op amp application

7/27/2019 Op Amp Application

http://slidepdf.com/reader/full/op-amp-application 1/15

1

OP-AMP APPLICATIONS

AGENDA:

Cascaded OP-AMP

Summing Amplifier Difference Amplifier

Voltage Buffer

Integrator Differentiator



2

REVIEW:

1

2

R R AV

iO V

R

RV

1

2

1 R Rin OL

CL specsout

CL

OL

specsout out

A A R

A A

R R )(

)(

1



3

REVIEW:

1

1

2

R

R AV

1

1

2

R RV V iO

)( specsin

CL

OLin R

A A R

)( specsout

OL

CL

out R A

A

R



4

2

2

2

in

i

f

O V R

RV

in

i

f

O V R

RV 1

Therefore the output voltage is equal to:

in

ii

f f

O V R R

R RV

2

2and the voltage gain is equal to:

2

2

ii

f f

V R R

R R A



5

2

2

1

1

2

2

1

1

21;*

)(0;*

i f

i f

O

f

Oii

f

V R RV

R RV

R

V V

R

V V

R

V V

I I I Anodeat

ground virtual V V V

more

http://c/Documents%20and%20Settings/windowsxp/Desktop/LECTURE/EP6E/CHAP20/MOVIES20.EXE




6

To avo id offs et errors place a R = R1||R2 ||RF to g roun d from the

non- inver t ing inp ut .



7

VOLTAGE BUFFER

The advantage of this

connection is that theload connected across oneoutput has no (or little)effect on the otheroutput. The outputs arebuffered or isolated fromeach other.



8

I2

I3

y

1.

11

0

;*

23

32

2

2

32

23

2

2

32

32

2

32

eqn R R

RV V

R

V

R R

R R

V

R

V

R RV

R

V

R

V V

I I

ynodeat

i y

i

y

i y

y yi



9

IF

I1

x

2.11

;*

1

1

1

1

1

1

eqn RV

RV

R RV

R

V V

R

V V I I

xnodeat

F

Oi

F

x

F

xO xi

F



10

IF

I1

x

2.11

1

1

1

eqn R

V

R

V

R RV

F

Oi

F

x

1

1

1

1

23

32

1

1

1

1

23

32

1

1

123

32

11

2.1.*

;;;sin

R

V

R R

R R

R R

RV

R

V

RV

RV

R R R R

R R RV

R

V

R

V

R R R R

RV

eqneqn substitute

V V V V V V ce

i

F

F

x

i

F

O

F

Oi

F

F

x

i

F

Oi

F x

i

x y y

1

1

1

1

23

32

R

RV

R

R R

R R

RV V F

i F

x

iO

1.

23

32 eqn

R R

RV V i y



11

If RF

= R3

and R2 = R1

12

1

ii F

O V V R RV

more





12

IC

Ii

X

dt vC R

V

dt

V d C

R

V

V but

dt

V V d C

R

V V

dt

dvC irecall

I I

X nodeat

iO

Oi

Oi

C C

C i

1

1

1

1

side bothintegrate

;

0

;

;

;

;

dt vC R

V iO

1

1



13

IF

IC

X

dt

V d C RV

R

V

dt

V d C

V but

R

V V

dt

V V d C

dt

dvC irecall

I I

X nodeat

i F O

F

Oi

F

Oi

C C

F C

;

0

;

;

;

;

dt

V d

C RV i

F O



14

THANK YOU!!! THANK YOU!!!



15

1. Calculate the output voltage of a cascaded circuit if the first

stage is a non-inverting amplifier and the 2nd and 3rd is aninverting op-amp having the following values for A1: Vin =

20μV, RF1=510KΩ and Ri1=18KΩ; for A2: RF2=680KΩ,

Ri2=22KΩ; for A3: RF3=750KΩ, Ri3=33KΩ. (Vo = 412 mV)

2. Calculate the output voltage for inverting summing amplifier having input resistances of 47KΩ and 12KΩ with an input

voltages of 40mV rms and 20 mV rms respectively. Use

feedback resistor value of 470KΩ. (-1.18V)

3. For a difference amplifier if RF=300KΩ with input resistor

value of R1=150KΩ(one terminal connected to inverting input)and R2=10KΩ(one terminal connected to non-inverting input)

and R3=10KΩ, determine output voltageVo.(Vo = -2.5V)

EXAMPLES: