1 9. design of multiple-input op. amp. circuits 9.1operational amplifiersoperational amplifiers...

1

9 . Design of Multiple-input Op. Amp. Circuits

9.1 Operational Amplifiers9.2 Basic Operational Amplifier Circuits9.3 Input resistance of Op. Amp. Amplifiers9.4 Multiple-Input Operational Amplifier Circuit9.5 Design of Multiple-Input Amplifiers

EE3601-09 Electronics Circuit Design

9.1 Operational Amplifiers

output

+Vcc

-Vcc

+Vcc

-Vcc

Inverting Input (-)

Non-InvInput (+)

Inverting Input (-)

output

Op. Amp. input Impedance Very high

Differential Amplifier

Emitter Constant current source

Collector Constant current source

Output Impedance very low

Push-pull Emitter follower at the output

Non Inverting Input (+)

Therefore Operational Amplifier has

(1) very high input impedance, Ri= (or) ii = 0

(2) very high voltage gain Av = 105 (or) v+ = v-

(3) very low output impedance R0 = 0

Amplifier voltage gain very high

+

-ii = 0

ii = 0

v+ = v-

R0= 0


1. INVERTING AMPLIFIER

aF

aO

VFo

aa

Fai

Fo

Fo

Faa

aa

a

RR

VVAR

VRVorII0I

RV

RVvIandR

VR

vVI

0vv

2. NON-INVERTING AMPLIFIER

aF

aoV

aF

ao

aoa

aF

Foa

aaFa

Foa

FoF

aa

aa

a

RR

VVAR

RVV

VVV

RR

RVV

RVIIBut

RVV

RVvIandR

VR

vI

vvV

11

0

vo

Ra

Va=

RF

v+

v-

Ia

IF

vo

RF

v+v-

Va

Ra Ia

IF

9.2 Basic Operational Amplifier Circuits


3. VOLTAGE FOLLOWER

4. INVERTING SUMMING AMPLIFIER

vo

RaVa

RF

v+

v-

IaIF

VbRb

VcRc

IbIc

voVa=v+

v- 1VVA

VVvv

aoV

ao

cc

bb

aa

FO

cc

bb

aacbaF

Fo

cc

ccc

bb

bb

b

aa

aaa

RV

RV

RVRV

RV

RV

RVIIIIR

Vand

RV

RvVIandR

VR

vVIand

RV

RvVI0vv


6. NON-INVERTING SUMMER

vo

Ra

RF

v+

v-

V1R1

V2R2

5. DIFFERENTIAL AMPLIFIER

vo

RaVa

RF

v+

v-

Ia

IF

V1

R1

R2

aF

a21

2aF

1o

Fo

aa

FaiF

oF

aa

a21

21

RRVRR

RRR1VV

havewevforngsubstitutiR

VvR

vVII0IButRVvI

andRvVIandRR

RVvv

aF

22

11

2121

aFo

aFFo

Fo

a

22

11

2121

22

11

2122

11

RR1R

VRV

RRRR

RR1VV

R1

R1VR

V0RV-V

RV-0

RV

RV

RRRRV

RV

RV

R1

R1V0R

V-VR

V-V


8. OP. AMP. INTEGRATOR

XofintegralYdtXCR1Y

dtRX

C1dtiC

1Y0vbutiIRX

Xv0vv

cccR

R

YX

R

v+

v-

iciR

C

vc

7. OP. AMP. DIFFERENTIATOR

YX

R

v+

v-

ic

IR

C

vc

XofaldifferentiYdtdXRCY

dtdXCdt

dvCIIRY-0

Xv0vvccR

c


ExamplesFind Vo of the following Op. Amp. circuits

vo

Ra

RF

v+

v-

V1

R1 RL

vov+

v-

V1

R1

vo

RF

v+

v-

V1

R1

Ra

R2

o1 VvvVresistorsbothincurrentnoisThere

aF

aoa

Faao

Faao

1

1iR1

RR1VVR

RRVV

RRRVVBut

Vvv0iI

aF

212

1o

2121

Faao

RR1RR

RVV

RRRVvRR

RVv


vo

RF

v+

v-

V1

R1

R2

vo

RF

v+

v-

V1

R1

R2RLIL

Find VFind Voo of the followings Op. Amp. circuits of the followings Op. Amp. circuits

Find IFind ILL of the followings Op. Amp. circuits of the followings Op. Amp. circuits

212

1o

2121

o

F

RRRVV

RRRVvVv

RincurrentnoisThere

212

L1

Lo

L21

21o

2121o

F

RRR

RV

RVIRR

RVV

RRRVvVv

RincurrentnoisThere


Inverting amplifiers Ri

Non-inverting amplifiers Ri

vo

RF

v+

v-

R1

Va

Ra

VbRb

vo

RF

v+

v-

V1R1

V2R2

V3R3

Ra

2133in

3122in

3211in

//RRR)V(atRV&VofRernalintifAlso

//RRR)V(atRV&VofRernalintifAlso//RRR)V(atRThen

V&VofRernalintIf(open)vatreistanceInput

0

0

0

12

31

32

bbin

aain

1

R)V(atRAlsoR)V(atRThen

0vvRincurrentnoisThere

9.3 Input resistance of Op. Amp. amplifiers


9.4 Multiple-Input Operational Amplifier Circuit

To derive a general design table for given Vo of Multiple-Input Op. Amp. Circuit:-

1. Find V- from the input voltages Va---Vm 2. Find V+ from the input voltages Va---Vn

3. Equate V+ and V- and find the Vo

4. Make equal dc path resistance at both input terminals of Op. Amp.

5. Assign coefficient and make design table

Multiple-Input Operational Amplifier circuitRx

R1R2Rn

V1V2Vn

RaRbRm

VaVbVm

Ry

RF

Vo


9.5 Design of Multiple-Input Amplifier


1. Find V- from the input voltages Va---Vm

RaRbRm

VaVbVm

Ry

RF

Vo

FAAo

FAFA

mm

bb

aa

FAFA

Fo

mm

bb

aa

Fo

mm

bb

aa

FAFYa

Fo

mm

bb

aa

YFmba

YFo

mm

bb

aa

RRRv

RRRR

Rv.......R

vRv

RRRR

Rv

Rv.......R

vRvv

Rv

Rv.......R

vRv

R1

R1vR

1R//....//R

1v

Rv

Rv.......R

vRv

R1

R1

R1.........R

1R1v

0Rv0

Rvv

Rvv.......R

vvR

vvisvatKCL


Xn21nn

22

11

nn

22

11

Xn21

nn

22

11

Xn21

Xnn

22

11

R//R....//R//RRv.......R

vRvv

Rv.......R

vRv

R//R.............//R//R1v

Rv.......R

vRv

R1

R1.........R

1R1v

0Rv

Rvv.......R

vvR

vvisvatKCL

2. Find V+ from the input voltages V1---Vn

Rx

R1

R2V+

Rn

V1V2Vn


3. Equate V+ and V- and find the Vo

Xn1AF

eq

Fmm

aaeq

nn

11

AFA

FAFA

mm

aa

Xn1AF

nn

11

FAFA

mm

aa

Xn1nn

11

AF

o

FAAo

FAFA

mm

aa

Xn1nn

11

FAAo

FAFA

mm

bb

aa

Xn21nn

22

11

R//R....//RRR1RWhere

RRv...R

vRRv...R

vR

RRRR

RRRv...R

vR//R....//RRR1R

v...Rv

RRRR

Rv...R

vR//R....//RRv...R

vRR1v

RRRv

RRRR

Rv...R

vR//R....//RRv...R

vRR

RvRR

RRRv.......R

vRvR//R....//R//RR

v.......Rv

Rv

vv


)Assume(ZRR

RR1YXThen

Y1RRXR

R,RbyallXlyingorR1Y1R

1R1XR

1YYYXXXtakingand

R1

R1YYR

1R1XXR

1R//R//R....//RR//R....//RThen

equalareinputs""and""bothatcestanresispathdcSince

xF

yF

yF

xF

FyFxF

man1

Fyma

Fxn1

F

FYmaXn

1

4. Design of the multiple coefficients


equationsdsubstitutetheofsummarythegivesbelowtabledesignamplifierSumming

equationabovethein0RR

RRsubstitute0Zthen,RRifIIICase

andequationabovethein0RRsubstitute0Zthen0R

R,RifIICase

andequationabovethein0RRsubstitute0Zthen0R

R,RifICase

RR

RRZor1,YXZowN

yF

xFyx

yF

yFy

xF

xFx

xF

yF

5. Design table

Case Z RY RXRi

(n-inv)Rj

(inv)I Z > 0 RF / Z RF / Xi RF / YjI I Z < 0 - RF / Z RF / Xi RF / YjI I I Z = 0 RF / Xi RF / Yj

vo

RF

VmRm

Va

Ra

VbRb

V1V2

RnVn

Rx

Ry

R1R2


Summary of Design Equations Multiple-Input Amplifier

According to Vo equation, multiple input op. amp. circuit is drawn first with positive term resistors plus standard RX at “n-inv” terminal then with negative term resistors plus standard RY at “inv” terminal. Add standard RF

vo

RF

V2

R2

V1

Rx

Ry

R1

Case Z RY RX Ri (n-inv) Rj (inv)II Z < 0 - RF / Z RF / Xi RF / Yj

1. Assume given VO = -10V2 + 5V1

2. Draw circuit with one resistor (R1) for +5V1 and one standard RX at “n-inv” terminal. 3. Continue with one resistor (R2) for –10V2 and one standard RY at “inv” terminal. 4. Add feedback resistor RF.

7. Find resistor ratio 10RR,5

RR,6RR,Rtable,From0ZAs F

2F

1F

XY

6

10

11051YXZFind10,Yis coeff.-and5,Xis.coeffV5VV 21o5. Find Z

6. Use Design Table


8. Find RF from required type and value of resistance at input terminal of the op. Amp. Assume required min 10k at Op. Amp. Input terminal.

kRFind10RRkthen,ratiosallof.minis10

RRAs FF

2F

2 10010

10RR,5

RR,6RR,R.minisresistorwhichfindNow F

2F

1F

XY

9. Now we can find RX , R1 and R2 by using RF=100k

k10100kR20k,5

100kR16.67k,6100kR 21X 10

10. Finally draw the circuit with designed values

vo

100k=RF

V210k=R2

V1

16.67k=Rx

Open Ry

20k=R1

VO = -10V2 + 5V1


Design Example-1

1011451YXZ14,Y5,X6V8VV4VV ba21o

H.W. 2.1 (LECTURE)Design an op. Amp. Circuit to implement the equation Vo=4V1+V2-8Va-6Vb .Design with lowest resistance at any input terminal of the op. Amp = 10k

6RR,8

RR,1RR,4

RR,10RR,Rtable,From0Z F

bFaF

2F

1F

XY

100kΩ10k10R10k10RRisresistanceLowest F

FX

Case Z RY RXRi

(n-inv)Rj

(inv)II Z < 0 - RF / Z RF / Xi RF / Yj

16.67k6100kR

,12.5k8100kR,&100k1

100kR

,25k4100kR,10k10

100kR

b

a2

1X

RF=100k

VO

Vb

Va

V1R1=25k

Ra=12.5kRb=16.67k

V2R2=100k

RX=10k

YR


Design an Op. Amp. Circuit with a minimum resistance 10k and the outputVO = -100VX +50VY.. Design with Superposition Method.

Design Example-2 (conventional method)

The circuit is differential Op. Amp. Circuit as the output contain one positive term and one negative term and derive Vo equation.

kRRkRTakekR,kRanyAs

kRRtakeandRRRandR

R

aFaa

aF

1000100101010

1021100 21

212

termnalVatk20&k20change,amp.opinputsof&bothatcetanresisdcequalmakeTo

Y

voVX

VY10k

10k

10k1000k

voVX

VY20k

20k

10k1000k

vo

RF

VXRa

VY

R2

R1

)given(VVVRRVRR

RRRV

thenVVif,theoremionSuperpositFrom

VRRR

RRV,VWhen

&VRRV,VWhen

XYXaF

YaFo

YX

YaFoX

XaFoY

100501

0

10

0

212

212


Design an Op. Amp. Circuit with a minimum resistance 10k and the outputVO = -100VX +50VY.. Design with Multiple input design method.

Design Example-3 (multiple input design method)

According to Vo equation, it is multiple input op. amp. circuit with one input at “inv” terminal (for -100VX) and another at “n-inv” terminal (for +50VY). Standard RX and RY are included first.

vo

RF

VXRa

VYRx

RyR1151100051YXZ100,Y50,XV10050VV XYo

Case Z RY RXRi

(n-inv)Rj


100R

YRR,50

RXRR,R

ZRR,R

table,From0ZF

1FaF

1F

1FF

XY 51

1000kΩ10k100R10k100RRisresistanceLowest F

Fa vo

VX

VY20k

20k

10k1000k

10k1001000kR,20k50

1000kR,20k1000kR a1X

51


EXAMPLE 2.5 (PAGE 51-TEXT)Design an op. Amp. Circuit to implement the equation Vo=4V1+V2-8Va-6Vb .Design with equal DC path resistance = 10k

Design Example-4

Case Z RY RX Ri (n-inv) Rj (inv)II Z < 0 - RF / Z RF / Xi RF / Yj

1011451YXZ14,Y5,Xb6Va8V2V14VoV

6FR

bR,8FR

aR,1FR

2R,4FR

1R,10FR

XR,YRtable,From0Z

150kΩ10k1)(1410k1YFR(2.41),equation49,pagefromOr

150kΩ10k15FRR15

R1

R4

R10

R1

RR10k1

spec.)red10kΩ0kΩ(reb//Ra//RYR2//R1//RXRpathDC

FFFF21X11

25k6150k

bR

,18.75k8150kaR,150k1

150k2R

,37.5k4150k

1R,15k10150k

XR

RF=150k

VO

Vb

Va

V1R1=37.5k

Ra=18.75k

Rb=25k

Ry=

V2R2=150k

RX=15k

YR


Design Example-5EXAMPLE 2.5 (PAGE 51-TEXT)

Design an op. Amp. Circuit to implement the equation Vo=4V1+V2-8Va-6Vb .Design with highest resistance of 200k at any input terminal except feedback resistor RF 1011451YXZ14,Y5,Xb6Va8V2V14VoV

Case Z RY RXRi

(n-inv)Rj


6FR

bR,8FR

aR,1FR

2R,4FR

1R,10FR

XR,YRtable,From0Z

200kΩFR200k1FR

2RisresistanceHighest

33.33k6200k

bR

,25k8200kaR,&200k1

200k2R

,50k4200k

1R,20k10200k

XR

R F=200k

V O

V b

V a

V 1R 1=50k

R a=25kR b=33.33k

R y=

V 2R 2=200kR X=20k

YR

1 9. design of multiple-input op. amp. circuits 9.1operational amplifiersoperational amplifiers...

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