TYPICAL DEVICE USING OP-AMPShome.konkuk.ac.kr/~keechan/ch04.pdf ·  · 2018-03-26We have already...

19
OPERATIONAL AMPLIFIERS Why do we study them at this point??? 1. OpAmps are very useful electronic components 2. We have already the tools to analyze practical circuits using OpAmps 3. The linear models for OpAmps include dependent sources TYPICAL DEVICE USING OP-AMPS 1 OP-AMP ASSEMBLED ON PRINTED CIRCUIT BOARD APEX PA03 PIN OUT FOR LM324 DIMENSIONAL DIAGRAM LM 324 LM324 DIP LMC6294 MAX4240 Unit: inch 2

Transcript of TYPICAL DEVICE USING OP-AMPShome.konkuk.ac.kr/~keechan/ch04.pdf ·  · 2018-03-26We have already...

OPERATIONAL AMPLIFIERS

Why do we study them at this point???

1. OpAmps are very useful electronic components

2. We have already the tools to analyze practical circuitsusing OpAmps

3. The linear models for OpAmps include dependent sources

TYPICAL DEVICE USING OP-AMPS

1

OP-AMP ASSEMBLED ON PRINTED CIRCUIT BOARD APEX PA03

PIN OUT FOR LM324

DIMENSIONAL DIAGRAM LM 324

LM324 DIP

LMC6294

MAX4240

Unit: inch

2

CIRCUIT SYMBOL FOR AN OP-AMP SHOWING POWER SUPPLIES

LINEAR MODEL OUTPUT RESISTANCEINPUT RESISTANCE

GAIN75

125

1010:

501:

1010:

A

R

R

O

i

TYPICAL VALUES

3

CIRCUIT WITH OPERATIONAL AMPLIFIER

DRIVING CIRCUIT

LOAD

OP-AMP

4

TRANSFER PLOTS FOR SOME COMERCIAL OP-AMPS

SATURATIONREGIONLINEAR

REGION

IDENTIFY SATURATION REGIONSOP-AMP IN SATURATION

5

CIRCUIT AND MODEL FOR UNITY GAINBUFFER

0 inOOis VAIRIRV :KVL

0 inOO VAIRoutV- :KVL

IRV iin : VARIABLEGCONTROLLIN

iOO

is

out

RARRV

V

1

1

SOLVING

1S

outO V

VA

WHY UNIT GAIN BUFFER?

BUFFERGAIN

6

A

)(0 vvAvR OO

THE IDEAL OP-AMP

iR

ARR iO ,,0IDEAL

i

i

7

THE SCHMITT TRIGGER

CCVRR

R

21

1

CCVRR

R

21

1

CCV

CCV

CCVR

R

2

1

CCVR

R

2

1

CCV

CCV

8

THE UNITY GAIN BUFFER – IDEAL OP-AMP ASSUMPTION

svv

vv

OUTv v

OUT Sv v

OUTv

1OUT

S

v

v

USING LINEAR (NON-IDEAL) OP-AMP MODEL WE OBTAINED

1

1

out

is

O O i

VRV

R A R

PERFORMANCE OF REAL OP-AMPS

Op-Amp BUFFER GAINLM324 0.99999LMC6492 0.99998MAX4240 0.99995

IDEAL OP-AMP ASSUMPTION YIELDS EXCELLENT APPROXIMATION!

9

WHY USE THE VOLTAGE FOLLOWER OR UNITY GAIN BUFFER?

svv

vv

vvO

SO vv

SO vv

THE VOLTAGE FOLLOWER ACTS AS BUFFER AMPLIFIER.

THE SOURCE SUPPLIES POWER THE SOURCE SUPPLIES NO POWER

THE VOLTAGE FOLLOWER ISOLATES ONECIRCUIT FROM ANOTHER.ESPECIALLY USEFUL IF THE SOURCE HASVERY LITTLE POWER.

CONNECTION WITHOUT BUFFER CONNECTION WITH BUFFER

10

LEARNING EXAMPLE

s

out

V

VG GAIN THE DETERMINE

0v

0 vvvAo

0v

0 iiRi

000

21

R

V

R

V outs

- v@KCL APPLY

0i1

2

R

R

V

VG

s

out

FOR COMPARISON, NEXT WE EXAMINE THE SAME CIRCUIT WITHOUT THE ASSUMPTION OF IDEALOP-AMP.

11

REPLACING OP-AMPS BY THEIR LINEAR MODEL

WE USE THIS EXAMPLE TO DEVELOPA PROCEDURE TO DETERMINE OP-AMPCIRCUITS USING THE LINEAR MODELS

12

1. Identify Op Amp nodes

v

v

ov

2. Redraw the circuit cutting outthe Op Amp

v

v

ov

3. Draw components of linear OpAmp (on circuit of step 2)

v

v

oviR

OR

( )A v v

4. Redraw as needed

2R

v

v

13

INVERTING AMPLIFIER: ANALYSIS OF NON IDEAL CASE

NODE ANALYSIS

CONTROLLING VARIABLE IN TERMS OF NODEVOLTAGES

USE LINEAR ALGEBRA

10,10

,108

5

Oi RR

A :AMP-OPTYPICAL 9996994.45,1 21

S

O

v

vkRkR 000.5

S

O

v

vA

14

0i

0 iiRi

vvA

0v

0v

KCL @ INVERTING TERMINAL

000

21

R

v

R

v OS

1

2

R

R

v

v

s

O

GAIN FOR NON-IDEAL CASE

THE IDEAL OP-AMP ASSUMPTION PROVIDES EXCELLENT APPROXIMATION.(UNLESS FORCED OTHERWISE WE WILL ALWAYS USE IT!)

SUMMARY COMPARISON: IDEAL OP-AMP AND NON-IDEAL CASE

NON-IDEAL CASE

REPLACE OP-AMP BY LINEAR MODELSOLVE THE RESULTING CIRCUIT WITHDEPENDENT SOURCES

15

LEARNING EXAMPLE: DIFFERENTIAL AMPLIFIER

THE OP-AMP IS DEFINED BY ITS 3 NODES. HENCE IT NEEDS 3 EQUATIONS.

THINK NODES!

KCL AT V_ AND V+ YIELD TWO EQUATIONS(INFINITE INPUT RESISTANCE IMPLIES THAT i-, i+ ARE KNOWN)

OUTPUT CURRENT IS NOT KNOWN

DON’T USE KCL AT OUTPUT NODE. GET THIRD EQUATION FROM INFINITEGAIN ASSUMPTION (v+ = v-).

16

LEARNING EXAMPLE: DIFFERENTIAL AMPLIFIER

NODES @ INVERTING TERMINAL

NODES @ NON INVERTING TERMINAL

IDEAL OP-AMP CONDITIONS

243

42

43

40 vRR

Rvv

RR

Rvi

1

2

1

1

21

1

2

1

2 11 vvR

R

R

Rv

R

Rv

R

RvO

)(, 121

21324 vv

R

RvRRRR O

17

LEARNING EXAMPLE: USE IDEAL OP-AMP Ov FIND

1v

1v

2v

2v

1ov

2ov

1mv

2mv

6 NODE EQUATIONS + 2 IDEAL OP-AMP

22

11

vv

vv

22

11

m

m

vv

vv

FINISH WITH INPUT NODE EQUATIONS…USE INFINTE GAIN ASSUMPTION

2211

vvvv

2211vvvv

mm

USE REMAINING NODE EQUATIONS

00:@1

2121

2

011

1

R

vv

R

vv

R

vvv o

G

m

00:@2

212

1

22

2

R

v

R

vv

R

vvv

G

o

m

ONLY UNKWONS ARE OUTPUT NODE VOLTAGES

1 VARIABLE REQUIREDFOR SOLVE

oovv

18

LEARNING EXTENSION AMP-OPIDEAL ASSUMEI FIND O.

Vv 12

VvAO 12

0 iRi

Vv 12

02

12

12

12:

kk

Vv o KCL@ VVo 84

mAk

VI o

O 4.810

19

“inverse voltage divider”

ii vR

RRvv

RR

Rv

1

2100

21

1

0i

INFINITE INPUT RESISTANCE

NONINVERTING AMPLIFIER - IDEAL OP-AMP

LEARNING EXTENSION

v

_vov

SET VOLTAGE 1vv

INFINITE GAIN ASSUMPTION

11 vvvv

R2

R1

ivv

0v

20

FIND GAIN AND INPUT RESISTANCE - NON IDEAL OP-AMP

NOW RE-DRAW CIRCUIT TO ENHANCECLARITY. THERE ARE ONLY TWO LOOPS.

DETERMINE EQUIVALENT CIRCUITUSING LINEAR MODEL FOR OP-AMP

v

v

Ov

iR

OR

( )A v v

COMPLETE EQUIVALENTFOR MESH ANALYSIS

Ov

iv v v

MESH 1

MESH 2

CONTROLLNG VARIABLE IN TERMS OF LOOPCURRENTS

)( 21122 iiRiRvO 21

INPUT RESISTANCE

1

1

i

vRin

GAIN

i

O

v

vG

MESH 1

MESH 2

CONTROLLNG VARIABLE IN TERMS OF LOOPCURRENTS

MATHEMATICAL MODEL

)( 21122 iiRiRvO

REPLACE AND PUT IN MATRIX FORM

0)(

)( 1

2

1

211

121 v

i

i

RRRRAR

RRR

Oi

0)(

)( 11

211

121

2

1 v

RRRRAR

RRR

i

i

Oi

THE FORMAL SOLUTION

)())(( 112121 RARRRRRRR iO

)()(

)(

211

121

RRRAR

RRRRAdj

i

O

0)()(

)(1 1

211

121

2

1 v

RRRAR

RRRR

i

i

i

O

THE SOLUTIONS

121

1 vRRR

i O

)( 1

2RAR

i i

???A

1121

1211

22111

))(()(

)(

vRARRR

vRRRR

iRRiRv

iO

O

iRARA 1 inR

1

21

1 R

RR

v

vG O

22

A SEMI-IDEAL OP-AMP MODEL

This is an intermediate model, more accurate than the ideal op-ampModel, but simpler than the linear model used so far.

, 0,i O O

R R A A

Non-inverting amplifier and semi-ideal model

0i i !!v v

Replacement Equation

( )O O e O

v A v A v v

v

v

e inv v

2 1

1

(as before)

( ) (replaces )

S

O

O S O

v v

R Rv v

R

A v v v v v

1

! 2

;1

O O

S O

v A R

v A R R

23

Si

Ov

R

+-

Sv

Sample Problem

Find the expression for Vo. Indicatewhere and how you are using the IdealOpAmp assumptions.

Set voltages? Svv Use infinite gain assumption Svv v

Use infinite input resistance assumptionand apply KCL to inverting input

0

R

vvi oS

SSo Rivv

0i

v

24

Si

Ov

R

Sample Problem

1. Locate nodes

2. Erase Op-Amp

3. Place linear model

4. Redraw if necessary

+-

ov

v

v R

RO

RiiS

A(v+ - v-)

TWO LOOPS. ONE CURRENT SOURCE. USE MESHES.

DRAW THE LINEAR EQUIVALENT CIRCUIT AND WRITE THE LOOP EQUATIONS

1i2i

MESH 1 sii 1

MESH 2 )()()( _22 vvAiRRiiR OSi

CONTROLLING VARIABLE )( 12_ iiRvv i

iR

OR

( )A v v

25

SV

OV ANDGAIN FIND

SVv

SVv _

0i

“INVERSE VOLTAGE DIVIDER”OV

SV2R

1R

SO Vk

kkV

1

1100

101S

O

V

VG

VVmVV OS 101.01

LEARNING EXTENSION

26

LEARNING EXAMPLE UNDER IDEAL CONDITIONS BOTH CIRCUITS SATISFY

1 28 4

OV V V

1 2If 1 2 , 2 3

dc supplis are 10

V V V V V V

V

DETERMINE IF BOTH IMPLEMENTATIONS PRODUCE THEFULL RANGE FOR THE OUTPUT

1 22

XV V V

1 21 2 , 2 3 1 2

XV V V V V V V V V OK!

XV

4O X

V V

1 2 4 8X O

V V V V V V !O

V OK

18

YV V

11 2 16 8

YV V V V V V

EXCEEDS SUPPLY VALUE.THIS OP-AMP SATURATES!

POOR IMPLEMENTATION

27

COMPARATOR CIRCUITS

Some REAL OpAmps requirea “pull up resistor.”

ZERO-CROSSING DETECTOR28

LEARNING BY APPLICATION OP-AMP BASED AMMETER

NON-INVERTING AMPLIFIER

1

21R

RG

IRV II

IRR

RGVV IIO

1

21

29

1 4 (design eq.)B

A

R

R

LEARNING EXAMPLE DC MOTOR CONTROL - REVISITED

CHOOSE NON-INVERTING AMPLIFIER (WITH POWEROP-AMP PA03)

Constraints: 20M

V VPower dissipationin amplifier 100mW

Simplifying assumptions: , 0i O

R R Significant power lossesOccur only in Ra, Rb

Worst case occurs when Vm=20

2(20 )100

MX

A B

VP mW

R R

4000

A BR R

3B

A

R

R

One solution: 3 , 1B A

R k R k

Standard values at 5%!

30

DESIGN EXAMPLE: INSTRUMENTATION AMPLIFIER

DESIGN SPECIFICATIONS

1 2

10OV

GV V

• “HIGH INPUT RESISTENCE”• “LOW POWER DISSIPATION”• OPERATE FROM 2 AA BATTERIES

ANALISIS OF PROPOSED CONFIGURATION

O X YV V V

1 2 1

1

@ : 0XV V V V

AR R

2 1 2

2

@ : 0YV V V V

BR R

1 2 2 1

1 2 1 21 1

O

R R R RV V V V V

R R R R

SIMPLIFY DESIGN BY MAKING 1 2R R

1

1 2

21 ( )

O

RV V V

R DESIGN EQUATION: 1

2 9R R

OV1

V

2V

G

USE LARGE RESISTORS FOR LOW POWER 1 2. ., 100 , 450e g R k R R k

1 2;

A BV V V V Infinite gain

MAX4240

31

DESIGN EXAMPLEDESIGN SPECIFICATION 10O

in

V

V

Power loss in resistors should not exceed

100mW 2 2in

V V V when

Design equationS:2

1

1 10O

in

V R

V R

1 2

2

1 2

(20 )100

R R

VP mW

R R

Max Vo is 20V

Solve design equations (by trial and error if necessary)

2 19R R

1 24R R k

1400R

32

DESIGN EXAMPLE IMPLEMENT THE OPERATION 1 20.9 0.1

OV V V

DESIGN CONSTRAINTS• AS FEW COMPONENTS AS POSSIBLE• MINIMIZE POWER DISSIPATED• USE RESISTORS NO LARGER THAN 10K

Given the function (weighted sumwith sign change) a basic weighted adder may work

ANALYSIS OF POSSIBLE SOLUTION

1 2

1 21

0

@ : 0O

V

V V VV

R R R

1 2

1 2

O

R RV V V

R R

2 1

2

9

0.1

R R

R

R

DESIGNEQUATIONS

2 1R R R

SOLVE DESIGN EQUATIONS USING TRIAL AND ERROR IF NECESSARY

210 ,5.6 ,...R k k

ANALYZE EACH SOLUTION FOR OTHER CONSTRAINTS AND FACTORS; e.g.DO WE USE ONLY STANDARD COMPONENTS?

33

DESIGN EXAMPLE DESIGN 4-20mA TO 0 – 5V CONVERTER

1. CONVERT CURRENT TO VOLTAGE USING A RESISTOR

I IV RICANNOT GIVE DESIRED RANGE!

2. CHOOSE RESISTOR TO PROVIDE THE 5V CHANGE … AND SHIFT LEVELS DOWN!

5 0312.5

0.020 0.004MAX MIN

MAX MIN

V VR

I I

1.25 6.25I

V MUST SHIFT DOWN BY 1.25V(SUBSTRACT 1.25 V)

2

1 2

I

RV V

R R 1

1 2

( )O SHIFT SHIFT

RV V V V

R R

2

1

O I SHIFT

RV V V

R V V

34

DOES NOT LOAD PHONOGRAPH

1000 OFTION AMPLIFICA ANPROVIDES IT

THATSO DETERMINE 12,RR

LEARNING BY DESIGN

)1)(1(1

2

1 R

R

V

VO

35

LEARNING EXAMPLE

UNITY GAINBUFFER

COMPARATOR CIRCUITS

TT eR 0227.045.57

ONLY ONE LEDIS ON AT ANYGIVEN TIME

36

MATLAB SIMULATION OF TEMPERATURE SENSOR

WE SHOW THE SEQUENCE OF MATLAB INSTRUCTIONS USED TO OBTAIN THE PLOTOF THE VOLTAGE AS FUNCTION OF THE TEMPERATURE

»T=[60:0.1:90]'; %define a column array of temperature values» RT=57.45*exp(-0.0227*T); %model of thermistor» RX=9.32; %computed resistance needed for voltage divider» VT=3*RX./(RX+RT); %voltage divider equation. Notice “./” to create output array» plot(T,VT, ‘mo’); %basic plotting instruction» title('OUTPUT OF TEMPERATURE SENSOR'); %proper graph labeling tools» xlabel('TEMPERATURE(DEG. FARENHEIT)')» ylabel('VOLTS')» legend('VOLTAGE V_T')

37

1vv

0i

1vv

KCL @ v_

EXAMPLE OF TRANSFER CURVE SHOWING SATURATION

THE TRANSFER CURVE

IN LINEAR RANGE

THIS SOURCE CREATES THE OFFSET

OFFSET

OUTPUT CANNOTEXCEED SUPPLY(10V)

38