Tutorial OpAmps

OpAmps Circuits Tutorial

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Question 1


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Question 2


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Question 3


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Question 4


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Question 5

An op-amp-based inverting integrator is measured at 100Hz to havean voltage gain of -100V/V. At what frequency is its gain reduced to-1V/V? What is the integrator time constant?

Solution:

The gain is given by:

that is

therefore

RCG

ω=ω 1)(

100200

1)200( =π

=πRC

Gπ

==τ20000

1RC

VVfRC

G /12

200001)( =π

π=ω

=ω Hzf 10000=


Question 5A

An op-amp-based inverting integrator is measured at 100Hz to havean voltage gain of -100V/V. At what frequency is its gain reduced to-1V/V? What is the integrator time constant?

Solution:

For integrator, the gain decays 20dB/decades. That is, when frequency increase by a factor, the gain decreases by the samefactor. Therefore, when the gain decrease from 100V/V by factorof 1/100 to -1V/V, the frequency should increase by 100 times. That is, at 10000Hz the gain will reduced to -1V/V.


Question 6

A differentiator uses an ideal op-amp, a 10K resistor, and a 0.01uf capacitor. What is the frequency fo at which its input and output sinewave signals have equal magnitude? What is the output signal forfor a 1-V p-p sine wave input with frequency equal to 10fo ?

Solution:The transmission function of differentiator is given by:

for

when f=10fo

ω−=×××ω−=ω−=

ω

−=ω − 0001.01001.0100001)( 6 jjRCj

Cj

RG

10001.0)( 00 =ω=ωG 100000 =ω π= 2/100000f

101000000001.0)1000010( =×=×G


Question 7

7. A weighted summer circuit using an ideal op-amp has three inputsusing 100K resistors and a feedback resistor of 50K. A signal v1 isconnected to two of the inputs, while a signal v2 is connected to thethird. Express v0 in terms of v1 and v2. If v1=3V, v2=-3V, what isv0?

Solution:

Vvv

vvv

vRR

vRR

vRR

v fff

5.1)5.13(21

10050

10050

10050

21

121

33

22

11

0

−=−−=

+−=

++−=

++−=


Question 8

Design an op-amp circuit to provide an output

Choose relatively low values of resistors but ones for which theinput current (for each source) does not exceed 0.1mA for 2-V input signals.

Solution:

The input resistors can be determined as:

+−= 210 213 vvv

1221

6213 RR

RR

andRR ff =⇒==

Ω=≥= KmAV

ivR 20

1.02

1

11 Ω=Ω=≥ KRK

mAVR f 60,120

1.02

2


Question 9

9For the difference amplifier use superposition to find v0 in termsof the input voltages v1 and v2:

Voltsttv ),10002sin(1.0)602sin(101 ×π−×π=

Voltsttv ),10002sin(1.0)602sin(102 ×π+×π=

Solution:

Disabling v1, the circuit is a non-inverting amplifier, therefore

2202 11)101( vvRRv =+=


Question 9 (cont.)

9

Solution:

Disabling v2, the circuit is an inverting amplifier, therefore

2101 1010 vvRRv −=−=

Therefore the total output should be:

2102010 1011 vvvvv −=+=

Voltsttv ),10002sin(1.2)602sin(0 ×π−×π=

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