loop gain by changing the ratio of R2 and R1
bull The closed-loop gain is (ideally) independent of op amp open-loop gain A (if A is large enough) and we can make it arbitrarily large or small and with the desired accuracy depending on the accuracy of the resistors
1048774
The terminal 1 is a virtual ground since terminal 2 is grounded
Inverting configuration
This is a classic example of what negative feedback does It takes an amplifier with very large gain and through negative feedback obtain a gain that is smaller stable and predictable In effect we have traded gain for accuracy This kind of trade off is common in electronic circuit designhellip as we will see more later
Negative feedback
1048774
Inverting configuration
Input ResistanceAssuming an ideal op amp (open-loop gain A = infinity) in the closed-loop inverting configuration the input resistance is R1
111 R
Rv
v
i
vR
i
iiin
Negative feedback
1
2
R
R
v
vA
i
of
Inverting configuration011
oi
in RRi
vR
We can model the closed-loop inverting amplifier (with A = infinite) with the following equivalent circuit using a voltage-controlled voltage sourcehellip
Negative feedback 1048774
29
Frequency-Gain Relationbull Ideally signals are amplified
from DC to the highest AC frequency
bull Practically bandwidth is limitedbull 741 family op-amp have an limit
bandwidth of few KHzbull Unity Gain frequency f1 the gain at
unitybull Cutoff frequency fc the gain drop by
3dB from dc gain Gd
(Voltage Gain)
(frequency)f1
Gd
0707Gd
fc0
1
GB Product f1 = Gd fc
20log(0707)=3dB
30
GB ProductExample Determine the cutoff frequency of an op-amp having a unit gain frequency f1 = 10 MHz and voltage differential gain Gd = 20VmV
Sol
Since f1 = 10 MHz
By using GB production equation
f1 = Gd fc
fc = f1 Gd = 10 MHz 20 VmV
= 10 106 20 103
= 500 Hz
(Voltage Gain)
(frequency)f1
Gd
0707Gd
fc0
1
10MHz
Hz
Op Amp transfer characteristic curve
saturation
)( vvAvAv io
active region
So far we have been looking at the amplification that can be achieved for relatively small (amplitude) signals For a fixed gain as we increase the input signal amplitude there is a limit to how large the output signal can be The output saturates as it approaches the positive andnegative power supply voltages In other words there is limited range across which the gain is linear vv Vvo
vv Vvo
Op Amp transfer characteristic curve
Applications of Op-AmpsFiltersTypesbullLow pass filterbullHigh pass filterbullBand pass filterbullCascading (2 or more filters connected together) R2
+
-
+
V0
__
+ Vcc
- Vcc
-+
R1
C
Low pass filter
Low pass filter Cutoff frequency
Low pass filter transfer function
Applications of Op-Ampsbull Electrocardiogram EKG Amplification
ndash Need to measure difference in voltage from lead 1 and lead 2ndash 60 Hz interference from electrical equipment
Applications of Op-Ampsbull Simple EKG circuit
ndash Uses differential amplifier to cancel common mode signal and amplify differential mode signal
bull Realistic EKG circuitndash Uses two non-inverting
amplifiers to first amplify voltage from each lead followed by differential amplifier
ndash Forms an ldquoinstrumentation amplifierrdquo
Strain Gauge
Use a Wheatstone bridge to determine the strain of an element by measuring the change in resistance of a
strain gauge
(No strain) Balanced Bridge R 1 = R 2
(Strain) Unbalanced Bridge R 1 ne R 2
Strain GaugeHalf-Bridge Arrangement
Using KCL at the inverting and non-inverting terminals of the op amp we find that ε ~ Vo = 2ΔR(Rf R2)
R + ΔR Rf
+
- +
V0
__
+ Vcc
- Vcc
-+
Rf
Vref
R
R - ΔR
R
Op amp used to amplify output from strain gauge
Applications of Op-Ampsbull Piezoelectric Transducerndash Used to measure force pressure accelerationndash Piezoelectric crystal generates an electric charge in
response to deformation
bull Use Charge Amplifierndash Just an integrator op-amp circuit
bullGoal is to have VSET = VOUT
bullRemember that VERROR = VSET ndash VSENSOR
bullOutput Process uses VERROR from the PID controller to adjust Vout such that it is ~VSET
P
I
D
Output Process
Sensor
VERRORVSET VOUT
VSENSOR
PID Controller ndash System Block Diagram
ApplicationsPID Controller ndash System Circuit Diagram
Source httpwwwecircuitcentercomCircuitsop_pidop_pidhtm
Calculates VERROR = -(VSET + VSENSOR)
Signal conditioning allows you to introduce a time delay which could
account for things like inertia
System to control
-VSENSOR
ApplicationsPID Controller ndash PID Controller Circuit Diagram
VERR
Adjust Change
Kp RP1 RP2
Ki RI CI
Kd RD CD
VERR PID
Applications of Op-Ampsbull Example of PI Control
Temperature Control
bull Thermal System we wish to automatically control the temperature of
bull Block Diagram of Control System
Applications of Op-Amps
bull Voltage Error Circuit
bull Proportional-Integral Control Circuit
bull Example of PI Control Temperature Control
44
Multiple Inputs(1) Kirchhoff node equation at V+
yields
(2) Kirchhoff node equation at V yields
(3) Setting V+ = Vndash yields
0V
0_
c
c
b
b
a
a
f
o
R
VV
R
VV
R
VV
R
VV
c
aj j
jf
c
c
b
b
a
afo R
VR
R
V
R
V
R
VRV
+
Rf
Va
VoRb
Ra
RcVb
Vc
Homework
1) Design a circuit to 50i
of V
VA
2) Find the vo=
Review Two fundamental Op Amp
Structure Af Input voltage ( )terminal
Feed back
( )terminal
Inverting Amp
_ _
Non inverting Amp
+ _
1 Inverting Amplifier Analysis
virtual ground
2 Non-Inverting Amplifier Analysis
17
Op-Amp Properties(1) Infinite Open Loop gain
- The gain without feedback- Equal to differential gain- Zero common-mode gain- Pratically Gd = 20000 to 200000
(2) Infinite Input impedance- Input current ii ~0A- T- in high-grade op-amp - m-A input current in low-grade op-
amp
(3) Zero Output Impedance- act as perfect internal voltage source- No internal resistance- Output impedance in series with load- Reducing output voltage to the load- Practically Rout ~ 20-100
+
V1
V2
Vo
+
Vo
i1~0
i2~0
+
Rout
VoRload
outload
loadoload RR
RVV
Op-Amp Buffer
Vout = VinIsolates loading effects
A
High output impedance
B
Low input impedance
Op-Amp Differentiator
20
Op-Amp Differentiator
RCdt
dVv io
+
RC
VoVi
0to t1 t2
0
to t1 t2
Op-Amp Integrator
Op-Amp Summing Amplifier
Op-Amp Differential Amplifier
If R1 = R2 and Rf = Rg
Closed-loop gain
2
21
21
21
R
RRv
RRARAv
v
IN
INo
Closed-loop gain Af=vovin
Suppose A=106 R1=9R R2=R
10INo vv
Gain
INf vA
)1(2
1
R
RAf
Closed-loop gain determined by resistor ratio insensitive to A temperature
Negative feedback
1048774
)1(2
1
R
R
v
vA
i
of
1
2
R
R
v
vA
i
of
bull We can adjust the closed-loop gain by changing the ratio of R2 and R1
bull The closed-loop gain is (ideally) independent of op amp open-loop gain A (if A is large enough) and we can make it arbitrarily large or small and with the desired accuracy depending on the accuracy of the resistors
1048774
The terminal 1 is a virtual ground since terminal 2 is grounded
Inverting configuration
This is a classic example of what negative feedback does It takes an amplifier with very large gain and through negative feedback obtain a gain that is smaller stable and predictable In effect we have traded gain for accuracy This kind of trade off is common in electronic circuit designhellip as we will see more later
Negative feedback
1048774
Inverting configuration
Input ResistanceAssuming an ideal op amp (open-loop gain A = infinity) in the closed-loop inverting configuration the input resistance is R1
111 R
Rv
v
i
vR
i
iiin
Negative feedback
1
2
R
R
v
vA
i
of
Inverting configuration011
oi
in RRi
vR
We can model the closed-loop inverting amplifier (with A = infinite) with the following equivalent circuit using a voltage-controlled voltage sourcehellip
Negative feedback 1048774
29
Frequency-Gain Relationbull Ideally signals are amplified
from DC to the highest AC frequency
bull Practically bandwidth is limitedbull 741 family op-amp have an limit
bandwidth of few KHzbull Unity Gain frequency f1 the gain at
unitybull Cutoff frequency fc the gain drop by
3dB from dc gain Gd
(Voltage Gain)
(frequency)f1
Gd
0707Gd
fc0
1
GB Product f1 = Gd fc
20log(0707)=3dB
30
GB ProductExample Determine the cutoff frequency of an op-amp having a unit gain frequency f1 = 10 MHz and voltage differential gain Gd = 20VmV
Sol
Since f1 = 10 MHz
By using GB production equation
f1 = Gd fc
fc = f1 Gd = 10 MHz 20 VmV
= 10 106 20 103
= 500 Hz
(Voltage Gain)
(frequency)f1
Gd
0707Gd
fc0
1
10MHz
Hz
Op Amp transfer characteristic curve
saturation
)( vvAvAv io
active region
So far we have been looking at the amplification that can be achieved for relatively small (amplitude) signals For a fixed gain as we increase the input signal amplitude there is a limit to how large the output signal can be The output saturates as it approaches the positive andnegative power supply voltages In other words there is limited range across which the gain is linear vv Vvo
vv Vvo
Op Amp transfer characteristic curve
Applications of Op-AmpsFiltersTypesbullLow pass filterbullHigh pass filterbullBand pass filterbullCascading (2 or more filters connected together) R2
+
-
+
V0
__
+ Vcc
- Vcc
-+
R1
C
Low pass filter
Low pass filter Cutoff frequency
Low pass filter transfer function
Applications of Op-Ampsbull Electrocardiogram EKG Amplification
ndash Need to measure difference in voltage from lead 1 and lead 2ndash 60 Hz interference from electrical equipment
Applications of Op-Ampsbull Simple EKG circuit
ndash Uses differential amplifier to cancel common mode signal and amplify differential mode signal
bull Realistic EKG circuitndash Uses two non-inverting
amplifiers to first amplify voltage from each lead followed by differential amplifier
ndash Forms an ldquoinstrumentation amplifierrdquo
Strain Gauge
Use a Wheatstone bridge to determine the strain of an element by measuring the change in resistance of a
strain gauge
(No strain) Balanced Bridge R 1 = R 2
(Strain) Unbalanced Bridge R 1 ne R 2
Strain GaugeHalf-Bridge Arrangement
Using KCL at the inverting and non-inverting terminals of the op amp we find that ε ~ Vo = 2ΔR(Rf R2)
R + ΔR Rf
+
- +
V0
__
+ Vcc
- Vcc
-+
Rf
Vref
R
R - ΔR
R
Op amp used to amplify output from strain gauge
Applications of Op-Ampsbull Piezoelectric Transducerndash Used to measure force pressure accelerationndash Piezoelectric crystal generates an electric charge in
response to deformation
bull Use Charge Amplifierndash Just an integrator op-amp circuit
bullGoal is to have VSET = VOUT
bullRemember that VERROR = VSET ndash VSENSOR
bullOutput Process uses VERROR from the PID controller to adjust Vout such that it is ~VSET
P
I
D
Output Process
Sensor
VERRORVSET VOUT
VSENSOR
PID Controller ndash System Block Diagram
ApplicationsPID Controller ndash System Circuit Diagram
Source httpwwwecircuitcentercomCircuitsop_pidop_pidhtm
Calculates VERROR = -(VSET + VSENSOR)
Signal conditioning allows you to introduce a time delay which could
account for things like inertia
System to control
-VSENSOR
ApplicationsPID Controller ndash PID Controller Circuit Diagram
VERR
Adjust Change
Kp RP1 RP2
Ki RI CI
Kd RD CD
VERR PID
Applications of Op-Ampsbull Example of PI Control
Temperature Control
bull Thermal System we wish to automatically control the temperature of
bull Block Diagram of Control System
Applications of Op-Amps
bull Voltage Error Circuit
bull Proportional-Integral Control Circuit
bull Example of PI Control Temperature Control
44
Multiple Inputs(1) Kirchhoff node equation at V+
yields
(2) Kirchhoff node equation at V yields
(3) Setting V+ = Vndash yields
0V
0_
c
c
b
b
a
a
f
o
R
VV
R
VV
R
VV
R
VV
c
aj j
jf
c
c
b
b
a
afo R
VR
R
V
R
V
R
VRV
+
Rf
Va
VoRb
Ra
RcVb
Vc
Homework
1) Design a circuit to 50i
of V
VA
2) Find the vo=
Review Two fundamental Op Amp
Structure Af Input voltage ( )terminal
Feed back
( )terminal
Inverting Amp
_ _
Non inverting Amp
+ _
2 Non-Inverting Amplifier Analysis
17
Op-Amp Properties(1) Infinite Open Loop gain
- The gain without feedback- Equal to differential gain- Zero common-mode gain- Pratically Gd = 20000 to 200000
(2) Infinite Input impedance- Input current ii ~0A- T- in high-grade op-amp - m-A input current in low-grade op-
amp
(3) Zero Output Impedance- act as perfect internal voltage source- No internal resistance- Output impedance in series with load- Reducing output voltage to the load- Practically Rout ~ 20-100
+
V1
V2
Vo
+
Vo
i1~0
i2~0
+
Rout
VoRload
outload
loadoload RR
RVV
Op-Amp Buffer
Vout = VinIsolates loading effects
A
High output impedance
B
Low input impedance
Op-Amp Differentiator
20
Op-Amp Differentiator
RCdt
dVv io
+
RC
VoVi
0to t1 t2
0
to t1 t2
Op-Amp Integrator
Op-Amp Summing Amplifier
Op-Amp Differential Amplifier
If R1 = R2 and Rf = Rg
Closed-loop gain
2
21
21
21
R
RRv
RRARAv
v
IN
INo
Closed-loop gain Af=vovin
Suppose A=106 R1=9R R2=R
10INo vv
Gain
INf vA
)1(2
1
R
RAf
Closed-loop gain determined by resistor ratio insensitive to A temperature
Negative feedback
1048774
)1(2
1
R
R
v
vA
i
of
1
2
R
R
v
vA
i
of
bull We can adjust the closed-loop gain by changing the ratio of R2 and R1
bull The closed-loop gain is (ideally) independent of op amp open-loop gain A (if A is large enough) and we can make it arbitrarily large or small and with the desired accuracy depending on the accuracy of the resistors
1048774
The terminal 1 is a virtual ground since terminal 2 is grounded
Inverting configuration
This is a classic example of what negative feedback does It takes an amplifier with very large gain and through negative feedback obtain a gain that is smaller stable and predictable In effect we have traded gain for accuracy This kind of trade off is common in electronic circuit designhellip as we will see more later
Negative feedback
1048774
Inverting configuration
Input ResistanceAssuming an ideal op amp (open-loop gain A = infinity) in the closed-loop inverting configuration the input resistance is R1
111 R
Rv
v
i
vR
i
iiin
Negative feedback
1
2
R
R
v
vA
i
of
Inverting configuration011
oi
in RRi
vR
We can model the closed-loop inverting amplifier (with A = infinite) with the following equivalent circuit using a voltage-controlled voltage sourcehellip
Negative feedback 1048774
29
Frequency-Gain Relationbull Ideally signals are amplified
from DC to the highest AC frequency
bull Practically bandwidth is limitedbull 741 family op-amp have an limit
bandwidth of few KHzbull Unity Gain frequency f1 the gain at
unitybull Cutoff frequency fc the gain drop by
3dB from dc gain Gd
(Voltage Gain)
(frequency)f1
Gd
0707Gd
fc0
1
GB Product f1 = Gd fc
20log(0707)=3dB
30
GB ProductExample Determine the cutoff frequency of an op-amp having a unit gain frequency f1 = 10 MHz and voltage differential gain Gd = 20VmV
Sol
Since f1 = 10 MHz
By using GB production equation
f1 = Gd fc
fc = f1 Gd = 10 MHz 20 VmV
= 10 106 20 103
= 500 Hz
(Voltage Gain)
(frequency)f1
Gd
0707Gd
fc0
1
10MHz
Hz
Op Amp transfer characteristic curve
saturation
)( vvAvAv io
active region
So far we have been looking at the amplification that can be achieved for relatively small (amplitude) signals For a fixed gain as we increase the input signal amplitude there is a limit to how large the output signal can be The output saturates as it approaches the positive andnegative power supply voltages In other words there is limited range across which the gain is linear vv Vvo
vv Vvo
Op Amp transfer characteristic curve
Applications of Op-AmpsFiltersTypesbullLow pass filterbullHigh pass filterbullBand pass filterbullCascading (2 or more filters connected together) R2
+
-
+
V0
__
+ Vcc
- Vcc
-+
R1
C
Low pass filter
Low pass filter Cutoff frequency
Low pass filter transfer function
Applications of Op-Ampsbull Electrocardiogram EKG Amplification
ndash Need to measure difference in voltage from lead 1 and lead 2ndash 60 Hz interference from electrical equipment
Applications of Op-Ampsbull Simple EKG circuit
ndash Uses differential amplifier to cancel common mode signal and amplify differential mode signal
bull Realistic EKG circuitndash Uses two non-inverting
amplifiers to first amplify voltage from each lead followed by differential amplifier
ndash Forms an ldquoinstrumentation amplifierrdquo
Strain Gauge
Use a Wheatstone bridge to determine the strain of an element by measuring the change in resistance of a
strain gauge
(No strain) Balanced Bridge R 1 = R 2
(Strain) Unbalanced Bridge R 1 ne R 2
Strain GaugeHalf-Bridge Arrangement
Using KCL at the inverting and non-inverting terminals of the op amp we find that ε ~ Vo = 2ΔR(Rf R2)
R + ΔR Rf
+
- +
V0
__
+ Vcc
- Vcc
-+
Rf
Vref
R
R - ΔR
R
Op amp used to amplify output from strain gauge
Applications of Op-Ampsbull Piezoelectric Transducerndash Used to measure force pressure accelerationndash Piezoelectric crystal generates an electric charge in
response to deformation
bull Use Charge Amplifierndash Just an integrator op-amp circuit
bullGoal is to have VSET = VOUT
bullRemember that VERROR = VSET ndash VSENSOR
bullOutput Process uses VERROR from the PID controller to adjust Vout such that it is ~VSET
P
I
D
Output Process
Sensor
VERRORVSET VOUT
VSENSOR
PID Controller ndash System Block Diagram
ApplicationsPID Controller ndash System Circuit Diagram
Source httpwwwecircuitcentercomCircuitsop_pidop_pidhtm
Calculates VERROR = -(VSET + VSENSOR)
Signal conditioning allows you to introduce a time delay which could
account for things like inertia
System to control
-VSENSOR
ApplicationsPID Controller ndash PID Controller Circuit Diagram
VERR
Adjust Change
Kp RP1 RP2
Ki RI CI
Kd RD CD
VERR PID
Applications of Op-Ampsbull Example of PI Control
Temperature Control
bull Thermal System we wish to automatically control the temperature of
bull Block Diagram of Control System
Applications of Op-Amps
bull Voltage Error Circuit
bull Proportional-Integral Control Circuit
bull Example of PI Control Temperature Control
44
Multiple Inputs(1) Kirchhoff node equation at V+
yields
(2) Kirchhoff node equation at V yields
(3) Setting V+ = Vndash yields
0V
0_
c
c
b
b
a
a
f
o
R
VV
R
VV
R
VV
R
VV
c
aj j
jf
c
c
b
b
a
afo R
VR
R
V
R
V
R
VRV
+
Rf
Va
VoRb
Ra
RcVb
Vc
Homework
1) Design a circuit to 50i
of V
VA
2) Find the vo=
Review Two fundamental Op Amp
Structure Af Input voltage ( )terminal
Feed back
( )terminal
Inverting Amp
_ _
Non inverting Amp
+ _
17
Op-Amp Properties(1) Infinite Open Loop gain
- The gain without feedback- Equal to differential gain- Zero common-mode gain- Pratically Gd = 20000 to 200000
(2) Infinite Input impedance- Input current ii ~0A- T- in high-grade op-amp - m-A input current in low-grade op-
amp
(3) Zero Output Impedance- act as perfect internal voltage source- No internal resistance- Output impedance in series with load- Reducing output voltage to the load- Practically Rout ~ 20-100
+
V1
V2
Vo
+
Vo
i1~0
i2~0
+
Rout
VoRload
outload
loadoload RR
RVV
Op-Amp Buffer
Vout = VinIsolates loading effects
A
High output impedance
B
Low input impedance
Op-Amp Differentiator
20
Op-Amp Differentiator
RCdt
dVv io
+
RC
VoVi
0to t1 t2
0
to t1 t2
Op-Amp Integrator
Op-Amp Summing Amplifier
Op-Amp Differential Amplifier
If R1 = R2 and Rf = Rg
Closed-loop gain
2
21
21
21
R
RRv
RRARAv
v
IN
INo
Closed-loop gain Af=vovin
Suppose A=106 R1=9R R2=R
10INo vv
Gain
INf vA
)1(2
1
R
RAf
Closed-loop gain determined by resistor ratio insensitive to A temperature
Negative feedback
1048774
)1(2
1
R
R
v
vA
i
of
1
2
R
R
v
vA
i
of
bull We can adjust the closed-loop gain by changing the ratio of R2 and R1
bull The closed-loop gain is (ideally) independent of op amp open-loop gain A (if A is large enough) and we can make it arbitrarily large or small and with the desired accuracy depending on the accuracy of the resistors
1048774
The terminal 1 is a virtual ground since terminal 2 is grounded
Inverting configuration
This is a classic example of what negative feedback does It takes an amplifier with very large gain and through negative feedback obtain a gain that is smaller stable and predictable In effect we have traded gain for accuracy This kind of trade off is common in electronic circuit designhellip as we will see more later
Negative feedback
1048774
Inverting configuration
Input ResistanceAssuming an ideal op amp (open-loop gain A = infinity) in the closed-loop inverting configuration the input resistance is R1
111 R
Rv
v
i
vR
i
iiin
Negative feedback
1
2
R
R
v
vA
i
of
Inverting configuration011
oi
in RRi
vR
We can model the closed-loop inverting amplifier (with A = infinite) with the following equivalent circuit using a voltage-controlled voltage sourcehellip
Negative feedback 1048774
29
Frequency-Gain Relationbull Ideally signals are amplified
from DC to the highest AC frequency
bull Practically bandwidth is limitedbull 741 family op-amp have an limit
bandwidth of few KHzbull Unity Gain frequency f1 the gain at
unitybull Cutoff frequency fc the gain drop by
3dB from dc gain Gd
(Voltage Gain)
(frequency)f1
Gd
0707Gd
fc0
1
GB Product f1 = Gd fc
20log(0707)=3dB
30
GB ProductExample Determine the cutoff frequency of an op-amp having a unit gain frequency f1 = 10 MHz and voltage differential gain Gd = 20VmV
Sol
Since f1 = 10 MHz
By using GB production equation
f1 = Gd fc
fc = f1 Gd = 10 MHz 20 VmV
= 10 106 20 103
= 500 Hz
(Voltage Gain)
(frequency)f1
Gd
0707Gd
fc0
1
10MHz
Hz
Op Amp transfer characteristic curve
saturation
)( vvAvAv io
active region
So far we have been looking at the amplification that can be achieved for relatively small (amplitude) signals For a fixed gain as we increase the input signal amplitude there is a limit to how large the output signal can be The output saturates as it approaches the positive andnegative power supply voltages In other words there is limited range across which the gain is linear vv Vvo
vv Vvo
Op Amp transfer characteristic curve
Applications of Op-AmpsFiltersTypesbullLow pass filterbullHigh pass filterbullBand pass filterbullCascading (2 or more filters connected together) R2
+
-
+
V0
__
+ Vcc
- Vcc
-+
R1
C
Low pass filter
Low pass filter Cutoff frequency
Low pass filter transfer function
Applications of Op-Ampsbull Electrocardiogram EKG Amplification
ndash Need to measure difference in voltage from lead 1 and lead 2ndash 60 Hz interference from electrical equipment
Applications of Op-Ampsbull Simple EKG circuit
ndash Uses differential amplifier to cancel common mode signal and amplify differential mode signal
bull Realistic EKG circuitndash Uses two non-inverting
amplifiers to first amplify voltage from each lead followed by differential amplifier
ndash Forms an ldquoinstrumentation amplifierrdquo
Strain Gauge
Use a Wheatstone bridge to determine the strain of an element by measuring the change in resistance of a
strain gauge
(No strain) Balanced Bridge R 1 = R 2
(Strain) Unbalanced Bridge R 1 ne R 2
Strain GaugeHalf-Bridge Arrangement
Using KCL at the inverting and non-inverting terminals of the op amp we find that ε ~ Vo = 2ΔR(Rf R2)
R + ΔR Rf
+
- +
V0
__
+ Vcc
- Vcc
-+
Rf
Vref
R
R - ΔR
R
Op amp used to amplify output from strain gauge
Applications of Op-Ampsbull Piezoelectric Transducerndash Used to measure force pressure accelerationndash Piezoelectric crystal generates an electric charge in
response to deformation
bull Use Charge Amplifierndash Just an integrator op-amp circuit
bullGoal is to have VSET = VOUT
bullRemember that VERROR = VSET ndash VSENSOR
bullOutput Process uses VERROR from the PID controller to adjust Vout such that it is ~VSET
P
I
D
Output Process
Sensor
VERRORVSET VOUT
VSENSOR
PID Controller ndash System Block Diagram
ApplicationsPID Controller ndash System Circuit Diagram
Source httpwwwecircuitcentercomCircuitsop_pidop_pidhtm
Calculates VERROR = -(VSET + VSENSOR)
Signal conditioning allows you to introduce a time delay which could
account for things like inertia
System to control
-VSENSOR
ApplicationsPID Controller ndash PID Controller Circuit Diagram
VERR
Adjust Change
Kp RP1 RP2
Ki RI CI
Kd RD CD
VERR PID
Applications of Op-Ampsbull Example of PI Control
Temperature Control
bull Thermal System we wish to automatically control the temperature of
bull Block Diagram of Control System
Applications of Op-Amps
bull Voltage Error Circuit
bull Proportional-Integral Control Circuit
bull Example of PI Control Temperature Control
44
Multiple Inputs(1) Kirchhoff node equation at V+
yields
(2) Kirchhoff node equation at V yields
(3) Setting V+ = Vndash yields
0V
0_
c
c
b
b
a
a
f
o
R
VV
R
VV
R
VV
R
VV
c
aj j
jf
c
c
b
b
a
afo R
VR
R
V
R
V
R
VRV
+
Rf
Va
VoRb
Ra
RcVb
Vc
Homework
1) Design a circuit to 50i
of V
VA
2) Find the vo=
Review Two fundamental Op Amp
Structure Af Input voltage ( )terminal
Feed back
( )terminal
Inverting Amp
_ _
Non inverting Amp
+ _
Op-Amp Buffer
Vout = VinIsolates loading effects
A
High output impedance
B
Low input impedance
Op-Amp Differentiator
20
Op-Amp Differentiator
RCdt
dVv io
+
RC
VoVi
0to t1 t2
0
to t1 t2
Op-Amp Integrator
Op-Amp Summing Amplifier
Op-Amp Differential Amplifier
If R1 = R2 and Rf = Rg
Closed-loop gain
2
21
21
21
R
RRv
RRARAv
v
IN
INo
Closed-loop gain Af=vovin
Suppose A=106 R1=9R R2=R
10INo vv
Gain
INf vA
)1(2
1
R
RAf
Closed-loop gain determined by resistor ratio insensitive to A temperature
Negative feedback
1048774
)1(2
1
R
R
v
vA
i
of
1
2
R
R
v
vA
i
of
bull We can adjust the closed-loop gain by changing the ratio of R2 and R1
bull The closed-loop gain is (ideally) independent of op amp open-loop gain A (if A is large enough) and we can make it arbitrarily large or small and with the desired accuracy depending on the accuracy of the resistors
1048774
The terminal 1 is a virtual ground since terminal 2 is grounded
Inverting configuration
This is a classic example of what negative feedback does It takes an amplifier with very large gain and through negative feedback obtain a gain that is smaller stable and predictable In effect we have traded gain for accuracy This kind of trade off is common in electronic circuit designhellip as we will see more later
Negative feedback
1048774
Inverting configuration
Input ResistanceAssuming an ideal op amp (open-loop gain A = infinity) in the closed-loop inverting configuration the input resistance is R1
111 R
Rv
v
i
vR
i
iiin
Negative feedback
1
2
R
R
v
vA
i
of
Inverting configuration011
oi
in RRi
vR
We can model the closed-loop inverting amplifier (with A = infinite) with the following equivalent circuit using a voltage-controlled voltage sourcehellip
Negative feedback 1048774
29
Frequency-Gain Relationbull Ideally signals are amplified
from DC to the highest AC frequency
bull Practically bandwidth is limitedbull 741 family op-amp have an limit
bandwidth of few KHzbull Unity Gain frequency f1 the gain at
unitybull Cutoff frequency fc the gain drop by
3dB from dc gain Gd
(Voltage Gain)
(frequency)f1
Gd
0707Gd
fc0
1
GB Product f1 = Gd fc
20log(0707)=3dB
30
GB ProductExample Determine the cutoff frequency of an op-amp having a unit gain frequency f1 = 10 MHz and voltage differential gain Gd = 20VmV
Sol
Since f1 = 10 MHz
By using GB production equation
f1 = Gd fc
fc = f1 Gd = 10 MHz 20 VmV
= 10 106 20 103
= 500 Hz
(Voltage Gain)
(frequency)f1
Gd
0707Gd
fc0
1
10MHz
Hz
Op Amp transfer characteristic curve
saturation
)( vvAvAv io
active region
So far we have been looking at the amplification that can be achieved for relatively small (amplitude) signals For a fixed gain as we increase the input signal amplitude there is a limit to how large the output signal can be The output saturates as it approaches the positive andnegative power supply voltages In other words there is limited range across which the gain is linear vv Vvo
vv Vvo
Op Amp transfer characteristic curve
Applications of Op-AmpsFiltersTypesbullLow pass filterbullHigh pass filterbullBand pass filterbullCascading (2 or more filters connected together) R2
+
-
+
V0
__
+ Vcc
- Vcc
-+
R1
C
Low pass filter
Low pass filter Cutoff frequency
Low pass filter transfer function
Applications of Op-Ampsbull Electrocardiogram EKG Amplification
ndash Need to measure difference in voltage from lead 1 and lead 2ndash 60 Hz interference from electrical equipment
Applications of Op-Ampsbull Simple EKG circuit
ndash Uses differential amplifier to cancel common mode signal and amplify differential mode signal
bull Realistic EKG circuitndash Uses two non-inverting
amplifiers to first amplify voltage from each lead followed by differential amplifier
ndash Forms an ldquoinstrumentation amplifierrdquo
Strain Gauge
Use a Wheatstone bridge to determine the strain of an element by measuring the change in resistance of a
strain gauge
(No strain) Balanced Bridge R 1 = R 2
(Strain) Unbalanced Bridge R 1 ne R 2
Strain GaugeHalf-Bridge Arrangement
Using KCL at the inverting and non-inverting terminals of the op amp we find that ε ~ Vo = 2ΔR(Rf R2)
R + ΔR Rf
+
- +
V0
__
+ Vcc
- Vcc
-+
Rf
Vref
R
R - ΔR
R
Op amp used to amplify output from strain gauge
Applications of Op-Ampsbull Piezoelectric Transducerndash Used to measure force pressure accelerationndash Piezoelectric crystal generates an electric charge in
response to deformation
bull Use Charge Amplifierndash Just an integrator op-amp circuit
bullGoal is to have VSET = VOUT
bullRemember that VERROR = VSET ndash VSENSOR
bullOutput Process uses VERROR from the PID controller to adjust Vout such that it is ~VSET
P
I
D
Output Process
Sensor
VERRORVSET VOUT
VSENSOR
PID Controller ndash System Block Diagram
ApplicationsPID Controller ndash System Circuit Diagram
Source httpwwwecircuitcentercomCircuitsop_pidop_pidhtm
Calculates VERROR = -(VSET + VSENSOR)
Signal conditioning allows you to introduce a time delay which could
account for things like inertia
System to control
-VSENSOR
ApplicationsPID Controller ndash PID Controller Circuit Diagram
VERR
Adjust Change
Kp RP1 RP2
Ki RI CI
Kd RD CD
VERR PID
Applications of Op-Ampsbull Example of PI Control
Temperature Control
bull Thermal System we wish to automatically control the temperature of
bull Block Diagram of Control System
Applications of Op-Amps
bull Voltage Error Circuit
bull Proportional-Integral Control Circuit
bull Example of PI Control Temperature Control
44
Multiple Inputs(1) Kirchhoff node equation at V+
yields
(2) Kirchhoff node equation at V yields
(3) Setting V+ = Vndash yields
0V
0_
c
c
b
b
a
a
f
o
R
VV
R
VV
R
VV
R
VV
c
aj j
jf
c
c
b
b
a
afo R
VR
R
V
R
V
R
VRV
+
Rf
Va
VoRb
Ra
RcVb
Vc
Homework
1) Design a circuit to 50i
of V
VA
2) Find the vo=
Review Two fundamental Op Amp
Structure Af Input voltage ( )terminal
Feed back
( )terminal
Inverting Amp
_ _
Non inverting Amp
+ _
Op-Amp Differentiator
20
Op-Amp Differentiator
RCdt
dVv io
+
RC
VoVi
0to t1 t2
0
to t1 t2
Op-Amp Integrator
Op-Amp Summing Amplifier
Op-Amp Differential Amplifier
If R1 = R2 and Rf = Rg
Closed-loop gain
2
21
21
21
R
RRv
RRARAv
v
IN
INo
Closed-loop gain Af=vovin
Suppose A=106 R1=9R R2=R
10INo vv
Gain
INf vA
)1(2
1
R
RAf
Closed-loop gain determined by resistor ratio insensitive to A temperature
Negative feedback
1048774
)1(2
1
R
R
v
vA
i
of
1
2
R
R
v
vA
i
of
bull We can adjust the closed-loop gain by changing the ratio of R2 and R1
bull The closed-loop gain is (ideally) independent of op amp open-loop gain A (if A is large enough) and we can make it arbitrarily large or small and with the desired accuracy depending on the accuracy of the resistors
1048774
The terminal 1 is a virtual ground since terminal 2 is grounded
Inverting configuration
This is a classic example of what negative feedback does It takes an amplifier with very large gain and through negative feedback obtain a gain that is smaller stable and predictable In effect we have traded gain for accuracy This kind of trade off is common in electronic circuit designhellip as we will see more later
Negative feedback
1048774
Inverting configuration
Input ResistanceAssuming an ideal op amp (open-loop gain A = infinity) in the closed-loop inverting configuration the input resistance is R1
111 R
Rv
v
i
vR
i
iiin
Negative feedback
1
2
R
R
v
vA
i
of
Inverting configuration011
oi
in RRi
vR
We can model the closed-loop inverting amplifier (with A = infinite) with the following equivalent circuit using a voltage-controlled voltage sourcehellip
Negative feedback 1048774
29
Frequency-Gain Relationbull Ideally signals are amplified
from DC to the highest AC frequency
bull Practically bandwidth is limitedbull 741 family op-amp have an limit
bandwidth of few KHzbull Unity Gain frequency f1 the gain at
unitybull Cutoff frequency fc the gain drop by
3dB from dc gain Gd
(Voltage Gain)
(frequency)f1
Gd
0707Gd
fc0
1
GB Product f1 = Gd fc
20log(0707)=3dB
30
GB ProductExample Determine the cutoff frequency of an op-amp having a unit gain frequency f1 = 10 MHz and voltage differential gain Gd = 20VmV
Sol
Since f1 = 10 MHz
By using GB production equation
f1 = Gd fc
fc = f1 Gd = 10 MHz 20 VmV
= 10 106 20 103
= 500 Hz
(Voltage Gain)
(frequency)f1
Gd
0707Gd
fc0
1
10MHz
Hz
Op Amp transfer characteristic curve
saturation
)( vvAvAv io
active region
So far we have been looking at the amplification that can be achieved for relatively small (amplitude) signals For a fixed gain as we increase the input signal amplitude there is a limit to how large the output signal can be The output saturates as it approaches the positive andnegative power supply voltages In other words there is limited range across which the gain is linear vv Vvo
vv Vvo
Op Amp transfer characteristic curve
Applications of Op-AmpsFiltersTypesbullLow pass filterbullHigh pass filterbullBand pass filterbullCascading (2 or more filters connected together) R2
+
-
+
V0
__
+ Vcc
- Vcc
-+
R1
C
Low pass filter
Low pass filter Cutoff frequency
Low pass filter transfer function
Applications of Op-Ampsbull Electrocardiogram EKG Amplification
ndash Need to measure difference in voltage from lead 1 and lead 2ndash 60 Hz interference from electrical equipment
Applications of Op-Ampsbull Simple EKG circuit
ndash Uses differential amplifier to cancel common mode signal and amplify differential mode signal
bull Realistic EKG circuitndash Uses two non-inverting
amplifiers to first amplify voltage from each lead followed by differential amplifier
ndash Forms an ldquoinstrumentation amplifierrdquo
Strain Gauge
Use a Wheatstone bridge to determine the strain of an element by measuring the change in resistance of a
strain gauge
(No strain) Balanced Bridge R 1 = R 2
(Strain) Unbalanced Bridge R 1 ne R 2
Strain GaugeHalf-Bridge Arrangement
Using KCL at the inverting and non-inverting terminals of the op amp we find that ε ~ Vo = 2ΔR(Rf R2)
R + ΔR Rf
+
- +
V0
__
+ Vcc
- Vcc
-+
Rf
Vref
R
R - ΔR
R
Op amp used to amplify output from strain gauge
Applications of Op-Ampsbull Piezoelectric Transducerndash Used to measure force pressure accelerationndash Piezoelectric crystal generates an electric charge in
response to deformation
bull Use Charge Amplifierndash Just an integrator op-amp circuit
bullGoal is to have VSET = VOUT
bullRemember that VERROR = VSET ndash VSENSOR
bullOutput Process uses VERROR from the PID controller to adjust Vout such that it is ~VSET
P
I
D
Output Process
Sensor
VERRORVSET VOUT
VSENSOR
PID Controller ndash System Block Diagram
ApplicationsPID Controller ndash System Circuit Diagram
Source httpwwwecircuitcentercomCircuitsop_pidop_pidhtm
Calculates VERROR = -(VSET + VSENSOR)
Signal conditioning allows you to introduce a time delay which could
account for things like inertia
System to control
-VSENSOR
ApplicationsPID Controller ndash PID Controller Circuit Diagram
VERR
Adjust Change
Kp RP1 RP2
Ki RI CI
Kd RD CD
VERR PID
Applications of Op-Ampsbull Example of PI Control
Temperature Control
bull Thermal System we wish to automatically control the temperature of
bull Block Diagram of Control System
Applications of Op-Amps
bull Voltage Error Circuit
bull Proportional-Integral Control Circuit
bull Example of PI Control Temperature Control
44
Multiple Inputs(1) Kirchhoff node equation at V+
yields
(2) Kirchhoff node equation at V yields
(3) Setting V+ = Vndash yields
0V
0_
c
c
b
b
a
a
f
o
R
VV
R
VV
R
VV
R
VV
c
aj j
jf
c
c
b
b
a
afo R
VR
R
V
R
V
R
VRV
+
Rf
Va
VoRb
Ra
RcVb
Vc
Homework
1) Design a circuit to 50i
of V
VA
2) Find the vo=
Review Two fundamental Op Amp
Structure Af Input voltage ( )terminal
Feed back
( )terminal
Inverting Amp
_ _
Non inverting Amp
+ _
20
Op-Amp Differentiator
RCdt
dVv io
+
RC
VoVi
0to t1 t2
0
to t1 t2
Op-Amp Integrator
Op-Amp Summing Amplifier
Op-Amp Differential Amplifier
If R1 = R2 and Rf = Rg
Closed-loop gain
2
21
21
21
R
RRv
RRARAv
v
IN
INo
Closed-loop gain Af=vovin
Suppose A=106 R1=9R R2=R
10INo vv
Gain
INf vA
)1(2
1
R
RAf
Closed-loop gain determined by resistor ratio insensitive to A temperature
Negative feedback
1048774
)1(2
1
R
R
v
vA
i
of
1
2
R
R
v
vA
i
of
bull We can adjust the closed-loop gain by changing the ratio of R2 and R1
bull The closed-loop gain is (ideally) independent of op amp open-loop gain A (if A is large enough) and we can make it arbitrarily large or small and with the desired accuracy depending on the accuracy of the resistors
1048774
The terminal 1 is a virtual ground since terminal 2 is grounded
Inverting configuration
This is a classic example of what negative feedback does It takes an amplifier with very large gain and through negative feedback obtain a gain that is smaller stable and predictable In effect we have traded gain for accuracy This kind of trade off is common in electronic circuit designhellip as we will see more later
Negative feedback
1048774
Inverting configuration
Input ResistanceAssuming an ideal op amp (open-loop gain A = infinity) in the closed-loop inverting configuration the input resistance is R1
111 R
Rv
v
i
vR
i
iiin
Negative feedback
1
2
R
R
v
vA
i
of
Inverting configuration011
oi
in RRi
vR
We can model the closed-loop inverting amplifier (with A = infinite) with the following equivalent circuit using a voltage-controlled voltage sourcehellip
Negative feedback 1048774
29
Frequency-Gain Relationbull Ideally signals are amplified
from DC to the highest AC frequency
bull Practically bandwidth is limitedbull 741 family op-amp have an limit
bandwidth of few KHzbull Unity Gain frequency f1 the gain at
unitybull Cutoff frequency fc the gain drop by
3dB from dc gain Gd
(Voltage Gain)
(frequency)f1
Gd
0707Gd
fc0
1
GB Product f1 = Gd fc
20log(0707)=3dB
30
GB ProductExample Determine the cutoff frequency of an op-amp having a unit gain frequency f1 = 10 MHz and voltage differential gain Gd = 20VmV
Sol
Since f1 = 10 MHz
By using GB production equation
f1 = Gd fc
fc = f1 Gd = 10 MHz 20 VmV
= 10 106 20 103
= 500 Hz
(Voltage Gain)
(frequency)f1
Gd
0707Gd
fc0
1
10MHz
Hz
Op Amp transfer characteristic curve
saturation
)( vvAvAv io
active region
So far we have been looking at the amplification that can be achieved for relatively small (amplitude) signals For a fixed gain as we increase the input signal amplitude there is a limit to how large the output signal can be The output saturates as it approaches the positive andnegative power supply voltages In other words there is limited range across which the gain is linear vv Vvo
vv Vvo
Op Amp transfer characteristic curve
Applications of Op-AmpsFiltersTypesbullLow pass filterbullHigh pass filterbullBand pass filterbullCascading (2 or more filters connected together) R2
+
-
+
V0
__
+ Vcc
- Vcc
-+
R1
C
Low pass filter
Low pass filter Cutoff frequency
Low pass filter transfer function
Applications of Op-Ampsbull Electrocardiogram EKG Amplification
ndash Need to measure difference in voltage from lead 1 and lead 2ndash 60 Hz interference from electrical equipment
Applications of Op-Ampsbull Simple EKG circuit
ndash Uses differential amplifier to cancel common mode signal and amplify differential mode signal
bull Realistic EKG circuitndash Uses two non-inverting
amplifiers to first amplify voltage from each lead followed by differential amplifier
ndash Forms an ldquoinstrumentation amplifierrdquo
Strain Gauge
Use a Wheatstone bridge to determine the strain of an element by measuring the change in resistance of a
strain gauge
(No strain) Balanced Bridge R 1 = R 2
(Strain) Unbalanced Bridge R 1 ne R 2
Strain GaugeHalf-Bridge Arrangement
Using KCL at the inverting and non-inverting terminals of the op amp we find that ε ~ Vo = 2ΔR(Rf R2)
R + ΔR Rf
+
- +
V0
__
+ Vcc
- Vcc
-+
Rf
Vref
R
R - ΔR
R
Op amp used to amplify output from strain gauge
Applications of Op-Ampsbull Piezoelectric Transducerndash Used to measure force pressure accelerationndash Piezoelectric crystal generates an electric charge in
response to deformation
bull Use Charge Amplifierndash Just an integrator op-amp circuit
bullGoal is to have VSET = VOUT
bullRemember that VERROR = VSET ndash VSENSOR
bullOutput Process uses VERROR from the PID controller to adjust Vout such that it is ~VSET
P
I
D
Output Process
Sensor
VERRORVSET VOUT
VSENSOR
PID Controller ndash System Block Diagram
ApplicationsPID Controller ndash System Circuit Diagram
Source httpwwwecircuitcentercomCircuitsop_pidop_pidhtm
Calculates VERROR = -(VSET + VSENSOR)
Signal conditioning allows you to introduce a time delay which could
account for things like inertia
System to control
-VSENSOR
ApplicationsPID Controller ndash PID Controller Circuit Diagram
VERR
Adjust Change
Kp RP1 RP2
Ki RI CI
Kd RD CD
VERR PID
Applications of Op-Ampsbull Example of PI Control
Temperature Control
bull Thermal System we wish to automatically control the temperature of
bull Block Diagram of Control System
Applications of Op-Amps
bull Voltage Error Circuit
bull Proportional-Integral Control Circuit
bull Example of PI Control Temperature Control
44
Multiple Inputs(1) Kirchhoff node equation at V+
yields
(2) Kirchhoff node equation at V yields
(3) Setting V+ = Vndash yields
0V
0_
c
c
b
b
a
a
f
o
R
VV
R
VV
R
VV
R
VV
c
aj j
jf
c
c
b
b
a
afo R
VR
R
V
R
V
R
VRV
+
Rf
Va
VoRb
Ra
RcVb
Vc
Homework
1) Design a circuit to 50i
of V
VA
2) Find the vo=
Review Two fundamental Op Amp
Structure Af Input voltage ( )terminal
Feed back
( )terminal
Inverting Amp
_ _
Non inverting Amp
+ _
Op-Amp Integrator
Op-Amp Summing Amplifier
Op-Amp Differential Amplifier
If R1 = R2 and Rf = Rg
Closed-loop gain
2
21
21
21
R
RRv
RRARAv
v
IN
INo
Closed-loop gain Af=vovin
Suppose A=106 R1=9R R2=R
10INo vv
Gain
INf vA
)1(2
1
R
RAf
Closed-loop gain determined by resistor ratio insensitive to A temperature
Negative feedback
1048774
)1(2
1
R
R
v
vA
i
of
1
2
R
R
v
vA
i
of
bull We can adjust the closed-loop gain by changing the ratio of R2 and R1
bull The closed-loop gain is (ideally) independent of op amp open-loop gain A (if A is large enough) and we can make it arbitrarily large or small and with the desired accuracy depending on the accuracy of the resistors
1048774
The terminal 1 is a virtual ground since terminal 2 is grounded
Inverting configuration
This is a classic example of what negative feedback does It takes an amplifier with very large gain and through negative feedback obtain a gain that is smaller stable and predictable In effect we have traded gain for accuracy This kind of trade off is common in electronic circuit designhellip as we will see more later
Negative feedback
1048774
Inverting configuration
Input ResistanceAssuming an ideal op amp (open-loop gain A = infinity) in the closed-loop inverting configuration the input resistance is R1
111 R
Rv
v
i
vR
i
iiin
Negative feedback
1
2
R
R
v
vA
i
of
Inverting configuration011
oi
in RRi
vR
We can model the closed-loop inverting amplifier (with A = infinite) with the following equivalent circuit using a voltage-controlled voltage sourcehellip
Negative feedback 1048774
29
Frequency-Gain Relationbull Ideally signals are amplified
from DC to the highest AC frequency
bull Practically bandwidth is limitedbull 741 family op-amp have an limit
bandwidth of few KHzbull Unity Gain frequency f1 the gain at
unitybull Cutoff frequency fc the gain drop by
3dB from dc gain Gd
(Voltage Gain)
(frequency)f1
Gd
0707Gd
fc0
1
GB Product f1 = Gd fc
20log(0707)=3dB
30
GB ProductExample Determine the cutoff frequency of an op-amp having a unit gain frequency f1 = 10 MHz and voltage differential gain Gd = 20VmV
Sol
Since f1 = 10 MHz
By using GB production equation
f1 = Gd fc
fc = f1 Gd = 10 MHz 20 VmV
= 10 106 20 103
= 500 Hz
(Voltage Gain)
(frequency)f1
Gd
0707Gd
fc0
1
10MHz
Hz
Op Amp transfer characteristic curve
saturation
)( vvAvAv io
active region
So far we have been looking at the amplification that can be achieved for relatively small (amplitude) signals For a fixed gain as we increase the input signal amplitude there is a limit to how large the output signal can be The output saturates as it approaches the positive andnegative power supply voltages In other words there is limited range across which the gain is linear vv Vvo
vv Vvo
Op Amp transfer characteristic curve
Applications of Op-AmpsFiltersTypesbullLow pass filterbullHigh pass filterbullBand pass filterbullCascading (2 or more filters connected together) R2
+
-
+
V0
__
+ Vcc
- Vcc
-+
R1
C
Low pass filter
Low pass filter Cutoff frequency
Low pass filter transfer function
Applications of Op-Ampsbull Electrocardiogram EKG Amplification
ndash Need to measure difference in voltage from lead 1 and lead 2ndash 60 Hz interference from electrical equipment
Applications of Op-Ampsbull Simple EKG circuit
ndash Uses differential amplifier to cancel common mode signal and amplify differential mode signal
bull Realistic EKG circuitndash Uses two non-inverting
amplifiers to first amplify voltage from each lead followed by differential amplifier
ndash Forms an ldquoinstrumentation amplifierrdquo
Strain Gauge
Use a Wheatstone bridge to determine the strain of an element by measuring the change in resistance of a
strain gauge
(No strain) Balanced Bridge R 1 = R 2
(Strain) Unbalanced Bridge R 1 ne R 2
Strain GaugeHalf-Bridge Arrangement
Using KCL at the inverting and non-inverting terminals of the op amp we find that ε ~ Vo = 2ΔR(Rf R2)
R + ΔR Rf
+
- +
V0
__
+ Vcc
- Vcc
-+
Rf
Vref
R
R - ΔR
R
Op amp used to amplify output from strain gauge
Applications of Op-Ampsbull Piezoelectric Transducerndash Used to measure force pressure accelerationndash Piezoelectric crystal generates an electric charge in
response to deformation
bull Use Charge Amplifierndash Just an integrator op-amp circuit
bullGoal is to have VSET = VOUT
bullRemember that VERROR = VSET ndash VSENSOR
bullOutput Process uses VERROR from the PID controller to adjust Vout such that it is ~VSET
P
I
D
Output Process
Sensor
VERRORVSET VOUT
VSENSOR
PID Controller ndash System Block Diagram
ApplicationsPID Controller ndash System Circuit Diagram
Source httpwwwecircuitcentercomCircuitsop_pidop_pidhtm
Calculates VERROR = -(VSET + VSENSOR)
Signal conditioning allows you to introduce a time delay which could
account for things like inertia
System to control
-VSENSOR
ApplicationsPID Controller ndash PID Controller Circuit Diagram
VERR
Adjust Change
Kp RP1 RP2
Ki RI CI
Kd RD CD
VERR PID
Applications of Op-Ampsbull Example of PI Control
Temperature Control
bull Thermal System we wish to automatically control the temperature of
bull Block Diagram of Control System
Applications of Op-Amps
bull Voltage Error Circuit
bull Proportional-Integral Control Circuit
bull Example of PI Control Temperature Control
44
Multiple Inputs(1) Kirchhoff node equation at V+
yields
(2) Kirchhoff node equation at V yields
(3) Setting V+ = Vndash yields
0V
0_
c
c
b
b
a
a
f
o
R
VV
R
VV
R
VV
R
VV
c
aj j
jf
c
c
b
b
a
afo R
VR
R
V
R
V
R
VRV
+
Rf
Va
VoRb
Ra
RcVb
Vc
Homework
1) Design a circuit to 50i
of V
VA
2) Find the vo=
Review Two fundamental Op Amp
Structure Af Input voltage ( )terminal
Feed back
( )terminal
Inverting Amp
_ _
Non inverting Amp
+ _
Op-Amp Summing Amplifier
Op-Amp Differential Amplifier
If R1 = R2 and Rf = Rg
Closed-loop gain
2
21
21
21
R
RRv
RRARAv
v
IN
INo
Closed-loop gain Af=vovin
Suppose A=106 R1=9R R2=R
10INo vv
Gain
INf vA
)1(2
1
R
RAf
Closed-loop gain determined by resistor ratio insensitive to A temperature
Negative feedback
1048774
)1(2
1
R
R
v
vA
i
of
1
2
R
R
v
vA
i
of
bull We can adjust the closed-loop gain by changing the ratio of R2 and R1
bull The closed-loop gain is (ideally) independent of op amp open-loop gain A (if A is large enough) and we can make it arbitrarily large or small and with the desired accuracy depending on the accuracy of the resistors
1048774
The terminal 1 is a virtual ground since terminal 2 is grounded
Inverting configuration
This is a classic example of what negative feedback does It takes an amplifier with very large gain and through negative feedback obtain a gain that is smaller stable and predictable In effect we have traded gain for accuracy This kind of trade off is common in electronic circuit designhellip as we will see more later
Negative feedback
1048774
Inverting configuration
Input ResistanceAssuming an ideal op amp (open-loop gain A = infinity) in the closed-loop inverting configuration the input resistance is R1
111 R
Rv
v
i
vR
i
iiin
Negative feedback
1
2
R
R
v
vA
i
of
Inverting configuration011
oi
in RRi
vR
We can model the closed-loop inverting amplifier (with A = infinite) with the following equivalent circuit using a voltage-controlled voltage sourcehellip
Negative feedback 1048774
29
Frequency-Gain Relationbull Ideally signals are amplified
from DC to the highest AC frequency
bull Practically bandwidth is limitedbull 741 family op-amp have an limit
bandwidth of few KHzbull Unity Gain frequency f1 the gain at
unitybull Cutoff frequency fc the gain drop by
3dB from dc gain Gd
(Voltage Gain)
(frequency)f1
Gd
0707Gd
fc0
1
GB Product f1 = Gd fc
20log(0707)=3dB
30
GB ProductExample Determine the cutoff frequency of an op-amp having a unit gain frequency f1 = 10 MHz and voltage differential gain Gd = 20VmV
Sol
Since f1 = 10 MHz
By using GB production equation
f1 = Gd fc
fc = f1 Gd = 10 MHz 20 VmV
= 10 106 20 103
= 500 Hz
(Voltage Gain)
(frequency)f1
Gd
0707Gd
fc0
1
10MHz
Hz
Op Amp transfer characteristic curve
saturation
)( vvAvAv io
active region
So far we have been looking at the amplification that can be achieved for relatively small (amplitude) signals For a fixed gain as we increase the input signal amplitude there is a limit to how large the output signal can be The output saturates as it approaches the positive andnegative power supply voltages In other words there is limited range across which the gain is linear vv Vvo
vv Vvo
Op Amp transfer characteristic curve
Applications of Op-AmpsFiltersTypesbullLow pass filterbullHigh pass filterbullBand pass filterbullCascading (2 or more filters connected together) R2
+
-
+
V0
__
+ Vcc
- Vcc
-+
R1
C
Low pass filter
Low pass filter Cutoff frequency
Low pass filter transfer function
Applications of Op-Ampsbull Electrocardiogram EKG Amplification
ndash Need to measure difference in voltage from lead 1 and lead 2ndash 60 Hz interference from electrical equipment
Applications of Op-Ampsbull Simple EKG circuit
ndash Uses differential amplifier to cancel common mode signal and amplify differential mode signal
bull Realistic EKG circuitndash Uses two non-inverting
amplifiers to first amplify voltage from each lead followed by differential amplifier
ndash Forms an ldquoinstrumentation amplifierrdquo
Strain Gauge
Use a Wheatstone bridge to determine the strain of an element by measuring the change in resistance of a
strain gauge
(No strain) Balanced Bridge R 1 = R 2
(Strain) Unbalanced Bridge R 1 ne R 2
Strain GaugeHalf-Bridge Arrangement
Using KCL at the inverting and non-inverting terminals of the op amp we find that ε ~ Vo = 2ΔR(Rf R2)
R + ΔR Rf
+
- +
V0
__
+ Vcc
- Vcc
-+
Rf
Vref
R
R - ΔR
R
Op amp used to amplify output from strain gauge
Applications of Op-Ampsbull Piezoelectric Transducerndash Used to measure force pressure accelerationndash Piezoelectric crystal generates an electric charge in
response to deformation
bull Use Charge Amplifierndash Just an integrator op-amp circuit
bullGoal is to have VSET = VOUT
bullRemember that VERROR = VSET ndash VSENSOR
bullOutput Process uses VERROR from the PID controller to adjust Vout such that it is ~VSET
P
I
D
Output Process
Sensor
VERRORVSET VOUT
VSENSOR
PID Controller ndash System Block Diagram
ApplicationsPID Controller ndash System Circuit Diagram
Source httpwwwecircuitcentercomCircuitsop_pidop_pidhtm
Calculates VERROR = -(VSET + VSENSOR)
Signal conditioning allows you to introduce a time delay which could
account for things like inertia
System to control
-VSENSOR
ApplicationsPID Controller ndash PID Controller Circuit Diagram
VERR
Adjust Change
Kp RP1 RP2
Ki RI CI
Kd RD CD
VERR PID
Applications of Op-Ampsbull Example of PI Control
Temperature Control
bull Thermal System we wish to automatically control the temperature of
bull Block Diagram of Control System
Applications of Op-Amps
bull Voltage Error Circuit
bull Proportional-Integral Control Circuit
bull Example of PI Control Temperature Control
44
Multiple Inputs(1) Kirchhoff node equation at V+
yields
(2) Kirchhoff node equation at V yields
(3) Setting V+ = Vndash yields
0V
0_
c
c
b
b
a
a
f
o
R
VV
R
VV
R
VV
R
VV
c
aj j
jf
c
c
b
b
a
afo R
VR
R
V
R
V
R
VRV
+
Rf
Va
VoRb
Ra
RcVb
Vc
Homework
1) Design a circuit to 50i
of V
VA
2) Find the vo=
Review Two fundamental Op Amp
Structure Af Input voltage ( )terminal
Feed back
( )terminal
Inverting Amp
_ _
Non inverting Amp
+ _
Op-Amp Differential Amplifier
If R1 = R2 and Rf = Rg
Closed-loop gain
2
21
21
21
R
RRv
RRARAv
v
IN
INo
Closed-loop gain Af=vovin
Suppose A=106 R1=9R R2=R
10INo vv
Gain
INf vA
)1(2
1
R
RAf
Closed-loop gain determined by resistor ratio insensitive to A temperature
Negative feedback
1048774
)1(2
1
R
R
v
vA
i
of
1
2
R
R
v
vA
i
of
bull We can adjust the closed-loop gain by changing the ratio of R2 and R1
bull The closed-loop gain is (ideally) independent of op amp open-loop gain A (if A is large enough) and we can make it arbitrarily large or small and with the desired accuracy depending on the accuracy of the resistors
1048774
The terminal 1 is a virtual ground since terminal 2 is grounded
Inverting configuration
This is a classic example of what negative feedback does It takes an amplifier with very large gain and through negative feedback obtain a gain that is smaller stable and predictable In effect we have traded gain for accuracy This kind of trade off is common in electronic circuit designhellip as we will see more later
Negative feedback
1048774
Inverting configuration
Input ResistanceAssuming an ideal op amp (open-loop gain A = infinity) in the closed-loop inverting configuration the input resistance is R1
111 R
Rv
v
i
vR
i
iiin
Negative feedback
1
2
R
R
v
vA
i
of
Inverting configuration011
oi
in RRi
vR
We can model the closed-loop inverting amplifier (with A = infinite) with the following equivalent circuit using a voltage-controlled voltage sourcehellip
Negative feedback 1048774
29
Frequency-Gain Relationbull Ideally signals are amplified
from DC to the highest AC frequency
bull Practically bandwidth is limitedbull 741 family op-amp have an limit
bandwidth of few KHzbull Unity Gain frequency f1 the gain at
unitybull Cutoff frequency fc the gain drop by
3dB from dc gain Gd
(Voltage Gain)
(frequency)f1
Gd
0707Gd
fc0
1
GB Product f1 = Gd fc
20log(0707)=3dB
30
GB ProductExample Determine the cutoff frequency of an op-amp having a unit gain frequency f1 = 10 MHz and voltage differential gain Gd = 20VmV
Sol
Since f1 = 10 MHz
By using GB production equation
f1 = Gd fc
fc = f1 Gd = 10 MHz 20 VmV
= 10 106 20 103
= 500 Hz
(Voltage Gain)
(frequency)f1
Gd
0707Gd
fc0
1
10MHz
Hz
Op Amp transfer characteristic curve
saturation
)( vvAvAv io
active region
So far we have been looking at the amplification that can be achieved for relatively small (amplitude) signals For a fixed gain as we increase the input signal amplitude there is a limit to how large the output signal can be The output saturates as it approaches the positive andnegative power supply voltages In other words there is limited range across which the gain is linear vv Vvo
vv Vvo
Op Amp transfer characteristic curve
Applications of Op-AmpsFiltersTypesbullLow pass filterbullHigh pass filterbullBand pass filterbullCascading (2 or more filters connected together) R2
+
-
+
V0
__
+ Vcc
- Vcc
-+
R1
C
Low pass filter
Low pass filter Cutoff frequency
Low pass filter transfer function
Applications of Op-Ampsbull Electrocardiogram EKG Amplification
ndash Need to measure difference in voltage from lead 1 and lead 2ndash 60 Hz interference from electrical equipment
Applications of Op-Ampsbull Simple EKG circuit
ndash Uses differential amplifier to cancel common mode signal and amplify differential mode signal
bull Realistic EKG circuitndash Uses two non-inverting
amplifiers to first amplify voltage from each lead followed by differential amplifier
ndash Forms an ldquoinstrumentation amplifierrdquo
Strain Gauge
Use a Wheatstone bridge to determine the strain of an element by measuring the change in resistance of a
strain gauge
(No strain) Balanced Bridge R 1 = R 2
(Strain) Unbalanced Bridge R 1 ne R 2
Strain GaugeHalf-Bridge Arrangement
Using KCL at the inverting and non-inverting terminals of the op amp we find that ε ~ Vo = 2ΔR(Rf R2)
R + ΔR Rf
+
- +
V0
__
+ Vcc
- Vcc
-+
Rf
Vref
R
R - ΔR
R
Op amp used to amplify output from strain gauge
Applications of Op-Ampsbull Piezoelectric Transducerndash Used to measure force pressure accelerationndash Piezoelectric crystal generates an electric charge in
response to deformation
bull Use Charge Amplifierndash Just an integrator op-amp circuit
bullGoal is to have VSET = VOUT
bullRemember that VERROR = VSET ndash VSENSOR
bullOutput Process uses VERROR from the PID controller to adjust Vout such that it is ~VSET
P
I
D
Output Process
Sensor
VERRORVSET VOUT
VSENSOR
PID Controller ndash System Block Diagram
ApplicationsPID Controller ndash System Circuit Diagram
Source httpwwwecircuitcentercomCircuitsop_pidop_pidhtm
Calculates VERROR = -(VSET + VSENSOR)
Signal conditioning allows you to introduce a time delay which could
account for things like inertia
System to control
-VSENSOR
ApplicationsPID Controller ndash PID Controller Circuit Diagram
VERR
Adjust Change
Kp RP1 RP2
Ki RI CI
Kd RD CD
VERR PID
Applications of Op-Ampsbull Example of PI Control
Temperature Control
bull Thermal System we wish to automatically control the temperature of
bull Block Diagram of Control System
Applications of Op-Amps
bull Voltage Error Circuit
bull Proportional-Integral Control Circuit
bull Example of PI Control Temperature Control
44
Multiple Inputs(1) Kirchhoff node equation at V+
yields
(2) Kirchhoff node equation at V yields
(3) Setting V+ = Vndash yields
0V
0_
c
c
b
b
a
a
f
o
R
VV
R
VV
R
VV
R
VV
c
aj j
jf
c
c
b
b
a
afo R
VR
R
V
R
V
R
VRV
+
Rf
Va
VoRb
Ra
RcVb
Vc
Homework
1) Design a circuit to 50i
of V
VA
2) Find the vo=
Review Two fundamental Op Amp
Structure Af Input voltage ( )terminal
Feed back
( )terminal
Inverting Amp
_ _
Non inverting Amp
+ _
Closed-loop gain
2
21
21
21
R
RRv
RRARAv
v
IN
INo
Closed-loop gain Af=vovin
Suppose A=106 R1=9R R2=R
10INo vv
Gain
INf vA
)1(2
1
R
RAf
Closed-loop gain determined by resistor ratio insensitive to A temperature
Negative feedback
1048774
)1(2
1
R
R
v
vA
i
of
1
2
R
R
v
vA
i
of
bull We can adjust the closed-loop gain by changing the ratio of R2 and R1
bull The closed-loop gain is (ideally) independent of op amp open-loop gain A (if A is large enough) and we can make it arbitrarily large or small and with the desired accuracy depending on the accuracy of the resistors
1048774
The terminal 1 is a virtual ground since terminal 2 is grounded
Inverting configuration
This is a classic example of what negative feedback does It takes an amplifier with very large gain and through negative feedback obtain a gain that is smaller stable and predictable In effect we have traded gain for accuracy This kind of trade off is common in electronic circuit designhellip as we will see more later
Negative feedback
1048774
Inverting configuration
Input ResistanceAssuming an ideal op amp (open-loop gain A = infinity) in the closed-loop inverting configuration the input resistance is R1
111 R
Rv
v
i
vR
i
iiin
Negative feedback
1
2
R
R
v
vA
i
of
Inverting configuration011
oi
in RRi
vR
We can model the closed-loop inverting amplifier (with A = infinite) with the following equivalent circuit using a voltage-controlled voltage sourcehellip
Negative feedback 1048774
29
Frequency-Gain Relationbull Ideally signals are amplified
from DC to the highest AC frequency
bull Practically bandwidth is limitedbull 741 family op-amp have an limit
bandwidth of few KHzbull Unity Gain frequency f1 the gain at
unitybull Cutoff frequency fc the gain drop by
3dB from dc gain Gd
(Voltage Gain)
(frequency)f1
Gd
0707Gd
fc0
1
GB Product f1 = Gd fc
20log(0707)=3dB
30
GB ProductExample Determine the cutoff frequency of an op-amp having a unit gain frequency f1 = 10 MHz and voltage differential gain Gd = 20VmV
Sol
Since f1 = 10 MHz
By using GB production equation
f1 = Gd fc
fc = f1 Gd = 10 MHz 20 VmV
= 10 106 20 103
= 500 Hz
(Voltage Gain)
(frequency)f1
Gd
0707Gd
fc0
1
10MHz
Hz
Op Amp transfer characteristic curve
saturation
)( vvAvAv io
active region
So far we have been looking at the amplification that can be achieved for relatively small (amplitude) signals For a fixed gain as we increase the input signal amplitude there is a limit to how large the output signal can be The output saturates as it approaches the positive andnegative power supply voltages In other words there is limited range across which the gain is linear vv Vvo
vv Vvo
Op Amp transfer characteristic curve
Applications of Op-AmpsFiltersTypesbullLow pass filterbullHigh pass filterbullBand pass filterbullCascading (2 or more filters connected together) R2
+
-
+
V0
__
+ Vcc
- Vcc
-+
R1
C
Low pass filter
Low pass filter Cutoff frequency
Low pass filter transfer function
Applications of Op-Ampsbull Electrocardiogram EKG Amplification
ndash Need to measure difference in voltage from lead 1 and lead 2ndash 60 Hz interference from electrical equipment
Applications of Op-Ampsbull Simple EKG circuit
ndash Uses differential amplifier to cancel common mode signal and amplify differential mode signal
bull Realistic EKG circuitndash Uses two non-inverting
amplifiers to first amplify voltage from each lead followed by differential amplifier
ndash Forms an ldquoinstrumentation amplifierrdquo
Strain Gauge
Use a Wheatstone bridge to determine the strain of an element by measuring the change in resistance of a
strain gauge
(No strain) Balanced Bridge R 1 = R 2
(Strain) Unbalanced Bridge R 1 ne R 2
Strain GaugeHalf-Bridge Arrangement
Using KCL at the inverting and non-inverting terminals of the op amp we find that ε ~ Vo = 2ΔR(Rf R2)
R + ΔR Rf
+
- +
V0
__
+ Vcc
- Vcc
-+
Rf
Vref
R
R - ΔR
R
Op amp used to amplify output from strain gauge
Applications of Op-Ampsbull Piezoelectric Transducerndash Used to measure force pressure accelerationndash Piezoelectric crystal generates an electric charge in
response to deformation
bull Use Charge Amplifierndash Just an integrator op-amp circuit
bullGoal is to have VSET = VOUT
bullRemember that VERROR = VSET ndash VSENSOR
bullOutput Process uses VERROR from the PID controller to adjust Vout such that it is ~VSET
P
I
D
Output Process
Sensor
VERRORVSET VOUT
VSENSOR
PID Controller ndash System Block Diagram
ApplicationsPID Controller ndash System Circuit Diagram
Source httpwwwecircuitcentercomCircuitsop_pidop_pidhtm
Calculates VERROR = -(VSET + VSENSOR)
Signal conditioning allows you to introduce a time delay which could
account for things like inertia
System to control
-VSENSOR
ApplicationsPID Controller ndash PID Controller Circuit Diagram
VERR
Adjust Change
Kp RP1 RP2
Ki RI CI
Kd RD CD
VERR PID
Applications of Op-Ampsbull Example of PI Control
Temperature Control
bull Thermal System we wish to automatically control the temperature of
bull Block Diagram of Control System
Applications of Op-Amps
bull Voltage Error Circuit
bull Proportional-Integral Control Circuit
bull Example of PI Control Temperature Control
44
Multiple Inputs(1) Kirchhoff node equation at V+
yields
(2) Kirchhoff node equation at V yields
(3) Setting V+ = Vndash yields
0V
0_
c
c
b
b
a
a
f
o
R
VV
R
VV
R
VV
R
VV
c
aj j
jf
c
c
b
b
a
afo R
VR
R
V
R
V
R
VRV
+
Rf
Va
VoRb
Ra
RcVb
Vc
Homework
1) Design a circuit to 50i
of V
VA
2) Find the vo=
Review Two fundamental Op Amp
Structure Af Input voltage ( )terminal
Feed back
( )terminal
Inverting Amp
_ _
Non inverting Amp
+ _
Negative feedback
1048774
)1(2
1
R
R
v
vA
i
of
1
2
R
R
v
vA
i
of
bull We can adjust the closed-loop gain by changing the ratio of R2 and R1
bull The closed-loop gain is (ideally) independent of op amp open-loop gain A (if A is large enough) and we can make it arbitrarily large or small and with the desired accuracy depending on the accuracy of the resistors
1048774
The terminal 1 is a virtual ground since terminal 2 is grounded
Inverting configuration
This is a classic example of what negative feedback does It takes an amplifier with very large gain and through negative feedback obtain a gain that is smaller stable and predictable In effect we have traded gain for accuracy This kind of trade off is common in electronic circuit designhellip as we will see more later
Negative feedback
1048774
Inverting configuration
Input ResistanceAssuming an ideal op amp (open-loop gain A = infinity) in the closed-loop inverting configuration the input resistance is R1
111 R
Rv
v
i
vR
i
iiin
Negative feedback
1
2
R
R
v
vA
i
of
Inverting configuration011
oi
in RRi
vR
We can model the closed-loop inverting amplifier (with A = infinite) with the following equivalent circuit using a voltage-controlled voltage sourcehellip
Negative feedback 1048774
29
Frequency-Gain Relationbull Ideally signals are amplified
from DC to the highest AC frequency
bull Practically bandwidth is limitedbull 741 family op-amp have an limit
bandwidth of few KHzbull Unity Gain frequency f1 the gain at
unitybull Cutoff frequency fc the gain drop by
3dB from dc gain Gd
(Voltage Gain)
(frequency)f1
Gd
0707Gd
fc0
1
GB Product f1 = Gd fc
20log(0707)=3dB
30
GB ProductExample Determine the cutoff frequency of an op-amp having a unit gain frequency f1 = 10 MHz and voltage differential gain Gd = 20VmV
Sol
Since f1 = 10 MHz
By using GB production equation
f1 = Gd fc
fc = f1 Gd = 10 MHz 20 VmV
= 10 106 20 103
= 500 Hz
(Voltage Gain)
(frequency)f1
Gd
0707Gd
fc0
1
10MHz
Hz
Op Amp transfer characteristic curve
saturation
)( vvAvAv io
active region
So far we have been looking at the amplification that can be achieved for relatively small (amplitude) signals For a fixed gain as we increase the input signal amplitude there is a limit to how large the output signal can be The output saturates as it approaches the positive andnegative power supply voltages In other words there is limited range across which the gain is linear vv Vvo
vv Vvo
Op Amp transfer characteristic curve
Applications of Op-AmpsFiltersTypesbullLow pass filterbullHigh pass filterbullBand pass filterbullCascading (2 or more filters connected together) R2
+
-
+
V0
__
+ Vcc
- Vcc
-+
R1
C
Low pass filter
Low pass filter Cutoff frequency
Low pass filter transfer function
Applications of Op-Ampsbull Electrocardiogram EKG Amplification
ndash Need to measure difference in voltage from lead 1 and lead 2ndash 60 Hz interference from electrical equipment
Applications of Op-Ampsbull Simple EKG circuit
ndash Uses differential amplifier to cancel common mode signal and amplify differential mode signal
bull Realistic EKG circuitndash Uses two non-inverting
amplifiers to first amplify voltage from each lead followed by differential amplifier
ndash Forms an ldquoinstrumentation amplifierrdquo
Strain Gauge
Use a Wheatstone bridge to determine the strain of an element by measuring the change in resistance of a
strain gauge
(No strain) Balanced Bridge R 1 = R 2
(Strain) Unbalanced Bridge R 1 ne R 2
Strain GaugeHalf-Bridge Arrangement
Using KCL at the inverting and non-inverting terminals of the op amp we find that ε ~ Vo = 2ΔR(Rf R2)
R + ΔR Rf
+
- +
V0
__
+ Vcc
- Vcc
-+
Rf
Vref
R
R - ΔR
R
Op amp used to amplify output from strain gauge
Applications of Op-Ampsbull Piezoelectric Transducerndash Used to measure force pressure accelerationndash Piezoelectric crystal generates an electric charge in
response to deformation
bull Use Charge Amplifierndash Just an integrator op-amp circuit
bullGoal is to have VSET = VOUT
bullRemember that VERROR = VSET ndash VSENSOR
bullOutput Process uses VERROR from the PID controller to adjust Vout such that it is ~VSET
P
I
D
Output Process
Sensor
VERRORVSET VOUT
VSENSOR
PID Controller ndash System Block Diagram
ApplicationsPID Controller ndash System Circuit Diagram
Source httpwwwecircuitcentercomCircuitsop_pidop_pidhtm
Calculates VERROR = -(VSET + VSENSOR)
Signal conditioning allows you to introduce a time delay which could
account for things like inertia
System to control
-VSENSOR
ApplicationsPID Controller ndash PID Controller Circuit Diagram
VERR
Adjust Change
Kp RP1 RP2
Ki RI CI
Kd RD CD
VERR PID
Applications of Op-Ampsbull Example of PI Control
Temperature Control
bull Thermal System we wish to automatically control the temperature of
bull Block Diagram of Control System
Applications of Op-Amps
bull Voltage Error Circuit
bull Proportional-Integral Control Circuit
bull Example of PI Control Temperature Control
44
Multiple Inputs(1) Kirchhoff node equation at V+
yields
(2) Kirchhoff node equation at V yields
(3) Setting V+ = Vndash yields
0V
0_
c
c
b
b
a
a
f
o
R
VV
R
VV
R
VV
R
VV
c
aj j
jf
c
c
b
b
a
afo R
VR
R
V
R
V
R
VRV
+
Rf
Va
VoRb
Ra
RcVb
Vc
Homework
1) Design a circuit to 50i
of V
VA
2) Find the vo=
Review Two fundamental Op Amp
Structure Af Input voltage ( )terminal
Feed back
( )terminal
Inverting Amp
_ _
Non inverting Amp
+ _
1048774
The terminal 1 is a virtual ground since terminal 2 is grounded
Inverting configuration
This is a classic example of what negative feedback does It takes an amplifier with very large gain and through negative feedback obtain a gain that is smaller stable and predictable In effect we have traded gain for accuracy This kind of trade off is common in electronic circuit designhellip as we will see more later
Negative feedback
1048774
Inverting configuration
Input ResistanceAssuming an ideal op amp (open-loop gain A = infinity) in the closed-loop inverting configuration the input resistance is R1
111 R
Rv
v
i
vR
i
iiin
Negative feedback
1
2
R
R
v
vA
i
of
Inverting configuration011
oi
in RRi
vR
We can model the closed-loop inverting amplifier (with A = infinite) with the following equivalent circuit using a voltage-controlled voltage sourcehellip
Negative feedback 1048774
29
Frequency-Gain Relationbull Ideally signals are amplified
from DC to the highest AC frequency
bull Practically bandwidth is limitedbull 741 family op-amp have an limit
bandwidth of few KHzbull Unity Gain frequency f1 the gain at
unitybull Cutoff frequency fc the gain drop by
3dB from dc gain Gd
(Voltage Gain)
(frequency)f1
Gd
0707Gd
fc0
1
GB Product f1 = Gd fc
20log(0707)=3dB
30
GB ProductExample Determine the cutoff frequency of an op-amp having a unit gain frequency f1 = 10 MHz and voltage differential gain Gd = 20VmV
Sol
Since f1 = 10 MHz
By using GB production equation
f1 = Gd fc
fc = f1 Gd = 10 MHz 20 VmV
= 10 106 20 103
= 500 Hz
(Voltage Gain)
(frequency)f1
Gd
0707Gd
fc0
1
10MHz
Hz
Op Amp transfer characteristic curve
saturation
)( vvAvAv io
active region
So far we have been looking at the amplification that can be achieved for relatively small (amplitude) signals For a fixed gain as we increase the input signal amplitude there is a limit to how large the output signal can be The output saturates as it approaches the positive andnegative power supply voltages In other words there is limited range across which the gain is linear vv Vvo
vv Vvo
Op Amp transfer characteristic curve
Applications of Op-AmpsFiltersTypesbullLow pass filterbullHigh pass filterbullBand pass filterbullCascading (2 or more filters connected together) R2
+
-
+
V0
__
+ Vcc
- Vcc
-+
R1
C
Low pass filter
Low pass filter Cutoff frequency
Low pass filter transfer function
Applications of Op-Ampsbull Electrocardiogram EKG Amplification
ndash Need to measure difference in voltage from lead 1 and lead 2ndash 60 Hz interference from electrical equipment
Applications of Op-Ampsbull Simple EKG circuit
ndash Uses differential amplifier to cancel common mode signal and amplify differential mode signal
bull Realistic EKG circuitndash Uses two non-inverting
amplifiers to first amplify voltage from each lead followed by differential amplifier
ndash Forms an ldquoinstrumentation amplifierrdquo
Strain Gauge
Use a Wheatstone bridge to determine the strain of an element by measuring the change in resistance of a
strain gauge
(No strain) Balanced Bridge R 1 = R 2
(Strain) Unbalanced Bridge R 1 ne R 2
Strain GaugeHalf-Bridge Arrangement
Using KCL at the inverting and non-inverting terminals of the op amp we find that ε ~ Vo = 2ΔR(Rf R2)
R + ΔR Rf
+
- +
V0
__
+ Vcc
- Vcc
-+
Rf
Vref
R
R - ΔR
R
Op amp used to amplify output from strain gauge
Applications of Op-Ampsbull Piezoelectric Transducerndash Used to measure force pressure accelerationndash Piezoelectric crystal generates an electric charge in
response to deformation
bull Use Charge Amplifierndash Just an integrator op-amp circuit
bullGoal is to have VSET = VOUT
bullRemember that VERROR = VSET ndash VSENSOR
bullOutput Process uses VERROR from the PID controller to adjust Vout such that it is ~VSET
P
I
D
Output Process
Sensor
VERRORVSET VOUT
VSENSOR
PID Controller ndash System Block Diagram
ApplicationsPID Controller ndash System Circuit Diagram
Source httpwwwecircuitcentercomCircuitsop_pidop_pidhtm
Calculates VERROR = -(VSET + VSENSOR)
Signal conditioning allows you to introduce a time delay which could
account for things like inertia
System to control
-VSENSOR
ApplicationsPID Controller ndash PID Controller Circuit Diagram
VERR
Adjust Change
Kp RP1 RP2
Ki RI CI
Kd RD CD
VERR PID
Applications of Op-Ampsbull Example of PI Control
Temperature Control
bull Thermal System we wish to automatically control the temperature of
bull Block Diagram of Control System
Applications of Op-Amps
bull Voltage Error Circuit
bull Proportional-Integral Control Circuit
bull Example of PI Control Temperature Control
44
Multiple Inputs(1) Kirchhoff node equation at V+
yields
(2) Kirchhoff node equation at V yields
(3) Setting V+ = Vndash yields
0V
0_
c
c
b
b
a
a
f
o
R
VV
R
VV
R
VV
R
VV
c
aj j
jf
c
c
b
b
a
afo R
VR
R
V
R
V
R
VRV
+
Rf
Va
VoRb
Ra
RcVb
Vc
Homework
1) Design a circuit to 50i
of V
VA
2) Find the vo=
Review Two fundamental Op Amp
Structure Af Input voltage ( )terminal
Feed back
( )terminal
Inverting Amp
_ _
Non inverting Amp
+ _
1048774
Inverting configuration
Input ResistanceAssuming an ideal op amp (open-loop gain A = infinity) in the closed-loop inverting configuration the input resistance is R1
111 R
Rv
v
i
vR
i
iiin
Negative feedback
1
2
R
R
v
vA
i
of
Inverting configuration011
oi
in RRi
vR
We can model the closed-loop inverting amplifier (with A = infinite) with the following equivalent circuit using a voltage-controlled voltage sourcehellip
Negative feedback 1048774
29
Frequency-Gain Relationbull Ideally signals are amplified
from DC to the highest AC frequency
bull Practically bandwidth is limitedbull 741 family op-amp have an limit
bandwidth of few KHzbull Unity Gain frequency f1 the gain at
unitybull Cutoff frequency fc the gain drop by
3dB from dc gain Gd
(Voltage Gain)
(frequency)f1
Gd
0707Gd
fc0
1
GB Product f1 = Gd fc
20log(0707)=3dB
30
GB ProductExample Determine the cutoff frequency of an op-amp having a unit gain frequency f1 = 10 MHz and voltage differential gain Gd = 20VmV
Sol
Since f1 = 10 MHz
By using GB production equation
f1 = Gd fc
fc = f1 Gd = 10 MHz 20 VmV
= 10 106 20 103
= 500 Hz
(Voltage Gain)
(frequency)f1
Gd
0707Gd
fc0
1
10MHz
Hz
Op Amp transfer characteristic curve
saturation
)( vvAvAv io
active region
So far we have been looking at the amplification that can be achieved for relatively small (amplitude) signals For a fixed gain as we increase the input signal amplitude there is a limit to how large the output signal can be The output saturates as it approaches the positive andnegative power supply voltages In other words there is limited range across which the gain is linear vv Vvo
vv Vvo
Op Amp transfer characteristic curve
Applications of Op-AmpsFiltersTypesbullLow pass filterbullHigh pass filterbullBand pass filterbullCascading (2 or more filters connected together) R2
+
-
+
V0
__
+ Vcc
- Vcc
-+
R1
C
Low pass filter
Low pass filter Cutoff frequency
Low pass filter transfer function
Applications of Op-Ampsbull Electrocardiogram EKG Amplification
ndash Need to measure difference in voltage from lead 1 and lead 2ndash 60 Hz interference from electrical equipment
Applications of Op-Ampsbull Simple EKG circuit
ndash Uses differential amplifier to cancel common mode signal and amplify differential mode signal
bull Realistic EKG circuitndash Uses two non-inverting
amplifiers to first amplify voltage from each lead followed by differential amplifier
ndash Forms an ldquoinstrumentation amplifierrdquo
Strain Gauge
Use a Wheatstone bridge to determine the strain of an element by measuring the change in resistance of a
strain gauge
(No strain) Balanced Bridge R 1 = R 2
(Strain) Unbalanced Bridge R 1 ne R 2
Strain GaugeHalf-Bridge Arrangement
Using KCL at the inverting and non-inverting terminals of the op amp we find that ε ~ Vo = 2ΔR(Rf R2)
R + ΔR Rf
+
- +
V0
__
+ Vcc
- Vcc
-+
Rf
Vref
R
R - ΔR
R
Op amp used to amplify output from strain gauge
Applications of Op-Ampsbull Piezoelectric Transducerndash Used to measure force pressure accelerationndash Piezoelectric crystal generates an electric charge in
response to deformation
bull Use Charge Amplifierndash Just an integrator op-amp circuit
bullGoal is to have VSET = VOUT
bullRemember that VERROR = VSET ndash VSENSOR
bullOutput Process uses VERROR from the PID controller to adjust Vout such that it is ~VSET
P
I
D
Output Process
Sensor
VERRORVSET VOUT
VSENSOR
PID Controller ndash System Block Diagram
ApplicationsPID Controller ndash System Circuit Diagram
Source httpwwwecircuitcentercomCircuitsop_pidop_pidhtm
Calculates VERROR = -(VSET + VSENSOR)
Signal conditioning allows you to introduce a time delay which could
account for things like inertia
System to control
-VSENSOR
ApplicationsPID Controller ndash PID Controller Circuit Diagram
VERR
Adjust Change
Kp RP1 RP2
Ki RI CI
Kd RD CD
VERR PID
Applications of Op-Ampsbull Example of PI Control
Temperature Control
bull Thermal System we wish to automatically control the temperature of
bull Block Diagram of Control System
Applications of Op-Amps
bull Voltage Error Circuit
bull Proportional-Integral Control Circuit
bull Example of PI Control Temperature Control
44
Multiple Inputs(1) Kirchhoff node equation at V+
yields
(2) Kirchhoff node equation at V yields
(3) Setting V+ = Vndash yields
0V
0_
c
c
b
b
a
a
f
o
R
VV
R
VV
R
VV
R
VV
c
aj j
jf
c
c
b
b
a
afo R
VR
R
V
R
V
R
VRV
+
Rf
Va
VoRb
Ra
RcVb
Vc
Homework
1) Design a circuit to 50i
of V
VA
2) Find the vo=
Review Two fundamental Op Amp
Structure Af Input voltage ( )terminal
Feed back
( )terminal
Inverting Amp
_ _
Non inverting Amp
+ _
1
2
R
R
v
vA
i
of
Inverting configuration011
oi
in RRi
vR
We can model the closed-loop inverting amplifier (with A = infinite) with the following equivalent circuit using a voltage-controlled voltage sourcehellip
Negative feedback 1048774
29
Frequency-Gain Relationbull Ideally signals are amplified
from DC to the highest AC frequency
bull Practically bandwidth is limitedbull 741 family op-amp have an limit
bandwidth of few KHzbull Unity Gain frequency f1 the gain at
unitybull Cutoff frequency fc the gain drop by
3dB from dc gain Gd
(Voltage Gain)
(frequency)f1
Gd
0707Gd
fc0
1
GB Product f1 = Gd fc
20log(0707)=3dB
30
GB ProductExample Determine the cutoff frequency of an op-amp having a unit gain frequency f1 = 10 MHz and voltage differential gain Gd = 20VmV
Sol
Since f1 = 10 MHz
By using GB production equation
f1 = Gd fc
fc = f1 Gd = 10 MHz 20 VmV
= 10 106 20 103
= 500 Hz
(Voltage Gain)
(frequency)f1
Gd
0707Gd
fc0
1
10MHz
Hz
Op Amp transfer characteristic curve
saturation
)( vvAvAv io
active region
So far we have been looking at the amplification that can be achieved for relatively small (amplitude) signals For a fixed gain as we increase the input signal amplitude there is a limit to how large the output signal can be The output saturates as it approaches the positive andnegative power supply voltages In other words there is limited range across which the gain is linear vv Vvo
vv Vvo
Op Amp transfer characteristic curve
Applications of Op-AmpsFiltersTypesbullLow pass filterbullHigh pass filterbullBand pass filterbullCascading (2 or more filters connected together) R2
+
-
+
V0
__
+ Vcc
- Vcc
-+
R1
C
Low pass filter
Low pass filter Cutoff frequency
Low pass filter transfer function
Applications of Op-Ampsbull Electrocardiogram EKG Amplification
ndash Need to measure difference in voltage from lead 1 and lead 2ndash 60 Hz interference from electrical equipment
Applications of Op-Ampsbull Simple EKG circuit
ndash Uses differential amplifier to cancel common mode signal and amplify differential mode signal
bull Realistic EKG circuitndash Uses two non-inverting
amplifiers to first amplify voltage from each lead followed by differential amplifier
ndash Forms an ldquoinstrumentation amplifierrdquo
Strain Gauge
Use a Wheatstone bridge to determine the strain of an element by measuring the change in resistance of a
strain gauge
(No strain) Balanced Bridge R 1 = R 2
(Strain) Unbalanced Bridge R 1 ne R 2
Strain GaugeHalf-Bridge Arrangement
Using KCL at the inverting and non-inverting terminals of the op amp we find that ε ~ Vo = 2ΔR(Rf R2)
R + ΔR Rf
+
- +
V0
__
+ Vcc
- Vcc
-+
Rf
Vref
R
R - ΔR
R
Op amp used to amplify output from strain gauge
Applications of Op-Ampsbull Piezoelectric Transducerndash Used to measure force pressure accelerationndash Piezoelectric crystal generates an electric charge in
response to deformation
bull Use Charge Amplifierndash Just an integrator op-amp circuit
bullGoal is to have VSET = VOUT
bullRemember that VERROR = VSET ndash VSENSOR
bullOutput Process uses VERROR from the PID controller to adjust Vout such that it is ~VSET
P
I
D
Output Process
Sensor
VERRORVSET VOUT
VSENSOR
PID Controller ndash System Block Diagram
ApplicationsPID Controller ndash System Circuit Diagram
Source httpwwwecircuitcentercomCircuitsop_pidop_pidhtm
Calculates VERROR = -(VSET + VSENSOR)
Signal conditioning allows you to introduce a time delay which could
account for things like inertia
System to control
-VSENSOR
ApplicationsPID Controller ndash PID Controller Circuit Diagram
VERR
Adjust Change
Kp RP1 RP2
Ki RI CI
Kd RD CD
VERR PID
Applications of Op-Ampsbull Example of PI Control
Temperature Control
bull Thermal System we wish to automatically control the temperature of
bull Block Diagram of Control System
Applications of Op-Amps
bull Voltage Error Circuit
bull Proportional-Integral Control Circuit
bull Example of PI Control Temperature Control
44
Multiple Inputs(1) Kirchhoff node equation at V+
yields
(2) Kirchhoff node equation at V yields
(3) Setting V+ = Vndash yields
0V
0_
c
c
b
b
a
a
f
o
R
VV
R
VV
R
VV
R
VV
c
aj j
jf
c
c
b
b
a
afo R
VR
R
V
R
V
R
VRV
+
Rf
Va
VoRb
Ra
RcVb
Vc
Homework
1) Design a circuit to 50i
of V
VA
2) Find the vo=
Review Two fundamental Op Amp
Structure Af Input voltage ( )terminal
Feed back
( )terminal
Inverting Amp
_ _
Non inverting Amp
+ _
29
Frequency-Gain Relationbull Ideally signals are amplified
from DC to the highest AC frequency
bull Practically bandwidth is limitedbull 741 family op-amp have an limit
bandwidth of few KHzbull Unity Gain frequency f1 the gain at
unitybull Cutoff frequency fc the gain drop by
3dB from dc gain Gd
(Voltage Gain)
(frequency)f1
Gd
0707Gd
fc0
1
GB Product f1 = Gd fc
20log(0707)=3dB
30
GB ProductExample Determine the cutoff frequency of an op-amp having a unit gain frequency f1 = 10 MHz and voltage differential gain Gd = 20VmV
Sol
Since f1 = 10 MHz
By using GB production equation
f1 = Gd fc
fc = f1 Gd = 10 MHz 20 VmV
= 10 106 20 103
= 500 Hz
(Voltage Gain)
(frequency)f1
Gd
0707Gd
fc0
1
10MHz
Hz
Op Amp transfer characteristic curve
saturation
)( vvAvAv io
active region
So far we have been looking at the amplification that can be achieved for relatively small (amplitude) signals For a fixed gain as we increase the input signal amplitude there is a limit to how large the output signal can be The output saturates as it approaches the positive andnegative power supply voltages In other words there is limited range across which the gain is linear vv Vvo
vv Vvo
Op Amp transfer characteristic curve
Applications of Op-AmpsFiltersTypesbullLow pass filterbullHigh pass filterbullBand pass filterbullCascading (2 or more filters connected together) R2
+
-
+
V0
__
+ Vcc
- Vcc
-+
R1
C
Low pass filter
Low pass filter Cutoff frequency
Low pass filter transfer function
Applications of Op-Ampsbull Electrocardiogram EKG Amplification
ndash Need to measure difference in voltage from lead 1 and lead 2ndash 60 Hz interference from electrical equipment
Applications of Op-Ampsbull Simple EKG circuit
ndash Uses differential amplifier to cancel common mode signal and amplify differential mode signal
bull Realistic EKG circuitndash Uses two non-inverting
amplifiers to first amplify voltage from each lead followed by differential amplifier
ndash Forms an ldquoinstrumentation amplifierrdquo
Strain Gauge
Use a Wheatstone bridge to determine the strain of an element by measuring the change in resistance of a
strain gauge
(No strain) Balanced Bridge R 1 = R 2
(Strain) Unbalanced Bridge R 1 ne R 2
Strain GaugeHalf-Bridge Arrangement
Using KCL at the inverting and non-inverting terminals of the op amp we find that ε ~ Vo = 2ΔR(Rf R2)
R + ΔR Rf
+
- +
V0
__
+ Vcc
- Vcc
-+
Rf
Vref
R
R - ΔR
R
Op amp used to amplify output from strain gauge
Applications of Op-Ampsbull Piezoelectric Transducerndash Used to measure force pressure accelerationndash Piezoelectric crystal generates an electric charge in
response to deformation
bull Use Charge Amplifierndash Just an integrator op-amp circuit
bullGoal is to have VSET = VOUT
bullRemember that VERROR = VSET ndash VSENSOR
bullOutput Process uses VERROR from the PID controller to adjust Vout such that it is ~VSET
P
I
D
Output Process
Sensor
VERRORVSET VOUT
VSENSOR
PID Controller ndash System Block Diagram
ApplicationsPID Controller ndash System Circuit Diagram
Source httpwwwecircuitcentercomCircuitsop_pidop_pidhtm
Calculates VERROR = -(VSET + VSENSOR)
Signal conditioning allows you to introduce a time delay which could
account for things like inertia
System to control
-VSENSOR
ApplicationsPID Controller ndash PID Controller Circuit Diagram
VERR
Adjust Change
Kp RP1 RP2
Ki RI CI
Kd RD CD
VERR PID
Applications of Op-Ampsbull Example of PI Control
Temperature Control
bull Thermal System we wish to automatically control the temperature of
bull Block Diagram of Control System
Applications of Op-Amps
bull Voltage Error Circuit
bull Proportional-Integral Control Circuit
bull Example of PI Control Temperature Control
44
Multiple Inputs(1) Kirchhoff node equation at V+
yields
(2) Kirchhoff node equation at V yields
(3) Setting V+ = Vndash yields
0V
0_
c
c
b
b
a
a
f
o
R
VV
R
VV
R
VV
R
VV
c
aj j
jf
c
c
b
b
a
afo R
VR
R
V
R
V
R
VRV
+
Rf
Va
VoRb
Ra
RcVb
Vc
Homework
1) Design a circuit to 50i
of V
VA
2) Find the vo=
Review Two fundamental Op Amp
Structure Af Input voltage ( )terminal
Feed back
( )terminal
Inverting Amp
_ _
Non inverting Amp
+ _
30
GB ProductExample Determine the cutoff frequency of an op-amp having a unit gain frequency f1 = 10 MHz and voltage differential gain Gd = 20VmV
Sol
Since f1 = 10 MHz
By using GB production equation
f1 = Gd fc
fc = f1 Gd = 10 MHz 20 VmV
= 10 106 20 103
= 500 Hz
(Voltage Gain)
(frequency)f1
Gd
0707Gd
fc0
1
10MHz
Hz
Op Amp transfer characteristic curve
saturation
)( vvAvAv io
active region
So far we have been looking at the amplification that can be achieved for relatively small (amplitude) signals For a fixed gain as we increase the input signal amplitude there is a limit to how large the output signal can be The output saturates as it approaches the positive andnegative power supply voltages In other words there is limited range across which the gain is linear vv Vvo
vv Vvo
Op Amp transfer characteristic curve
Applications of Op-AmpsFiltersTypesbullLow pass filterbullHigh pass filterbullBand pass filterbullCascading (2 or more filters connected together) R2
+
-
+
V0
__
+ Vcc
- Vcc
-+
R1
C
Low pass filter
Low pass filter Cutoff frequency
Low pass filter transfer function
Applications of Op-Ampsbull Electrocardiogram EKG Amplification
ndash Need to measure difference in voltage from lead 1 and lead 2ndash 60 Hz interference from electrical equipment
Applications of Op-Ampsbull Simple EKG circuit
ndash Uses differential amplifier to cancel common mode signal and amplify differential mode signal
bull Realistic EKG circuitndash Uses two non-inverting
amplifiers to first amplify voltage from each lead followed by differential amplifier
ndash Forms an ldquoinstrumentation amplifierrdquo
Strain Gauge
Use a Wheatstone bridge to determine the strain of an element by measuring the change in resistance of a
strain gauge
(No strain) Balanced Bridge R 1 = R 2
(Strain) Unbalanced Bridge R 1 ne R 2
Strain GaugeHalf-Bridge Arrangement
Using KCL at the inverting and non-inverting terminals of the op amp we find that ε ~ Vo = 2ΔR(Rf R2)
R + ΔR Rf
+
- +
V0
__
+ Vcc
- Vcc
-+
Rf
Vref
R
R - ΔR
R
Op amp used to amplify output from strain gauge
Applications of Op-Ampsbull Piezoelectric Transducerndash Used to measure force pressure accelerationndash Piezoelectric crystal generates an electric charge in
response to deformation
bull Use Charge Amplifierndash Just an integrator op-amp circuit
bullGoal is to have VSET = VOUT
bullRemember that VERROR = VSET ndash VSENSOR
bullOutput Process uses VERROR from the PID controller to adjust Vout such that it is ~VSET
P
I
D
Output Process
Sensor
VERRORVSET VOUT
VSENSOR
PID Controller ndash System Block Diagram
ApplicationsPID Controller ndash System Circuit Diagram
Source httpwwwecircuitcentercomCircuitsop_pidop_pidhtm
Calculates VERROR = -(VSET + VSENSOR)
Signal conditioning allows you to introduce a time delay which could
account for things like inertia
System to control
-VSENSOR
ApplicationsPID Controller ndash PID Controller Circuit Diagram
VERR
Adjust Change
Kp RP1 RP2
Ki RI CI
Kd RD CD
VERR PID
Applications of Op-Ampsbull Example of PI Control
Temperature Control
bull Thermal System we wish to automatically control the temperature of
bull Block Diagram of Control System
Applications of Op-Amps
bull Voltage Error Circuit
bull Proportional-Integral Control Circuit
bull Example of PI Control Temperature Control
44
Multiple Inputs(1) Kirchhoff node equation at V+
yields
(2) Kirchhoff node equation at V yields
(3) Setting V+ = Vndash yields
0V
0_
c
c
b
b
a
a
f
o
R
VV
R
VV
R
VV
R
VV
c
aj j
jf
c
c
b
b
a
afo R
VR
R
V
R
V
R
VRV
+
Rf
Va
VoRb
Ra
RcVb
Vc
Homework
1) Design a circuit to 50i
of V
VA
2) Find the vo=
Review Two fundamental Op Amp
Structure Af Input voltage ( )terminal
Feed back
( )terminal
Inverting Amp
_ _
Non inverting Amp
+ _
Op Amp transfer characteristic curve
saturation
)( vvAvAv io
active region
So far we have been looking at the amplification that can be achieved for relatively small (amplitude) signals For a fixed gain as we increase the input signal amplitude there is a limit to how large the output signal can be The output saturates as it approaches the positive andnegative power supply voltages In other words there is limited range across which the gain is linear vv Vvo
vv Vvo
Op Amp transfer characteristic curve
Applications of Op-AmpsFiltersTypesbullLow pass filterbullHigh pass filterbullBand pass filterbullCascading (2 or more filters connected together) R2
+
-
+
V0
__
+ Vcc
- Vcc
-+
R1
C
Low pass filter
Low pass filter Cutoff frequency
Low pass filter transfer function
Applications of Op-Ampsbull Electrocardiogram EKG Amplification
ndash Need to measure difference in voltage from lead 1 and lead 2ndash 60 Hz interference from electrical equipment
Applications of Op-Ampsbull Simple EKG circuit
ndash Uses differential amplifier to cancel common mode signal and amplify differential mode signal
bull Realistic EKG circuitndash Uses two non-inverting
amplifiers to first amplify voltage from each lead followed by differential amplifier
ndash Forms an ldquoinstrumentation amplifierrdquo
Strain Gauge
Use a Wheatstone bridge to determine the strain of an element by measuring the change in resistance of a
strain gauge
(No strain) Balanced Bridge R 1 = R 2
(Strain) Unbalanced Bridge R 1 ne R 2
Strain GaugeHalf-Bridge Arrangement
Using KCL at the inverting and non-inverting terminals of the op amp we find that ε ~ Vo = 2ΔR(Rf R2)
R + ΔR Rf
+
- +
V0
__
+ Vcc
- Vcc
-+
Rf
Vref
R
R - ΔR
R
Op amp used to amplify output from strain gauge
Applications of Op-Ampsbull Piezoelectric Transducerndash Used to measure force pressure accelerationndash Piezoelectric crystal generates an electric charge in
response to deformation
bull Use Charge Amplifierndash Just an integrator op-amp circuit
bullGoal is to have VSET = VOUT
bullRemember that VERROR = VSET ndash VSENSOR
bullOutput Process uses VERROR from the PID controller to adjust Vout such that it is ~VSET
P
I
D
Output Process
Sensor
VERRORVSET VOUT
VSENSOR
PID Controller ndash System Block Diagram
ApplicationsPID Controller ndash System Circuit Diagram
Source httpwwwecircuitcentercomCircuitsop_pidop_pidhtm
Calculates VERROR = -(VSET + VSENSOR)
Signal conditioning allows you to introduce a time delay which could
account for things like inertia
System to control
-VSENSOR
ApplicationsPID Controller ndash PID Controller Circuit Diagram
VERR
Adjust Change
Kp RP1 RP2
Ki RI CI
Kd RD CD
VERR PID
Applications of Op-Ampsbull Example of PI Control
Temperature Control
bull Thermal System we wish to automatically control the temperature of
bull Block Diagram of Control System
Applications of Op-Amps
bull Voltage Error Circuit
bull Proportional-Integral Control Circuit
bull Example of PI Control Temperature Control
44
Multiple Inputs(1) Kirchhoff node equation at V+
yields
(2) Kirchhoff node equation at V yields
(3) Setting V+ = Vndash yields
0V
0_
c
c
b
b
a
a
f
o
R
VV
R
VV
R
VV
R
VV
c
aj j
jf
c
c
b
b
a
afo R
VR
R
V
R
V
R
VRV
+
Rf
Va
VoRb
Ra
RcVb
Vc
Homework
1) Design a circuit to 50i
of V
VA
2) Find the vo=
Review Two fundamental Op Amp
Structure Af Input voltage ( )terminal
Feed back
( )terminal
Inverting Amp
_ _
Non inverting Amp
+ _
So far we have been looking at the amplification that can be achieved for relatively small (amplitude) signals For a fixed gain as we increase the input signal amplitude there is a limit to how large the output signal can be The output saturates as it approaches the positive andnegative power supply voltages In other words there is limited range across which the gain is linear vv Vvo
vv Vvo
Op Amp transfer characteristic curve
Applications of Op-AmpsFiltersTypesbullLow pass filterbullHigh pass filterbullBand pass filterbullCascading (2 or more filters connected together) R2
+
-
+
V0
__
+ Vcc
- Vcc
-+
R1
C
Low pass filter
Low pass filter Cutoff frequency
Low pass filter transfer function
Applications of Op-Ampsbull Electrocardiogram EKG Amplification
ndash Need to measure difference in voltage from lead 1 and lead 2ndash 60 Hz interference from electrical equipment
Applications of Op-Ampsbull Simple EKG circuit
ndash Uses differential amplifier to cancel common mode signal and amplify differential mode signal
bull Realistic EKG circuitndash Uses two non-inverting
amplifiers to first amplify voltage from each lead followed by differential amplifier
ndash Forms an ldquoinstrumentation amplifierrdquo
Strain Gauge
Use a Wheatstone bridge to determine the strain of an element by measuring the change in resistance of a
strain gauge
(No strain) Balanced Bridge R 1 = R 2
(Strain) Unbalanced Bridge R 1 ne R 2
Strain GaugeHalf-Bridge Arrangement
Using KCL at the inverting and non-inverting terminals of the op amp we find that ε ~ Vo = 2ΔR(Rf R2)
R + ΔR Rf
+
- +
V0
__
+ Vcc
- Vcc
-+
Rf
Vref
R
R - ΔR
R
Op amp used to amplify output from strain gauge
Applications of Op-Ampsbull Piezoelectric Transducerndash Used to measure force pressure accelerationndash Piezoelectric crystal generates an electric charge in
response to deformation
bull Use Charge Amplifierndash Just an integrator op-amp circuit
bullGoal is to have VSET = VOUT
bullRemember that VERROR = VSET ndash VSENSOR
bullOutput Process uses VERROR from the PID controller to adjust Vout such that it is ~VSET
P
I
D
Output Process
Sensor
VERRORVSET VOUT
VSENSOR
PID Controller ndash System Block Diagram
ApplicationsPID Controller ndash System Circuit Diagram
Source httpwwwecircuitcentercomCircuitsop_pidop_pidhtm
Calculates VERROR = -(VSET + VSENSOR)
Signal conditioning allows you to introduce a time delay which could
account for things like inertia
System to control
-VSENSOR
ApplicationsPID Controller ndash PID Controller Circuit Diagram
VERR
Adjust Change
Kp RP1 RP2
Ki RI CI
Kd RD CD
VERR PID
Applications of Op-Ampsbull Example of PI Control
Temperature Control
bull Thermal System we wish to automatically control the temperature of
bull Block Diagram of Control System
Applications of Op-Amps
bull Voltage Error Circuit
bull Proportional-Integral Control Circuit
bull Example of PI Control Temperature Control
44
Multiple Inputs(1) Kirchhoff node equation at V+
yields
(2) Kirchhoff node equation at V yields
(3) Setting V+ = Vndash yields
0V
0_
c
c
b
b
a
a
f
o
R
VV
R
VV
R
VV
R
VV
c
aj j
jf
c
c
b
b
a
afo R
VR
R
V
R
V
R
VRV
+
Rf
Va
VoRb
Ra
RcVb
Vc
Homework
1) Design a circuit to 50i
of V
VA
2) Find the vo=
Review Two fundamental Op Amp
Structure Af Input voltage ( )terminal
Feed back
( )terminal
Inverting Amp
_ _
Non inverting Amp
+ _
Applications of Op-AmpsFiltersTypesbullLow pass filterbullHigh pass filterbullBand pass filterbullCascading (2 or more filters connected together) R2
+
-
+
V0
__
+ Vcc
- Vcc
-+
R1
C
Low pass filter
Low pass filter Cutoff frequency
Low pass filter transfer function
Applications of Op-Ampsbull Electrocardiogram EKG Amplification
ndash Need to measure difference in voltage from lead 1 and lead 2ndash 60 Hz interference from electrical equipment
Applications of Op-Ampsbull Simple EKG circuit
ndash Uses differential amplifier to cancel common mode signal and amplify differential mode signal
bull Realistic EKG circuitndash Uses two non-inverting
amplifiers to first amplify voltage from each lead followed by differential amplifier
ndash Forms an ldquoinstrumentation amplifierrdquo
Strain Gauge
Use a Wheatstone bridge to determine the strain of an element by measuring the change in resistance of a
strain gauge
(No strain) Balanced Bridge R 1 = R 2
(Strain) Unbalanced Bridge R 1 ne R 2
Strain GaugeHalf-Bridge Arrangement
Using KCL at the inverting and non-inverting terminals of the op amp we find that ε ~ Vo = 2ΔR(Rf R2)
R + ΔR Rf
+
- +
V0
__
+ Vcc
- Vcc
-+
Rf
Vref
R
R - ΔR
R
Op amp used to amplify output from strain gauge
Applications of Op-Ampsbull Piezoelectric Transducerndash Used to measure force pressure accelerationndash Piezoelectric crystal generates an electric charge in
response to deformation
bull Use Charge Amplifierndash Just an integrator op-amp circuit
bullGoal is to have VSET = VOUT
bullRemember that VERROR = VSET ndash VSENSOR
bullOutput Process uses VERROR from the PID controller to adjust Vout such that it is ~VSET
P
I
D
Output Process
Sensor
VERRORVSET VOUT
VSENSOR
PID Controller ndash System Block Diagram
ApplicationsPID Controller ndash System Circuit Diagram
Source httpwwwecircuitcentercomCircuitsop_pidop_pidhtm
Calculates VERROR = -(VSET + VSENSOR)
Signal conditioning allows you to introduce a time delay which could
account for things like inertia
System to control
-VSENSOR
ApplicationsPID Controller ndash PID Controller Circuit Diagram
VERR
Adjust Change
Kp RP1 RP2
Ki RI CI
Kd RD CD
VERR PID
Applications of Op-Ampsbull Example of PI Control
Temperature Control
bull Thermal System we wish to automatically control the temperature of
bull Block Diagram of Control System
Applications of Op-Amps
bull Voltage Error Circuit
bull Proportional-Integral Control Circuit
bull Example of PI Control Temperature Control
44
Multiple Inputs(1) Kirchhoff node equation at V+
yields
(2) Kirchhoff node equation at V yields
(3) Setting V+ = Vndash yields
0V
0_
c
c
b
b
a
a
f
o
R
VV
R
VV
R
VV
R
VV
c
aj j
jf
c
c
b
b
a
afo R
VR
R
V
R
V
R
VRV
+
Rf
Va
VoRb
Ra
RcVb
Vc
Homework
1) Design a circuit to 50i
of V
VA
2) Find the vo=
Review Two fundamental Op Amp
Structure Af Input voltage ( )terminal
Feed back
( )terminal
Inverting Amp
_ _
Non inverting Amp
+ _
Applications of Op-Ampsbull Electrocardiogram EKG Amplification
ndash Need to measure difference in voltage from lead 1 and lead 2ndash 60 Hz interference from electrical equipment
Applications of Op-Ampsbull Simple EKG circuit
ndash Uses differential amplifier to cancel common mode signal and amplify differential mode signal
bull Realistic EKG circuitndash Uses two non-inverting
amplifiers to first amplify voltage from each lead followed by differential amplifier
ndash Forms an ldquoinstrumentation amplifierrdquo
Strain Gauge
Use a Wheatstone bridge to determine the strain of an element by measuring the change in resistance of a
strain gauge
(No strain) Balanced Bridge R 1 = R 2
(Strain) Unbalanced Bridge R 1 ne R 2
Strain GaugeHalf-Bridge Arrangement
Using KCL at the inverting and non-inverting terminals of the op amp we find that ε ~ Vo = 2ΔR(Rf R2)
R + ΔR Rf
+
- +
V0
__
+ Vcc
- Vcc
-+
Rf
Vref
R
R - ΔR
R
Op amp used to amplify output from strain gauge
Applications of Op-Ampsbull Piezoelectric Transducerndash Used to measure force pressure accelerationndash Piezoelectric crystal generates an electric charge in
response to deformation
bull Use Charge Amplifierndash Just an integrator op-amp circuit
bullGoal is to have VSET = VOUT
bullRemember that VERROR = VSET ndash VSENSOR
bullOutput Process uses VERROR from the PID controller to adjust Vout such that it is ~VSET
P
I
D
Output Process
Sensor
VERRORVSET VOUT
VSENSOR
PID Controller ndash System Block Diagram
ApplicationsPID Controller ndash System Circuit Diagram
Source httpwwwecircuitcentercomCircuitsop_pidop_pidhtm
Calculates VERROR = -(VSET + VSENSOR)
Signal conditioning allows you to introduce a time delay which could
account for things like inertia
System to control
-VSENSOR
ApplicationsPID Controller ndash PID Controller Circuit Diagram
VERR
Adjust Change
Kp RP1 RP2
Ki RI CI
Kd RD CD
VERR PID
Applications of Op-Ampsbull Example of PI Control
Temperature Control
bull Thermal System we wish to automatically control the temperature of
bull Block Diagram of Control System
Applications of Op-Amps
bull Voltage Error Circuit
bull Proportional-Integral Control Circuit
bull Example of PI Control Temperature Control
44
Multiple Inputs(1) Kirchhoff node equation at V+
yields
(2) Kirchhoff node equation at V yields
(3) Setting V+ = Vndash yields
0V
0_
c
c
b
b
a
a
f
o
R
VV
R
VV
R
VV
R
VV
c
aj j
jf
c
c
b
b
a
afo R
VR
R
V
R
V
R
VRV
+
Rf
Va
VoRb
Ra
RcVb
Vc
Homework
1) Design a circuit to 50i
of V
VA
2) Find the vo=
Review Two fundamental Op Amp
Structure Af Input voltage ( )terminal
Feed back
( )terminal
Inverting Amp
_ _
Non inverting Amp
+ _
Applications of Op-Ampsbull Simple EKG circuit
ndash Uses differential amplifier to cancel common mode signal and amplify differential mode signal
bull Realistic EKG circuitndash Uses two non-inverting
amplifiers to first amplify voltage from each lead followed by differential amplifier
ndash Forms an ldquoinstrumentation amplifierrdquo
Strain Gauge
Use a Wheatstone bridge to determine the strain of an element by measuring the change in resistance of a
strain gauge
(No strain) Balanced Bridge R 1 = R 2
(Strain) Unbalanced Bridge R 1 ne R 2
Strain GaugeHalf-Bridge Arrangement
Using KCL at the inverting and non-inverting terminals of the op amp we find that ε ~ Vo = 2ΔR(Rf R2)
R + ΔR Rf
+
- +
V0
__
+ Vcc
- Vcc
-+
Rf
Vref
R
R - ΔR
R
Op amp used to amplify output from strain gauge
Applications of Op-Ampsbull Piezoelectric Transducerndash Used to measure force pressure accelerationndash Piezoelectric crystal generates an electric charge in
response to deformation
bull Use Charge Amplifierndash Just an integrator op-amp circuit
bullGoal is to have VSET = VOUT
bullRemember that VERROR = VSET ndash VSENSOR
bullOutput Process uses VERROR from the PID controller to adjust Vout such that it is ~VSET
P
I
D
Output Process
Sensor
VERRORVSET VOUT
VSENSOR
PID Controller ndash System Block Diagram
ApplicationsPID Controller ndash System Circuit Diagram
Source httpwwwecircuitcentercomCircuitsop_pidop_pidhtm
Calculates VERROR = -(VSET + VSENSOR)
Signal conditioning allows you to introduce a time delay which could
account for things like inertia
System to control
-VSENSOR
ApplicationsPID Controller ndash PID Controller Circuit Diagram
VERR
Adjust Change
Kp RP1 RP2
Ki RI CI
Kd RD CD
VERR PID
Applications of Op-Ampsbull Example of PI Control
Temperature Control
bull Thermal System we wish to automatically control the temperature of
bull Block Diagram of Control System
Applications of Op-Amps
bull Voltage Error Circuit
bull Proportional-Integral Control Circuit
bull Example of PI Control Temperature Control
44
Multiple Inputs(1) Kirchhoff node equation at V+
yields
(2) Kirchhoff node equation at V yields
(3) Setting V+ = Vndash yields
0V
0_
c
c
b
b
a
a
f
o
R
VV
R
VV
R
VV
R
VV
c
aj j
jf
c
c
b
b
a
afo R
VR
R
V
R
V
R
VRV
+
Rf
Va
VoRb
Ra
RcVb
Vc
Homework
1) Design a circuit to 50i
of V
VA
2) Find the vo=
Review Two fundamental Op Amp
Structure Af Input voltage ( )terminal
Feed back
( )terminal
Inverting Amp
_ _
Non inverting Amp
+ _
Strain Gauge
Use a Wheatstone bridge to determine the strain of an element by measuring the change in resistance of a
strain gauge
(No strain) Balanced Bridge R 1 = R 2
(Strain) Unbalanced Bridge R 1 ne R 2
Strain GaugeHalf-Bridge Arrangement
Using KCL at the inverting and non-inverting terminals of the op amp we find that ε ~ Vo = 2ΔR(Rf R2)
R + ΔR Rf
+
- +
V0
__
+ Vcc
- Vcc
-+
Rf
Vref
R
R - ΔR
R
Op amp used to amplify output from strain gauge
Applications of Op-Ampsbull Piezoelectric Transducerndash Used to measure force pressure accelerationndash Piezoelectric crystal generates an electric charge in
response to deformation
bull Use Charge Amplifierndash Just an integrator op-amp circuit
bullGoal is to have VSET = VOUT
bullRemember that VERROR = VSET ndash VSENSOR
bullOutput Process uses VERROR from the PID controller to adjust Vout such that it is ~VSET
P
I
D
Output Process
Sensor
VERRORVSET VOUT
VSENSOR
PID Controller ndash System Block Diagram
ApplicationsPID Controller ndash System Circuit Diagram
Source httpwwwecircuitcentercomCircuitsop_pidop_pidhtm
Calculates VERROR = -(VSET + VSENSOR)
Signal conditioning allows you to introduce a time delay which could
account for things like inertia
System to control
-VSENSOR
ApplicationsPID Controller ndash PID Controller Circuit Diagram
VERR
Adjust Change
Kp RP1 RP2
Ki RI CI
Kd RD CD
VERR PID
Applications of Op-Ampsbull Example of PI Control
Temperature Control
bull Thermal System we wish to automatically control the temperature of
bull Block Diagram of Control System
Applications of Op-Amps
bull Voltage Error Circuit
bull Proportional-Integral Control Circuit
bull Example of PI Control Temperature Control
44
Multiple Inputs(1) Kirchhoff node equation at V+
yields
(2) Kirchhoff node equation at V yields
(3) Setting V+ = Vndash yields
0V
0_
c
c
b
b
a
a
f
o
R
VV
R
VV
R
VV
R
VV
c
aj j
jf
c
c
b
b
a
afo R
VR
R
V
R
V
R
VRV
+
Rf
Va
VoRb
Ra
RcVb
Vc
Homework
1) Design a circuit to 50i
of V
VA
2) Find the vo=
Review Two fundamental Op Amp
Structure Af Input voltage ( )terminal
Feed back
( )terminal
Inverting Amp
_ _
Non inverting Amp
+ _
Strain GaugeHalf-Bridge Arrangement
Using KCL at the inverting and non-inverting terminals of the op amp we find that ε ~ Vo = 2ΔR(Rf R2)
R + ΔR Rf
+
- +
V0
__
+ Vcc
- Vcc
-+
Rf
Vref
R
R - ΔR
R
Op amp used to amplify output from strain gauge
Applications of Op-Ampsbull Piezoelectric Transducerndash Used to measure force pressure accelerationndash Piezoelectric crystal generates an electric charge in
response to deformation
bull Use Charge Amplifierndash Just an integrator op-amp circuit
bullGoal is to have VSET = VOUT
bullRemember that VERROR = VSET ndash VSENSOR
bullOutput Process uses VERROR from the PID controller to adjust Vout such that it is ~VSET
P
I
D
Output Process
Sensor
VERRORVSET VOUT
VSENSOR
PID Controller ndash System Block Diagram
ApplicationsPID Controller ndash System Circuit Diagram
Source httpwwwecircuitcentercomCircuitsop_pidop_pidhtm
Calculates VERROR = -(VSET + VSENSOR)
Signal conditioning allows you to introduce a time delay which could
account for things like inertia
System to control
-VSENSOR
ApplicationsPID Controller ndash PID Controller Circuit Diagram
VERR
Adjust Change
Kp RP1 RP2
Ki RI CI
Kd RD CD
VERR PID
Applications of Op-Ampsbull Example of PI Control
Temperature Control
bull Thermal System we wish to automatically control the temperature of
bull Block Diagram of Control System
Applications of Op-Amps
bull Voltage Error Circuit
bull Proportional-Integral Control Circuit
bull Example of PI Control Temperature Control
44
Multiple Inputs(1) Kirchhoff node equation at V+
yields
(2) Kirchhoff node equation at V yields
(3) Setting V+ = Vndash yields
0V
0_
c
c
b
b
a
a
f
o
R
VV
R
VV
R
VV
R
VV
c
aj j
jf
c
c
b
b
a
afo R
VR
R
V
R
V
R
VRV
+
Rf
Va
VoRb
Ra
RcVb
Vc
Homework
1) Design a circuit to 50i
of V
VA
2) Find the vo=
Review Two fundamental Op Amp
Structure Af Input voltage ( )terminal
Feed back
( )terminal
Inverting Amp
_ _
Non inverting Amp
+ _
Applications of Op-Ampsbull Piezoelectric Transducerndash Used to measure force pressure accelerationndash Piezoelectric crystal generates an electric charge in
response to deformation
bull Use Charge Amplifierndash Just an integrator op-amp circuit
bullGoal is to have VSET = VOUT
bullRemember that VERROR = VSET ndash VSENSOR
bullOutput Process uses VERROR from the PID controller to adjust Vout such that it is ~VSET
P
I
D
Output Process
Sensor
VERRORVSET VOUT
VSENSOR
PID Controller ndash System Block Diagram
ApplicationsPID Controller ndash System Circuit Diagram
Source httpwwwecircuitcentercomCircuitsop_pidop_pidhtm
Calculates VERROR = -(VSET + VSENSOR)
Signal conditioning allows you to introduce a time delay which could
account for things like inertia
System to control
-VSENSOR
ApplicationsPID Controller ndash PID Controller Circuit Diagram
VERR
Adjust Change
Kp RP1 RP2
Ki RI CI
Kd RD CD
VERR PID
Applications of Op-Ampsbull Example of PI Control
Temperature Control
bull Thermal System we wish to automatically control the temperature of
bull Block Diagram of Control System
Applications of Op-Amps
bull Voltage Error Circuit
bull Proportional-Integral Control Circuit
bull Example of PI Control Temperature Control
44
Multiple Inputs(1) Kirchhoff node equation at V+
yields
(2) Kirchhoff node equation at V yields
(3) Setting V+ = Vndash yields
0V
0_
c
c
b
b
a
a
f
o
R
VV
R
VV
R
VV
R
VV
c
aj j
jf
c
c
b
b
a
afo R
VR
R
V
R
V
R
VRV
+
Rf
Va
VoRb
Ra
RcVb
Vc
Homework
1) Design a circuit to 50i
of V
VA
2) Find the vo=
Review Two fundamental Op Amp
Structure Af Input voltage ( )terminal
Feed back
( )terminal
Inverting Amp
_ _
Non inverting Amp
+ _
bullGoal is to have VSET = VOUT
bullRemember that VERROR = VSET ndash VSENSOR
bullOutput Process uses VERROR from the PID controller to adjust Vout such that it is ~VSET
P
I
D
Output Process
Sensor
VERRORVSET VOUT
VSENSOR
PID Controller ndash System Block Diagram
ApplicationsPID Controller ndash System Circuit Diagram
Source httpwwwecircuitcentercomCircuitsop_pidop_pidhtm
Calculates VERROR = -(VSET + VSENSOR)
Signal conditioning allows you to introduce a time delay which could
account for things like inertia
System to control
-VSENSOR
ApplicationsPID Controller ndash PID Controller Circuit Diagram
VERR
Adjust Change
Kp RP1 RP2
Ki RI CI
Kd RD CD
VERR PID
Applications of Op-Ampsbull Example of PI Control
Temperature Control
bull Thermal System we wish to automatically control the temperature of
bull Block Diagram of Control System
Applications of Op-Amps
bull Voltage Error Circuit
bull Proportional-Integral Control Circuit
bull Example of PI Control Temperature Control
44
Multiple Inputs(1) Kirchhoff node equation at V+
yields
(2) Kirchhoff node equation at V yields
(3) Setting V+ = Vndash yields
0V
0_
c
c
b
b
a
a
f
o
R
VV
R
VV
R
VV
R
VV
c
aj j
jf
c
c
b
b
a
afo R
VR
R
V
R
V
R
VRV
+
Rf
Va
VoRb
Ra
RcVb
Vc
Homework
1) Design a circuit to 50i
of V
VA
2) Find the vo=
Review Two fundamental Op Amp
Structure Af Input voltage ( )terminal
Feed back
( )terminal
Inverting Amp
_ _
Non inverting Amp
+ _
ApplicationsPID Controller ndash System Circuit Diagram
Source httpwwwecircuitcentercomCircuitsop_pidop_pidhtm
Calculates VERROR = -(VSET + VSENSOR)
Signal conditioning allows you to introduce a time delay which could
account for things like inertia
System to control
-VSENSOR
ApplicationsPID Controller ndash PID Controller Circuit Diagram
VERR
Adjust Change
Kp RP1 RP2
Ki RI CI
Kd RD CD
VERR PID
Applications of Op-Ampsbull Example of PI Control
Temperature Control
bull Thermal System we wish to automatically control the temperature of
bull Block Diagram of Control System
Applications of Op-Amps
bull Voltage Error Circuit
bull Proportional-Integral Control Circuit
bull Example of PI Control Temperature Control
44
Multiple Inputs(1) Kirchhoff node equation at V+
yields
(2) Kirchhoff node equation at V yields
(3) Setting V+ = Vndash yields
0V
0_
c
c
b
b
a
a
f
o
R
VV
R
VV
R
VV
R
VV
c
aj j
jf
c
c
b
b
a
afo R
VR
R
V
R
V
R
VRV
+
Rf
Va
VoRb
Ra
RcVb
Vc
Homework
1) Design a circuit to 50i
of V
VA
2) Find the vo=
Review Two fundamental Op Amp
Structure Af Input voltage ( )terminal
Feed back
( )terminal
Inverting Amp
_ _
Non inverting Amp
+ _
ApplicationsPID Controller ndash PID Controller Circuit Diagram
VERR
Adjust Change
Kp RP1 RP2
Ki RI CI
Kd RD CD
VERR PID
Applications of Op-Ampsbull Example of PI Control
Temperature Control
bull Thermal System we wish to automatically control the temperature of
bull Block Diagram of Control System
Applications of Op-Amps
bull Voltage Error Circuit
bull Proportional-Integral Control Circuit
bull Example of PI Control Temperature Control
44
Multiple Inputs(1) Kirchhoff node equation at V+
yields
(2) Kirchhoff node equation at V yields
(3) Setting V+ = Vndash yields
0V
0_
c
c
b
b
a
a
f
o
R
VV
R
VV
R
VV
R
VV
c
aj j
jf
c
c
b
b
a
afo R
VR
R
V
R
V
R
VRV
+
Rf
Va
VoRb
Ra
RcVb
Vc
Homework
1) Design a circuit to 50i
of V
VA
2) Find the vo=
Review Two fundamental Op Amp
Structure Af Input voltage ( )terminal
Feed back
( )terminal
Inverting Amp
_ _
Non inverting Amp
+ _
Applications of Op-Ampsbull Example of PI Control
Temperature Control
bull Thermal System we wish to automatically control the temperature of
bull Block Diagram of Control System
Applications of Op-Amps
bull Voltage Error Circuit
bull Proportional-Integral Control Circuit
bull Example of PI Control Temperature Control
44
Multiple Inputs(1) Kirchhoff node equation at V+
yields
(2) Kirchhoff node equation at V yields
(3) Setting V+ = Vndash yields
0V
0_
c
c
b
b
a
a
f
o
R
VV
R
VV
R
VV
R
VV
c
aj j
jf
c
c
b
b
a
afo R
VR
R
V
R
V
R
VRV
+
Rf
Va
VoRb
Ra
RcVb
Vc
Homework
1) Design a circuit to 50i
of V
VA
2) Find the vo=
Review Two fundamental Op Amp
Structure Af Input voltage ( )terminal
Feed back
( )terminal
Inverting Amp
_ _
Non inverting Amp
+ _
Applications of Op-Amps
bull Voltage Error Circuit
bull Proportional-Integral Control Circuit
bull Example of PI Control Temperature Control
44
Multiple Inputs(1) Kirchhoff node equation at V+
yields
(2) Kirchhoff node equation at V yields
(3) Setting V+ = Vndash yields
0V
0_
c
c
b
b
a
a
f
o
R
VV
R
VV
R
VV
R
VV
c
aj j
jf
c
c
b
b
a
afo R
VR
R
V
R
V
R
VRV
+
Rf
Va
VoRb
Ra
RcVb
Vc
Homework
1) Design a circuit to 50i
of V
VA
2) Find the vo=
Review Two fundamental Op Amp
Structure Af Input voltage ( )terminal
Feed back
( )terminal
Inverting Amp
_ _
Non inverting Amp
+ _
44
Multiple Inputs(1) Kirchhoff node equation at V+
yields
(2) Kirchhoff node equation at V yields
(3) Setting V+ = Vndash yields
0V
0_
c
c
b
b
a
a
f
o
R
VV
R
VV
R
VV
R
VV
c
aj j
jf
c
c
b
b
a
afo R
VR
R
V
R
V
R
VRV
+
Rf
Va
VoRb
Ra
RcVb
Vc
Homework
1) Design a circuit to 50i
of V
VA
2) Find the vo=
Review Two fundamental Op Amp
Structure Af Input voltage ( )terminal
Feed back
( )terminal
Inverting Amp
_ _
Non inverting Amp
+ _
Homework
1) Design a circuit to 50i
of V
VA
2) Find the vo=
Review Two fundamental Op Amp
Structure Af Input voltage ( )terminal
Feed back
( )terminal
Inverting Amp
_ _
Non inverting Amp
+ _
Review Two fundamental Op Amp
Structure Af Input voltage ( )terminal
Feed back
( )terminal
Inverting Amp
_ _
Non inverting Amp
+ _
