tutorial opamps
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OpAmps Circuits Tutorial
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Question 1
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Q1- Solution
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Q1 - Solution
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Question 2
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Q2 - Solution
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Q2 - Solution
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Q2 - Solution
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Q2 - Solution
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Question 3
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Q3 - Solution
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Q3 - Solution
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Q3 - Solution
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Q3 - Solution
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Q3 - Solution
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Question 4
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Q4 - Solution
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Q4 - Solution
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Q4 - Solution
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Q4 - Solution
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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=
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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.
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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
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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
−=−−=
+−=
++−=
++−=
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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
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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 =+=
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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=