CIRCUITS WITH OPERATIONAL AMPLIFIERSWhy 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

COMMERCIAL PACKAGING OF TYPICAL OP-AMP

OP-AMP ASSEMBLED ON PRINTED CIRCUIT BOARD

LMC 6294 DIP

PIN OUT FOR LM324

DIMENSIONAL DIAGRAM LM 324

CIRCUIT SYMBOLFOR AN OP-AMP

LINEAR MODEL OUTPUT RESISTANCE

INPUT RESISTANCE

GAIN

75

125

1010:

501:

1010:

A

R

R

O

i

TYPICAL VALUES

CIRCUIT WITH OPERATIONAL AMPLIFIER

DRIVING CIRCUIT

LOAD

OP-AMP

MANUFACTURER PART No A Ri[MOhm] Ro[Ohm]National LM324 100,000 1 20National LMC6492 50,000 10 150Maxim MAX4240 20,000 45 160

COMMERCIAL OP-AMPS AND THEIR MODEL VALUES

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

PERFORMANCE OF REAL OP-AMPSOp-Amp BUFFER GAINLM324 0.99999LMC6492 0.9998MAX4240 0.99995

A

)(0 vvAvR OO

THE IDEAL OP-AMP

iR

ARR iO ,,0IDEAL

i

i

THE VOLTAGE FOLLOWER OR UNITY GAIN BUFFER

svv

vv

vvO

SO vv

CONNECTION WITHOUT BUFFER CONNECTION WITH BUFFER

SO vv

THE VOLTAGE FOLLOWER ACTS AS BUFFER AMPLIFIER

THE SOURCE SUPPLIES POWERTHE SOURCE SUPPLIES NO POWER

THE VOLTAGE FOLLOWER ISOLATES ONECIRCUIT FROM ANOTHERESPECIALLY USEFUL IF THE SOURCE HASVERY LITTLE POWER

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

0i

1

2

R

R

V

VG

s

out

NEXT WE EXAMINE THE SAME CIRCUITWITHOUT THE ASSUMPTION OF IDEALOP-AMP

REPLACING OP-AMPS BY THEIR LINEAR MODELLABEL THE NODES FOR TRACKING

DRAW THE LINEAR EQUIVALENT FOR OP-AMP

CONNECT THE EXTERNAL COMPONENTS

b - a

b - d

REDRAW CIRCUIT FOR INCREASED CLARITY

2R

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

0i

0 iiRi

vvA

0v

0v

KCL @ INVERTING TERMINAL

000

21

R

v

R

v OS

1

2

R

R

v

v

s

O

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

GAIN FOR NON-IDEAL CASE

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

IDEAL OP-AMP ASSUMPTIONSNON-IDEAL CASE

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

LEARNING EXAMPLE: DIFFERENTIAL AMPLIFIER

KCL @ INVERTING TERMINAL

KCL @ 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

LEARNING EXAMPLE: USE IDEAL OP-AMP

Ov FIND

2v

Which voltages are set? 2211 , vvvv What voltages are also known due to infinite gain assumption?

1v

2v

Use now the infinite resistance assumption

0i

CAUTION: There could be currents flowing OUT of the OpAmps

1v

2v

av

THE CIRCUIT REDUCES TO THIS determined be to aO vv ,

KCL@v1

KCL@v2

Solve for vo

11 vvvv INFINITE GAIN ASSUMPTION

0i

“inverse voltage divider”

ii vR

RRvv

RR

Rv

1

2100

21

1

INFINITE INPUT RESISTANCE

NONINVERTING AMPLIFIER - IDEAL OP-AMP

SET VOLTAGE 1vv

R 2

R 1

ivv

0vLEARNING EXTENSION

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 oO 4.8

10

FIND GAIN AND INPUT RESISTANCE - NON IDEAL OP-AMP

iR

v

v

OvOR

)( vvA

THE OP-AMP

iR

v

v

OvOR

)( vvA1v

ADD THE INPUT VOLTAGE SOURCE

iR

v

v

OvOR

)( vvA1v

1R

2R

AND THE EXTERNAL RESISTORS

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

DETERMINE EQUIVALENT CIRCUITUSING LINEAR MODEL FOR OP-AMP

iR

v

v

OvOR

)( vvA1v

1R

2R

COMPLETE EQUIVALENT CIRCUIT

NON-INVERTING OP-AMP. NON IDEAL OP-AMP

MESH 1

MESH 2

CONTROLLNG VARIABLE IN TERMS OF LOOPCURRENTS

COMPLETE EQUIVALENTFOR MESH ANALYSIS

Ov

)( 21122 iiRiRvO

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

)( 12

RARi i

???A

1121

1211

22111

))(()(

)(

vRARRR

vRRRR

iRRiRv

iO

O

iRARA 1 inR

1

21

1 R

RR

v

vG O

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

Si

Ov

R

Sample Problem

1. Locate nodes

2. Place the nodes in linear circuit model

3. Place linear model

+-

ov

v

v

R i

R O

4. Place remaining components

+-

ov

v

v R

R O

R iiS

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

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

COMPARATOR CIRCUITS

Some REAL OpAmps requirea “pull up resistor.”

ZERO-CROSSING DETECTOR

LEARNING BY APPLICATION OP-AMP BASED AMMETER

NON-INVERTING AMPLIFIER

1

21R

RG

IRV II

IRR

RGVV IIO

1

21

DOES NOT LOAD PHONOGRAPH

1000 OF TION AMPLIFICA ANPROVIDES IT

THAT SO DETERMINE 12 ,RR

LEARNING BY DESIGN

)1)(1(1

2

1 R

R

V

VO

LEARNING EXAMPLE

UNITY GAINBUFFER

COMPARATOR CIRCUITS

TT eR 0227.045.57

ONLY ONE LEDIS ON AT ANYGIVEN TIME

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')

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)