Lecture 10 Op-Amp CircuitsAmp Circuits - Courses | … 10 Op-Amp CircuitsAmp Circuits ELECTRICAL...

Lecture 10Op Amp CircuitsOp-Amp Circuits

ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.

Characteristics of Ideal Op pAmps

Infinite gain for the differential input signal

Zero gain for the common mode input signalZero gain for the common-mode input signal

Infinite input impedances

Zero output impedance

Infinite bandwidth



Real Versus Ideal Op Amp

Parameter Typical Range Ideal ValuesOpen-loop gain A 10^5-10^8 ∞Input resistance, Ri 10^5 to 10^13Ω ∞ ΩOutput resistance, Ro 10 to 100Ω 0 Ω


741 Op-Amp


741 Op-Amp

Current Mirrors

Output Stage

Differential Amplifier

Class A Amplifier


SUMMING POINTSUMMING-POINT CONSTRAINT

Operational amplifiers are almost alwaysOperational amplifiers are almost always used with negative feedback, in which part of the output signal is returned to the input inthe output signal is returned to the input in opposition to the source signal.


In a negative feedback system, the ideal op-amp output voltage attains the value neededamp output voltage attains the value needed to force the differential input voltage and input current to zero. We call this fact thecurrent to zero. We call this fact the summing-point constraint.


Ideal op amp circuits areIdeal op-amp circuits are analyzed by the following steps:

1. Verify that negative feedback is present.

2. Assume that the differential input voltage and the input current of the op amp are forced to zero. (This is the summing-point constraint.)


3 Apply standard circuit analysis principles3. Apply standard circuit-analysis principles, such as Kirchhoff’s laws and Ohm’s law, to solve for the quantities of interestsolve for the quantities of interest.


The Basic Inverter

decreasesvvv xox →<<→> 00


Applying the Summing Point Constraint

20 RvA

vvvi

vi oinoino −−


1

2

1211

22 Rv

ARRR

iR

iin

ov

inoino −==→=→===

Inverting Amplifierg

1Riv

Zin

inin == 0

1

2 =→−= outinout ZvRRV


in 1

Summing Amplifierg


Summing AmplifiergiA

iF

iB

++ BABA VViiiViVi

⎟⎟⎞

⎜⎜⎛

+→+−−

+=+===

BABAoo

B

B

A

ABAF

B

BB

A

AA

VVRVVvvi

RRiii

Ri

Ri

0


⎟⎟⎠

⎜⎜⎝

+−=→+===B

B

A

AFo

B

B

A

A

F

o

F

oF RR

RvRRRR

i

Summing Amplifierg

VRA

iA

−−

==

tt

B

BB

A

AA

vvRvi

Rvi

0

V = 0

VA

Vout

RfVB

iout

iB

+−

→+

==

BAout

F

out

F

outout

vvviii

Rv

Rv

i0

VBRB

⎟⎟⎞

⎜⎜⎛

+−=

+=→+=

BAF

BAFBAout

vvRv

RRRiii

⎟⎟⎠

⎜⎜⎝

+=BA

Fout RRRv


Summing Amplifierg

Input resistance seen by vA = RA

Input resistance seen by vB = RB

Since the output voltage does not depend on the load resistance


RL, the output impedance is zero.

Exercise 14.2(a)

Fi d i i i i dFind i1, i2, ix, io and vo:

111

11 mA

kV

Rv

i in =Ω

==

1011011

1

2

12

1

mAVkvRvi

mAii

o −=⎟⎞

⎜⎛ Ω

−=⎟⎟⎞

⎜⎜⎛−==

==

1011110

10111

22

2

kRmAmAmAiiiiii

mAVkk

vRRR

i

oxox

inLL

o

Ω

−=−−=−=→=+

=⎟⎠

⎜⎝ ΩΩ

=⎟⎟⎠

⎜⎜⎝

==


)1(1

1010)1)(10(2

2 Vkkv

RRVkmAv ino Ω

Ω−=−=−=Ω−=

Exercise 14.2(b)v

VvmAvi

imAkVi

550

515

2

21

−=→=−

=

==Ω

=

vvv

mAkV

kvi

VvmAk

i

5

515

10

551

3

2

=Ω

=Ω

−=

=→=Ω

=


Vvk

vk

vvmAmAmAiii o

oo 1515

11055324 −=→

Ω−−

=Ω

−==+=+=

Exercise 14.3


Exercise 14.3

20110

20 vvo −=

Inverting SummingInverting amplifier

Summingamplifier

22 201 vvv⎟⎞

⎜⎛ ⎞⎛

⎟⎞

⎜⎛


212

12 24

101020

10120

101020 vv

kvv

kk

kv

kv

kv oout −=⎟⎟

⎠

⎞⎜⎜⎝

⎛Ω

+⎟⎠⎞

⎜⎝⎛−

ΩΩ−=⎟

⎠

⎞⎜⎝

⎛Ω

+Ω

Ω−=

Positive FeedbackPositive Feedback

With positive feedback the opfeedback, the op amp’s input and output voltagesoutput voltages increase in magnitude until the outputuntil the output voltage reaches one of its extremes.


of its extremes.

Non-inverting Amplifierg

⎟⎞

⎜⎛+

=→= 10

RRRR

vvv ini

⎟⎟⎞

⎜⎜⎛

+

⎟⎟⎠

⎞⎜⎜⎝

⎛+=

+=→

+=

2

1

2

1

21

21

11

1

1

RvA

vRRv

RRRvv

RRRv

o

ininoo


⎟⎟⎠

⎜⎜⎝

+==1

21Rv

Ain

ov

Non-inverting Amplifierg

−0 vi in

−=

=1

vvi

Ri

outinout

inin

iin iout

Vin0 Vout

R1 R2 −=

−→=

2

Rvv

Rv

ii

Ri

outininoutin

out

+=212

21

Rv

Rv

Rv

RR

ininout

⎟⎟⎠

⎞⎜⎜⎝

⎛+=

1

2

212

1RRvv

RRR

inout


⎠⎝ 1

NONINVERTING AMPLIFIERS

Under the ideal-op-amp assumption, the p p ,noninverting amplifier is an ideal pvoltage amplifier having infinite inputg presistance and zero output resistance.

21RR

vvA o

v +==


p1in Rv

Voltage Followerg

1011 2 ++Rv

A o 1111

=∞

+=+==Rv

A vin


Exercise 14.4


Exercise 14.4

i1 = 0

i2 = 0

inv

+v

inv

+v

ininin vRivv =+=+

vo = vin

in

iiR

vvi

==

=−

= +

12

1

0

0


ininout vvRiv =+= 2

Exercise 14.4

0

0

inino vvRRv −=−=


Voltage-to-Current Converterg


Voltage-to-Current Converterg

vinin

F

ino R

vi =


F

Exercise 14.6

vin

i1i1

v2

i2

i3

1 =Rv

i in

2

212

211

21111212

11

1)(1

)()(

⎟⎟⎞

⎜⎜⎛ +

+⎥⎤

⎢⎡

++++

+=+=+=

RRvRR

vvvviii

RRRv

RRiRiRiv

R

inininin

in

21

21

1

221

1232

1

21

121

1111

2

1213

1)(

1)(

⎟⎟⎠

⎞⎜⎜⎝

⎛ ++++=+=

⎟⎟⎠

⎜⎜⎝

+=⎥⎦

⎢⎣

++=+=+=

RRR

RRv

RRRv

Rivv

RRRR

RRRRRiii

inino

inininin


2

1

2

1

2

1

2

1

2

1

2

1

2

1

2 311 ⎟⎟⎠

⎞⎜⎜⎝

⎛++=++++=

RR

RR

RR

RR

RR

RR

RR

vv

i

o

Design of Simple AmplifiersDesign of Simple Amplifiers

Amplifier design using op amps mainly consists of selecting a suitable circuitconfiguration and values for the feedback resistors.


If the resistances are too small, an impractical amount of current and power will be needed to operate the amplifier.


Very large resistance may be unstable in y g yvalue and lead to stray coupling of undesired signals.g


Example 14.3Want the voltage gain to be -10 ± 5 percent:

Varying resistance

%5102 ±Rvout Need R1>>RS so that variability in


%5101

2 ±=+

−=RRv Ss

out 1 S yRS is a small percentage change

Example 14.3

Choose R1 = 100Rs = 50kΩChoose R1 100Rs 50kΩ

R1+RS min = 50kΩ

R +R max = 50 5kΩR1+RS max = 50.5kΩ

%15.0505.50minmax ==Ω−Ω

=− kkRR


50501 ΩkR

Example 14.3

T t th i f 10 h R 10R 500kΩTo get the gain of 10, choose R2 ≈ 10R1 = 500kΩ

Since R1, RS, R2 can all vary, use 1% tolerance resistors:

R1 = 49.9kΩ ± 499Ω

R2 = 499kΩ ± 4.99kΩ


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