Lesson #5 Chapter 2 - New Jersey Institute of Technologyjoelsd/electronics/classwork/...BME 372...

BME 372 Electronics I –J.Schesser

138

Operational Amplifiers

Lesson #5Chapter 2


139

Operational Amplifiers

• An operational Amplifier is an ideal differential with the following characteristics:– Infinite input impedance– Infinite gain for the differential signal– Zero gain for the common-mode signal– Zero output impedance– Infinite Bandwidth vo-

+

v1

v2

Inverting input

Non-Inverting input


140

Operational Amplifier Feedback• Operational Amplifiers are used with negative feedback• Feedback is a way to return a portion of the output of an

amplifier to the input– Negative Feedback: returned output opposes the source signal– Positive Feedback: returned output aids the source signal

• For Negative Feedback– In an Op-amp, the negative feedback returns a fraction of the

output to the inverting input terminal forcing the differential input to zero.

– Since the Op-amp is ideal and has infinite gain, the differential input will exactly be zero. This is called a virtual short circuit

– Since the input impedance is infinite the current flowing into the input is also zero.

– These latter two points are called the summing-point constraint.


141

Operational Amplifier Analysis Using the Summing Point Constraint

• In order to analyze Op-amps, the following steps should be followed:

1. Verify that negative feedback is present2. Assume that the voltage and current at the

input of the Op-amp are both zero (Summing-point Constraint

3. Apply standard circuit analyses techniques such as Kirchhoff’s Laws, Nodal or Mesh Analysis to solve for the quantities of interest.


142

Example: Inverting Amplifier1. Verify Negative Feedback: Note that a

portion of vo is fed back via R2 to the inverting input. So if vi increases and, therefore, increases vo, the portion of vo fed back will then have the affect of reducing vi (i.e., negative feedback).

2. Use the summing point constraint.

3. Use KVL at the inverting input node for both the branch connected to the source and the branch connected to the output

RLvin

R2

R1

vi

vo

-

+

0i1=vin/R1

i2

inverted) :input the toopposite isoutput that the(note oft independen is which -

zero is since 0constraintpoint -summing the todue

constraintpoint -summing the todue zero is since 0

1

2

220

21

11

Lin

i

iin

RvRR

vRivii

vRiv

1

1

11

1

RiRi

ivZ in

in

-+ +

+

-

-


143

Op-amp

• Because we assumed that the Op-amp was ideal, we found that with negative feedback we can achieve a gain which is:

1. Independent of the load2. Dependent only on values of the circuit

parameter3. We can choose the gain of our amplifier by

proper selection of resistors.


144

Another Example: Inverting Amplifier1. Verify Negative Feedback:

2. Use the summing point constraint.

3. Use KVL at the inverting input node for the branch connected to the source and KCL & KVL at the node where the 3 resistors are connectedRLvin

R2

R1

vivo

-

+

0i1=vin/R1

i2

3344

234

33223322

21

11

zero is since 0constraintpoint -summing the todue

constraintpoint -summing the todue zero is since 0

iRiRviii

vRiRiRiRivii

vRiv

o

ii

iin

R4

R3

i4

i3


145

Another Example: Continued

RLvin

R2

R1

vivo

-

+

0i1=vin/R1

i2

3344

234

3322

21

11

0

iRiRviii

RiRiii

Riv

o

in

R4

R3

i4

i3

11

1

2

1

4

13

42

1

2

1

4

13

42

13

23

113

243344

113

22

3

2234

122111

23

233322

)(

)(

)1(

)1()1(

;0

RivRZ

RR

RR

RRRR

vvA

RR

RR

RRRRv

Rv

RRRv

RRRRRiRiRv

vRRR

RiRRiii

RviiiRiv

iRRiRiRi

ininin

in

ov

in

inino

in

inin

Why would we use this design over the simpler one?

1. Same gain but with smaller values of resistance

2. Higher gain


146

Non-inverting Amp1. First check: negative feedback?2. Next apply, summing point constraint3. Use circuit analysis

+

--vo

+

--

RLR2

R1

+

--

vi

+

--

vf

io

Iin= 0

+

-

vin

1

1 2

2 1 2

1 1

0

;

1

Since 0;

in i f f f

f o in

ov

in

inin in

in

v v v v v

Rv v vR Rv R R RAv R R

vi Zi

Note: 1. The gain is always greater than one2. The output has the same sign as the

input


147

Non-inverting Amp Special CaseWhat happens if R2 = 0?

+

--vo

+

--

RL

+

--

vi

io

Iin= 0

+

-

vin

in

ininin

in

o

inoof

fffiin

ivZi

RR

vvAv

vvvR

Rv

vvvvv

;0 Since

10

;0

0

1

1

1

1

This is a unity gain amplifier and is also called a voltage follower.


148

Some Practical Issues when Designing Op-amps• Since ratios of resistor values determines the gain,

choosing the proper resistor values is crucial– Too small means large currents drawn– Too large yields another set of problems

• Open Loop Gain is not constant but a function of frequency

• Non-linearities of the amplifier– Voltage clipping– Slew rate

• DC imperfections– Offsets– Bias Currents


149

Selecting Resistor Values

• Let say we want a gain of 10. This means that R2 = 9R1.

• If we chose R1=1Ω, then for a 10 volt output, there will be 1 A flowing threw R1 and R2.

• THIS IS DANGEROUS!!!!• On the other hand if R1=10 MΩ, then

there may be unwanted effects due to pickup of induced signals

• Therefore choosing values between 100Ω and 1MΩ is optimum

+

--vo

+

--

R2

R1

+

--

vi

+

--

vf

io

Iin= 0

+

-

vin

1

21RRAv


150

Frequency Issues• Let’s assume that the open loop gain of our Op-amp

is a function of frequency.

( )1 ( / )

where is the open-loop gain for 0, is called the open-loop break

frequency since when

then ( ) / 2 or at the half power point.

Note that when ,

oOLOL

BOL

oOL

BOL

BOL

OL oOL

oOL BOL

AA fj f f

A ff

f f

A f A

f A f A

( ) 1 This is will define an important relationship for the amplifierwhen feedback is used

OL f

20 log |AOL( f )| dB20 log |A0OL|

fBOL


151

Frequency Issues

1 1

1 2 1 2

1 2 2

1 1

( )1 ( / )

Using phasors:

; where

; since

1

(Note for the ideal Op-amp 1and 1

which what we e

oOLOL

BOL

f O O

Oin i O O O i OL

OL

O OLCL

in OL

OL

CL

AA fj f f

R RR R R R

AA

AAA

AR R RA

R R

V V V

VV V V V V V

VV

xpected.)

+

--Vo

+

--

R2

R1

+

--

Vi

+

--

Vf

io

Iin= 0

+

-

Vin

( ) ( ) 1 ( / )( )1 ( ) 1 ( ) 1

1 ( / )

1 ( / )1 ( / ) 1 ( / )1 ( / ) 1 ( / )

1

1 ( ) 1 ( )(1 )

oOL

OL OL BOLCL

oOLOL OL

BOL

oOL

BOL oOL

BOL oOL oOL BOL

BOL BOL

oOL

oOL oCL

BOL oOL BCL

AA f A f j f fA f AA f A f

j f fA

j f f Aj f f A A j f fj f f j f f

AA A

f fj jf A f


152

Frequency Issues

A interesting factor now comes to light:

(1 )1

The gain bandwidth product is constant!!!

oOLoCL BCL BOL oOL oOL BOL

oOL

AA f f A A fA

+

--Vo

+

--

R2

R1

+

--

Vi

+

--

Vf

io

Iin= 0

+

-

Vin20 log |AOL( f )| dB

fBOL fBOL(1+AoOL)

20 log |A0OL|

20 log |A0CL|20 log |AOL( f )| dB

Where:

and (1 )1

oOLoCL BCL BOL oOL

oOL

AA f f AA


153

Example• For an amplifier with Open-loop dc gain of 100

dB with Open-loop breakpoint of 40 Hz the Bandwidth product for β=1, 0.1, 0.01

β AOCL AOCL(dB) fBCL

1 0.9999 0 4 MHz

0.1 9.9990 20 400 kHz

0.01 99.90 40 40 kHz


154

Voltage clipping

• The Op-amp has a limit as to how large an output voltage and current can be produced. (See figure 2.28)

• For example for the circuit shown it has the following limitations: ±12 V and ±25 mAa) If the load resistor is 10k Ω, what is

the maximum input voltage which can be handled without clipping

b) Repeat for 100 Ω

VVVV

VVmARR

VR

VI

mARR

VR

VIb

in

o

ooo

L

oo

o

L

oo

61.0444.2 44.2

1000300010025

reached. is tageoutput vol maximum beforeoccur llcurrent wioutput maximum case, For this

12310003000

1210012)

max

max

maxmax

21

maxmaxmax

21

maxmaxmax

+

--Vo

+

--

3k

1k

+

--

Vi

+

--

Vf

io

Iin= 0

+

-

Vin

VV

mARR

VR

VIa

A

in

o

L

oo

vCL

3412 reached. iscurrent output maximum before

occur willtageoutput vol maximum case, For this2.4

1000300012

1012)

41/31

max

421

maxmaxmax

-20

-10

0

10

20

0 5 10 15 20


155

Slew Rate

• Slew Rate is a phenomenon which occurs when the Op-Amp can not keep up the change in the input.

• Therefore, we identify the maximum rate of change of the Op-amp as the Slew Rate -SR


156

DC imperfections

• We saw that we have to provide DC voltages to an amplifier in order to provide it with the power to support amplification.– For a differential amplifier which must handle both positive and

negative voltages• The process of designing this DC circuitry is called biasing• As a result, biasing currents flow through the amplifier

which affects its performance. • In particular, a voltage during to the biasing will appear at

the output without any input signal.• These extra voltage can be due to:

– Bias currents flowing in the feedback circuitry– Bias current differentials – Voltage offsets due to the fact that the Op-amp circuitry is not

ideal


157

Special Amplifiers

• Summer (Homework Problem)• Instrumentation Amplifier

– Uses 3 Op-amps– One as a differential amplifier– Two Non-inverting Amps using for providing

gain


158

Medical Instrumentation Amplifier

Differential Amplifier

+

--

R2

R1

+

-

v2

+

--

R2

R1

+

-

v1

R3

R4

+

-

vo

R5

R6

Non-inverting Amplifier



159


+

--

R2

R1

+

-

v2

+

--

R2

R1

+

-

v1

R3

R4

vo

R5

R6




v2D

v1D

+

-


160

Medical Instrumentation Amplifier2

6 5

2

6 6 5 5

5 62

6 6 5 5

41

3 4

5 62 41

6 3 4 6 5 5

5 5 6 41 2

6 5 3 4

5 6 4

5 3 4

51 2

6

1 1( )

( )

( )

( )

Chose 1

( )

D x x o

oDx

oDx

y D x i x

oDD

o D D

o D D

v v v vR R

vv vR R R R

R R vv vR R R R

Rv v v v vR R

R R vv R vR R R R R R

R R R Rv v vR R R R

R R RR R R

Rv v vR

R3

R4

vo

R5

R6


v2D

v1D

vx

+

-

4

3

5

6

3564

45356454

43

4

5

65

1 Chose

RR

RR

RRRRRRRRRRRR

RRR

RRR

vi=0

vy


161


+

--

R2

R1

+

-

v2

+

--

R2

R1

+

-

v1

R3

R4

+

-

vo

R5

R6





162


+

--

R2

R1

+

-

v2

+

--

R2

R1

+

-

v1 Non-inverting Amplifier


v2D

v1DvA

AD

AD

AD

AD

vRRv

RRRv

vRRv

RRRv

vRRv

RRRv

Rvv

Rvv

1

21

1

211

1

22

1

212

1

22

12

22

1

2

2

22

Likewise

)11(


163


))(1())((

)]([

)(

21

1

2

6

521

1

21

6

5

1

22

1

21

1

21

1

21

6

5

1

21

1

211

1

22

1

212

21

6

5

vvRR

RRvv

RRR

RRv

vRRv

RRRv

RRv

RRR

RRv

vRRv

RRRv

vRRv

RRRv

vvRRv

o

AAo

AD

AD

DDo

+

--R2

R1

+

-

v2

+

--

R2

R1+

-

v1

R3

R4

vo

R5

R6




+

-

-

+


164

Wheatstone Bridge as a sensor

R3

R4

vo

R2

R1v2D

v1D

+

-rArB

rDrC

V


165

Wheatstone Bridge

Using Thevinin's Theorem on the Wheatstone BridgeLeft side Right side

V2 rB

rB rC

V;r2 rBrC

rB rC

V1 rA

rA rD

V;r1 rArD

rA rD

rArB

rDrC

V

V2 V1

rArB

rDrC

V

V2 V1rA

rB

rDrC

V

V2

V1

r1+V1

--

r2+V2

--


166

Wheatstone Bridge

VBridge V2 V1 ( rB

rB rC

rA

rA rD

)V

When bridge is balanced rB

rB rC

rA

rA rD

rBrD rArC rA

rD

rB

rC

(nominally, rA rB rC rD )and VBridge 0

rArB

rDrC

V

V2 V1


167

Wheatstone Bridge

V2 vx

R1 r2

vx vo

R2

V2

R1 r2

vx ( 1R1 r2

1R2

) vo

R2

V2

R1 r2

vx (R1 R2 r2

(R1 r2 )R2

) vo

R2

vx vy R4

R3 r1 R4

V1

V2

R1 r2

R4

R3 r1 R4

V1(R1 R2 r2

(R1 r2 )R2

) vo

R2

V2

R1 r2

R4

R3 r1 R4

(R1 R2 r2

R2

)V1

(R1 r2 )vo

R2

vo R2

R1 r2

[(R4

R3 r1 R4

)(R1 R2 r2

R2

)V1 V2 ]

Chose R1 R2 r2

R2

R4

R3 r1 R4

1

vo R2

R1 r2

(V1 V2 ) R2

R1 r2

(rA

rA rD

rB

rB rC

)V

R3

R4

vo

R2

R1 +

-

r1+V1

--

r2+V2

--

vx


168

Wheatstone Bridge

R3

R4

vo

R2

R1 +

-

r1+V1

--

r2+V2

--

Chose R1 R2 r2

R2

R4

R3 r1 R4

1

R1 R2 r2

R2

R3 r1 R4

R4

R1 R2 rBrC

rB rC

R2

R3

rArD

rA rD

R4

R4

Hard to deal with; then let's make sure that r2 R1 R2 and r1 R3 R4

R1 R2

R2

R3 R4

R4

R2R3 R2R4 R4R1 R4R2 R2R3 R4R1

R4

R3

R2

R1

vo R2

R1

(rA

rA rD

rB

rB rC

)V

Note that the output depends on the unbalance of the bridge and when the bridge is balanced vo 0

Gain is still R2

R1

vx


169

Integrators and Differentiators

t

in

t

o

in

dxxvRC

dxxiC

v

tiR

tvti

002

21

)(1)(1

)()()(

vin

C

R

vi

vo

-

+

0i2i1

vin

R

C

vi

vo

-

+

0i2i1

dttdvRCRtiv

tidt

tCdvti

ino

in

)()(

)()()(

2

21


170

Frequency Analysis

vin

vivo

-

+

0i1=vin/Z1

i2

1in

o

Lin1

o

1

11in

ZZ

VV

ZVZZ

ZIVII

ZIV

2

2

22

2

- )()(

oft independen is which )(-

zero )(virtually is since 0)()(constraintpoint -summing the todue )()(

zero )(virtually is since 0)()()(

jj

j

vjjjj

vjjj

i

i

ZLvin

C

R

vivo

-+

0i1=vin/Z1 i2

ZLvin

R

C

vivo

-+

0i1=vin/Z1 i2

integratoran 1- )()( 2

RCjjj

1in

o

ZZ

VV tordifferenia a - )(

)( 2 RCjjj

1in

o

ZZ

VV

Z2

Z1


171

Frequency Response

vin

CR

vi

vo

-

+

0i2i1

vin

RC

vi

vo

-

+

0i2i1

1- )()( 2

RCjjj

1in

o

ZZ

VV

RCjjj

1in

o

ZZ

VV 2-

)()(

ω

ω


172

Frequency Response

vin

C2R1

vi

vo

-

+

0i2i1

vin

R2

C1vi

vo

-

+

0i2

i1

R2 12 2 21 12

1 1 2 2 1 1 1

1 1 tan ( )(1 ) 1 ( )

out

in

V Z R R C RV Z R j C R R C R

0

500

1,000

1,500

2,000

2,500

1.E+00 1.E+02 1.E+04 1.E+06 1.E+08HZ

R1 51 C1 10mf R2 100k

R1

Vout

Vin

Z2

Z1

R2

R1

jC1R1

(1 jC1R1)

R2

R1

C1R1

1 (C1R1)2

2 tan1(C1R1)

0

500

1,000

1,500

2,000

2,500

1.E+00 1.E+02 1.E+04 1.E+06 1.E+08HZ

R1 51 C1 10mf R2 100k


173

Integrators and Differentiators• Integrators and Differentiators are used in analog

computers• An analog computer solves a differential equations• By using Integrators and Differentiators one can

“program” a particular differential equation to be solved• Usually only integrators are used since the gain of a

differentiator occurs at high frequencies while the opposite is true for the integrator.

• Since the frequency response of an real Op-amp attenuates high frequencies, using a differentiator conflicts with the characteristics with a real Op-amp

• Also noise (high frequencies) are amplified by differentiators


174

R-Wave Detector


175

Positive Feedback• What happens to our Op-amp circuit if positive

feedback is used?

• Even if vin = 0 and there is a slight voltage at vi then – vo = AOLvi will increase vi– vo will grow even larger– Eventually this will reach an extreme since there is not

infinite energy in the circuit

RLvin

R

R

vivo

+

-

iini1

i2

) (21)(

21

gain loopopen theis where

0

impedanceinput infinite hasAmplifier since ;00

:input at the KCL Using

21

iOLinoini

OLiOLo

oiini

in

in

vAvvvv

AvAvR

vvR

vvi

iii


176

Positive Feedback• Then our positive feedback design will operate

between its positive and negative extremes, say ±5V

RLvin

R

R

vivo

+

-

iini1

i21 ( )2

For to be at 5 , 01 1( ) ( 5) 02 2

This is true as long as -5For to be at 5 , 0

1 1( ) ( 5) 02 2

This is true as long as 5

i in o

o i

i in o in

in

o i

i in o in

in

v v v

v V v

v v v v

v Vv V v

v v v v

v V

-5 +5

-5

+5

Vin

Vo


177

Homework• Probs 2.2, 2.5, 2.6, 2.10, 2.22, 2.24, 2.25,

2.28• Calculate and plot the output vs frequency

for these circuits. R1=1k, R2=3k, C=1μf Use Matlab to perform the plot

vin

C

R1vi

vo

-

+

R2

vin

CR1vi

vo

-

+

R2

vin

C

R1vi

vo

-

+

R2

vin

C

R1

vi

vo

-

+

R2


178

Homework

rA= R-ΔR

rB = R+ΔR

rD=R+ΔRrC=

R-ΔR

V

V2 V1

Wheatstone Bridge - Strain Gauge• A strain gauge shown here is used with a difference amplifier. Calculate the

amplifier output signal as a function of ΔR and difference amplifier resistors. Assume that ΔR<<R.

• For the strain gauge calculate the value of each of the difference amplifier resistors for a value of R=10Ω and a percent change in R of ±10% if the output of the amplifier is less than ±5 volts. Assume that the bridge is powered with 5 volts.


179

HomeworkCalculate and plot the output vs frequency for

this circuits. R1=1k, R2=3k, C=1μf. Use Matlab to perform the plot.

vin

C

R1

vi

vo

+

-

R2

C

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