Download Tutorial OpAmps

OpAmps Circuits
Tutorial
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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Question 3
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Question 4
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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 have
an 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:
G (ω) =
1
ωRC
that is
G (200π) =
1
= 100
200πRC
τ = RC =
1
20000π
therefore
G (ω) =
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1
20000π
=
= 1V / V
ωRC
2πf
f = 10000 Hz
Question 5A
An op-amp-based inverting integrator is measured at 100Hz to have
an 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 same
factor. Therefore, when the gain decrease from 100V/V by factor
of 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 sine
wave signals have equal magnitude? What is the output signal for
for a 1-V p-p sine wave input with frequency equal to 10fo ?
Solution:
The transmission function of differentiator is given by:
G (ω) = −
for
G (ω0 ) = 0.0001ω0 = 1
when f=10fo
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R
= − jωRC = − jω ×10000 × 0.01×10 −6 = − j 0.0001ω
1
j ωC
ω0 = 10000
f 0 = 10000 / 2π
G (10 × 10000) = 0.0001 × 100000 = 10
Question 7
7.
A weighted summer circuit using an ideal op-amp has three inputs
using 100K resistors and a feedback resistor of 50K. A signal v1 is
connected to two of the inputs, while a signal v2 is connected to the
third. Express v0 in terms of v1 and v2. If v1=3V, v2=-3V, what is
v0?
Solution:
Rf 
Rf
 Rf
v3 
v2 +
v1 +
v0 = −
R3 
R2
 R1
50
50 
 50
= −
v1 +
v2 +
v1 
100
100 
 100
1 

= − v1 + v2  = −(3 − 1.5) = −1.5V
2 

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Question 8
1 

Design an op-amp circuit to provide an output v0 = − 3v1 + v2 
2 

Choose relatively low values of resistors but ones for which the
input current (for each source) does not exceed 0.1mA for 2-V
input signals.
Solution:
The input resistors can be determined as:
Rf
Rf
1
= 3 and
=
⇒ R2 = 6 R1
R1
R2 2
R1 =
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v1
2V
≥
= 20 KΩ
i1 0.1mA
R2 ≥
2V
= 120 KΩ,
0.1mA
R f = 60 KΩ
Question 9
9
For the difference amplifier use superposition to find v0 in terms
of the input voltages v1 and v2:
v1 = 10 sin( 2π × 60t ) − 0.1sin( 2π × 1000t ),Volts
v2 = 10 sin( 2π × 60t ) + 0.1sin( 2π × 1000t ), Volts
Solution:
Disabling v1, the circuit is a non-inverting amplifier, therefore
v02 = (1 +
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10 R
)v2 = 11v2
R
Question 9 (cont.)
9
Solution:
Disabling v2, the circuit is an inverting amplifier, therefore
v01 = −
10 R
v1 = −10v2
R
Therefore the total output should be:
v0 = v01 + v02 = 11v1 − 10v2
v0 = sin( 2π × 60t ) − 2.1sin( 2π × 1000t ), Volts
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