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Lesson 3: Operational Amplifier Circuits in

Analog Control ET 438a

Automatic Control Systems Technology

lesson3et438a.pptx 1

LEARNING OBJECTIVES

lesson3et438a.pptx 2

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.

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Ideal OP AMP Characteristics

lesson3et438a.pptx 3

Zout =0 Zero output resistance

Infinite Zin

Infinite voltage gain, Av

Bandwidth Infinite Gain constant for all f

No offset voltage, Vo=0 with Vd=0

Iin

Instant recovery from saturation

Iin=0 due to infinite Zin

Non-Ideal OP AMPs

lesson3et438a.pptx 4

OP AMPs are voltage amplifiers designed originally for use in analog computers

Schematic symbol for non-ideal OP AMP

Two inputs: V1 = inverting input V2 = non inverting input

OP AMPs are direct coupled (dc) amplifiers that amplify both ac and dc signals simultaneously. Requires bipolar supplies.

+

- V1

V2

Vo

+V

-V

V2>0, Vo>0

+Vin -V0

+V0 +Vin

V1>0, Vo

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Fundamental OP AMP Circuits

lesson3et438a.pptx 5

Inverting Voltage Amplifier Non-Inverting Voltage Amplifier

AV R

R

f

in

Vo

V

R

Ri

f

in

Vo Limited by saturation Large Av causes Vo = ±V

V 1 + o

V

R

Ri

f

in

A 1 + V

R

R

f

in

Rin = Rin of OP AMP Av has minimum value of 1

Voltage Follower Circuit

lesson3et438a.pptx 6

Voltage follower (Impedance buffer) circuit used to reduce circuit loading. (Has a high Zin and low Zout)

Zo Zin

+V

-V

Characteristics Practical Circuit (LM741) Ideal Av = 1 Av = 1 Zin = 1 MW Zin = infinite Zo = 10 W Zo = 0

Circuit causes minimum loading on previous stage

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Example 3-1 Buffered Voltage Divider Circuit

lesson3et438a.pptx 7

Voltage divider formula only valid for infinite load resistance

5k = RL

Find Vo under load

10k

5k

+12 Vdc

Vo

V 2.4V 12k 10k 5.2

k 5.2V

k 5.2k 5 ||k 5 k 5 ||R

resistor loadWith

V 0.4V

V 12k 10k 5

k 5V

V load No

oL

L

0

o

o

WW

W

WWWW

WW

W

ANS

Example 3-1 Buffered Voltage Divider Circuit (1)

lesson3et438a.pptx 8

Add impedance buffer

+12 Vdc

5k 5k = RL

10k

Vo

Find Vo with load and OP AMP buffer Assume LM741 with Ri=1 MW and Ro =10 W AV= 1 so Vin = Vo

V 987.3VV

V 3.987V 12k 10 4975

4975V

4975k 5M 1

k 5M 1R

RM 1 ||k 5 M 1 ||R

resistor loadWith

oin

in

eq

eqL

WW

W

WWW

WW

WWW

Rin

Ro

ANS

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Electronic Addition and Subtraction

lesson3et438a.pptx 9

Inverting Summing Amplifier

Ideal circuit

Find total output using superposition

Gain v1 R

Rf

1

Gain v2 R

Rf

2

Gain v3 R

Rf

3

Total output v Rv

R

v

R

v

Rf0

1

1

2

2

3

3

Output is inverted sum of v1, v2, and v3

Electronic Addition and Subtraction

lesson3et438a.pptx 10

Improved circuit (non-ideal OP AMP)

Rc = R1 || R2 || R3 || Rf

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

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Electronic Addition and Subtraction

lesson3et438a.pptx 11

Non-inverting Summing Amp

Assuming R1 = R2 = R3

3

vvv

R

R1v 321f0

For any number, n, inputs.

Assuming R1=R2=R3=...Rn

n

v.....vv

R

R1v n21f0

n input average

Example 3-2 Non-inverting Averager

lesson3et438a.pptx 12

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

3

vvv

R

R1v 321f0

5667.53

8.037.05.0

k1

k91v0

ANS

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Example 3-3:Inverting Amplifier

lesson3et438a.pptx 13

For the circuit shown find Vo with Vin = 2 Vdc

5k

10k

Example 3-3: Solution (2)

lesson3et438a.pptx 14

5k

2.5k

Let Rf= 2.5 kW and find Av and Vo for Vi= 2.0 Vdc

Reduces input voltage

Output is smaller than input. Circuit divides input by 2 (0.5 = ½)

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Example 3-4 Non-Inverting Amp (1)

lesson3et438a.pptx 15

50k 100k +15

-15

Find Vo and Av given values of R and Vi = -1 Vdc. Assume non-ideal OP AMP with power supply values of ±15 Vdc

No sign change

Note: Rin of non-inverting OP AMP is infinite (Ideally). Circuit will not load previous stage significantly

Rin

Example 3-4 Solution (2)

lesson3et438a.pptx 16

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

This value can’t be achieved since the OP AMP saturates between 13-15 Vdc. Power supplies limit output. Ac signals distorted (clipping)

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Example 3-5 Inverting Summing Amplifier

lesson3et438a.pptx 17

Find Vo given v1 = 0.2 Vdc v2 = 0.1 Vdc v3 = -0.1 Vdc 25k

15k

10k

100k

-15

+15

Gains

Summing Amplifier Applications

lesson3et438a.pptx 18

Letting R1, R2 and R3 be potentiometers produces an audio

mixer

When R1=R2=R3 Output voltage is the average of the input

values

Vo

V3

V2

R2

R3

R4

V1

-15V

15V

+

R1

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Example 3-6: Averaging Sensor Signals

lesson3et438a.pptx 19

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.

50k

50k

50k

Example 3-6 Solution (1)

lesson3et438a.pptx 20

To average let R1=R2=R3=50 kW

3

3

2

2

1

1f0

R

v

R

v

R

vRVSumming equation

Since R1=R2=R3

Find relationship for average with 3 inputs and gain of -1

3

vvv1V 321ave

3213211

faveo vvv

3

1vvv

R

RVV

3211

f

1

3

1

2

1

1f0 vvv

R

R

R

v

R

v

R

vRV

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Example 3-6 Solution (2)

lesson3et438a.pptx 21

3

RR

3

1

R

R

1f

1

f

Complete Algebra to find value of Rf

Make equation more general by letting n be the number of inputs and Av be the desired gain factor.

n211

f

1

n

1

2

1

1f0 v ...vv

R

R

R

v ...

R

v

R

vRV

n

v ...vvAV n21vave

lesson3et438a.pptx 22

Example 3-6 Solution (3)

n21v

n21

1

faveo v ...vv

n

Av ...vv

R

RVV

Equate the OP AMP output and the average formula

n

RAR

n

A

R

R

1vf

v

1

f

Use this formula

Find sensor output voltages using temperature and gain value

V 5.0F 50mV/F 10V F 50T 11

V 45.0F 45mV/F 10V F 45T 22

V 40.0F 40mV/F 10V F 40T 33

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Example 3-6 Solution (4)

lesson3et438a.pptx 23

Find Rf and Vo for a gain of -1 and R1=50 kW, n3

WW

670,163

000,501

n

RAR 1vf

V 45.040.045.050.0k50

k67.16vvv

R

R321

1

f

45.0

3

40.045.050.01Vave

Use average formula to check output

ANS

ANS

Checks

Example 3-6 Solution (5)

lesson3et438a.pptx 24

Find Rf and Vo for a gain of -5 and R1=50 kW, n=3

25.2

3

40.045.050.05V

V 25.240.045.050.0k50

k33.83vvv

R

RV

330,833

000,505

n

RAR

ave

321

1

fo

1vf

WW

ANS

ANS

Note: both values of Rf are not standard values. Use potentiometer and closest standard value then calibrate circuit to get desired output

82 k 5 kPractical Rf

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Differential Voltage Amplifier

lesson3et438a.pptx 25

Input/output Formula

To simplify let R1 = R3 R2 = R4

V (V Vo 2 1

R

R2

1)

Amplifiers the difference between + - terminals

Differential Voltage Amplifier Circuit

1

1

22

1

4

34

12o V

R

RV

R

R

RR

RRV

Differential Amplifier Applications

lesson3et438a.pptx 26

Can use voltage differential amp to generate an error signal

r

m

-

+ e

Block diagram

r

e m

setpoint

Measured value

error

e=r-m

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End Lesson 3: Operational Amplifier Circuits in Analog Control

ET 438a Automatic Control Systems Technology

lesson3et438a.pptx 27