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Lesson 3: Operational
Amplifier Circuits in
Analog Control
ET 438a
Automatic Control Systems Technology
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After this presentation you will be able to:




List the characteristics of an ideal Operational Amplifier (OP
AMP) circuit.
Identify and utilize fundamental OP AMP circuits to amplify
and signals.
Use OP AMP circuits to reduce inter-stage loading effects in
sensor circuits.
Average sensor signals using OP AMP circuits.
LEARNING OBJECTIVES
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Zout =0
Zero output resistance
No offset voltage,
Vo=0 with Vd=0
Iin
Infinite
Zin
Infinite voltage gain,
Av
Bandwidth Infinite
Gain constant for all f
Instant recovery
from saturation
Iin=0 due to
infinite Zin
Ideal OP AMP Characteristics
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OP AMPs are voltage amplifiers designed originally for use in
analog computers
OP AMPs are direct coupled (dc) amplifiers that amplify both ac and
dc signals simultaneously. Requires bipolar supplies.
+V
+Vin V1
Vo
+Vin V2
+
-V0
V1>0, Vo<0
+V0
V2>0, Vo>0
-V
Schematic symbol for non-ideal OP AMP
Two inputs:
V1 = inverting input
V2 = non inverting input
Non-Ideal OP AMPs
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Inverting Voltage Amplifier
AV 
R f
Rin
 R 
Vo  Vi  f 
 Rin 
Vo Limited by saturation
Large Av causes Vo = ±V
Non-Inverting Voltage Amplifier

Rf 
A V  1 +

Rin 


Rf 
Vo  Vi  1 +

Rin 

Rin = Rin of OP AMP
Av has minimum value of 1
Fundamental OP AMP Circuits
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Voltage follower (Impedance buffer) circuit used to reduce circuit
loading. (Has a high Zin and low Zout)
+V
Circuit causes
minimum
loading on
previous
stage
Zin
Zo
-V
Characteristics
Practical Circuit (LM741)
Av = 1
Zin = 1 MW
Zo = 10 W
Ideal
Av = 1
Zin = infinite
Zo = 0
Voltage Follower Circuit
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Voltage divider formula only valid for infinite load resistance
Find Vo under load
+12 Vdc
10k
No load Vo
5k = RL
Vo
5k
5 kW


Vo  
 12 V
 5 kW  10 kW 
V0  4.0 V
With load resistor
R L || 5 kW  5 kW || 5 kW  2.5 kW
2.5 kW


VoL  
 12 V  2.4 V
2
.
5
k
W  10 kW 

ANS
Example 3-1 Buffered Voltage Divider
Circuit
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Add impedance buffer
+12 Vdc
10k
5k
Rin
Ro
5k = RL
7
Find Vo with load and OP AMP buffer
Assume LM741 with Ri=1 MW and
Ro =10 W AV= 1 so Vin = Vo
With load resistor
R L || 1 MW  5 kW || 1 MW  R eq
Vo
1 MW  5 kW
 4975 W
1 MW  5 kW
4975 W


Vin  
 12 V  3.987 V
4975
W

10
k
W


ANS
Vin  Vo  3.987 V
R eq 
Example 3-1 Buffered Voltage
Divider Circuit (1)
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Inverting Summing Amplifier
Find total output using
superposition
Ideal
circuit
Gain v1
Rf
R1
Gain v2
Rf
R2
Gain v3
Total output
Rf
R3
v
v
v 
v 0  Rf  1  2  3 
 R1 R2 R3 
Output is inverted sum of v1, v2, and v3
Electronic Addition and Subtraction
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Improved circuit (non-ideal OP AMP)
Practical OP AMP chips require
bias currents to operate.
Unequal R values at each input
cause voltage differences that
produce output errors.
Bias
compensation R
Rc = R1 || R2 || R3 || Rf
Electronic Addition and Subtraction
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Non-inverting Summing Amp
Assuming R1 = R2 = R3
 R  v  v 2  v3 
v 0  1  f  1

R 
3


For any number, n, inputs.
 R  v  v 2  .....  v n  n input
v 0  1  f  1
 average
R 
n


Assuming R1=R2=R3=...Rn
Electronic Addition and Subtraction
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For circuit shown n=3
R1 = R2 = R3=56k
Rf = 9k R = 1k
V1 = 0.5 Vdc
V2 = 0.37 Vdc
V3 = 0.8 Vdc Find Vo
 R  v  v 2  v3 
v 0  1  f  1

R 
3


 9k  0.5  0.37  0.8 
v 0  1  
  5.5667
3
 1k 

ANS
Example 3-2 Non-inverting Averager
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For the circuit shown find Vo with Vin = 2 Vdc
10k
5k
Example 3-3:Inverting Amplifier
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2.5k
Let Rf= 2.5 kW and find
Av and Vo for Vi= 2.0 Vdc
5k
Reduces input voltage
Output is smaller than input. Circuit divides input by 2
(0.5 = ½)
Example 3-3: Solution (2)
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Find Vo and Av given values of R and Vi = -1 Vdc. Assume nonideal OP AMP with power supply values of ±15 Vdc
50k
100k
+15
Rin
No sign change
-15
Note: Rin of non-inverting OP
AMP is infinite (Ideally). Circuit
will not load previous stage
significantly
Example 3-4 Non-Inverting Amp (1)
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50k
100k
+15
15
Vin rises to 6 Vdc. What is
Vo? Assume a non-ideal OP
AMP with given power supply
values of ±15 Vdc
-15
This value can’t be achieved since the OP AMP saturates
between 13-15 Vdc. Power supplies limit output. Ac signals
distorted (clipping)
Example 3-4 Solution (2)
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Find Vo given v1 = 0.2 Vdc v2 = 0.1 Vdc
v3 = -0.1 Vdc
25k
15k
100k
+15
10k
-15
Gains
Example 3-5 Inverting Summing
Amplifier
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Letting R1, R2 and R3
be potentiometers
produces an audio
mixer
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R1
V1
R2
V2
R3
15V
Vo
+
V3
R4
-15V
When R1=R2=R3
Output voltage is the
average of the input
values
Summing Amplifier Applications
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50k
50k
50k
LM34 - temperature sensors. Gain = 10 mV/F
T1 = 50 F T2 = 45 F T3= 40 F
Average the temperature using a gain of -1 and -5. Find the
value of Rf and Vo for each gain value.
Example 3-6: Averaging Sensor Signals
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To average let R1=R2=R3=50 kW
Summing equation
Since R1=R2=R3
v
v 
v
V0  R f  1  2  3 
 R1 R 2 R 3 
v
R 
v 
v
V0  R f  1  2  3    f v1  v 2  v3 
R
R
R
1
1
 1
 R1 
Find relationship for average with 3 inputs and gain of -1
 v  v 2  v3 
Vave  1  1

3


R 
1
Vo  Vave   f v1  v 2  v3    v1  v 2  v3 
 3
 R1 
Example 3-6 Solution (1)
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Complete Algebra to
find value of Rf
R 
1
  f    
3
 R1 
R
Rf  1
3
Make equation more general by letting n be the number of inputs
and Av be the desired gain factor.
v
R 
v
v 
V0  R f  1  2 ...  n    f v1  v 2 ...  v n 
R1 
 R1 R1
 R1 
 v  v 2 ...  v n 
Vave  A v   1

n


Example 3-6 Solution (2)
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Equate the OP AMP output and the average formula
R 
A 
Vo  Vave   f v1  v 2 ...  v n    v v1  v 2 ...  v n 
 n 
 R1 
R 
A 
  f    v 
 n 
 R1 
A R
Rf  v 1
n
Use this formula
Find sensor output voltages using temperature and gain value
T1  50 F V1  10 mV/F 50 F  0.5 V
T2  45 F V2  10 mV/F 45 F  0.45 V
 
Example 3-6 Solution (3)
T3  40 F V3  10 mV/F  40 F  0.40 V
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Find Rf and Vo for a gain of -1 and R1=50 kW, n3
Rf 
A v  R1 1 50,000 W

 16,670 W
n
3
ANS
R 
 16.67k 
 f v1  v 2  v3   
0.50  0.45  0.40  0.45 V
 50k 
 R1 
Use average
formula to
check output
 0.50  0.45  0.40 
Vave  1 
  0.45
3


ANS
Checks
Example 3-6 Solution (4)
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Find Rf and Vo for a gain of -5 and R1=50 kW, n=3
A v  R 1 5  50,000 W
ANS

 83,330 W
n
3
R 
 83.33k 
Vo   f v1  v 2  v 3   
0.50  0.45  0.40  2.25 V
 50k 
 R1 
Rf 
ANS
 0.50  0.45  0.40 
Vave  5  
  2.25
3


Note: both values of Rf are not standard values. Use potentiometer
and closest standard value then calibrate circuit to get desired output
Practical Rf
82 k
5k
Example 3-6 Solution (5)
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Differential Voltage Amplifier Circuit
Input/output Formula
 R  R1   R 4 
R 
  
  V2   2   V1
Vo   2
 R1 
 R 4  R 3   R1 
To simplify let
R1 = R3 R2 = R4
R 
Vo   2  (V2  V1)
 R1 
Amplifiers the difference between
+ - terminals
Differential Voltage Amplifier
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Can use voltage differential amp to generate an error
signal
Measured
value
e
r +
m
Block diagram
m
e
r
error
setpoint
e=r-m
Differential Amplifier Applications
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End Lesson 3: Operational Amplifier
Circuits in Analog Control
ET 438a
Automatic Control Systems Technology
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