Opera&onal  Amplifier  Part1  

ESc201  :    Introduc&on  to  Electronics  L18

rhegde Dept. of Electrical Engineering

IIT Kanpur

1  

Amplifier Design requires specialized knowledge

R1

R2 CE

CC

RC

RE

CB

VS

RL

vo

VCC

CE Amplifier

Vin1

-12V

20K2K

0.1mA 10mA

+12V

137K

Q1a Q1b

Q2Q3

It is not possible for every user to design his/her own amplifier !

2  

Why can’t we have experts design and implement amplifiers and make it available to everybody else !

Although this is done, it does not satisfy all the users due to diverse requirements

Design

Custom

Standard

Semi-Custom

Semi-custom: partially competed design which is customized by the user

Opamp is a good illustration of the advantages of semi-custom approach 3  

X

v1

v2

1 2 0v v− =

short X

i 0i =

open

Can something be both a short as well as open circuit ?

4  

-An amplifier that is sensitive to difference in input voltages and insensitive to what is common.

vovin1vin2

1 2

1 22

id in in

in inic

v v vv vv

= −

+=

o d id cm icv A v A v= +

: Differential mode gain: Common mode gain

d

cm

AA

d cmA A>>

Common Mode Rejection Ratio: dcm

ACMRRA

=

Difference Amplifier

5  

Whatever is common is rejected and whatever is different is amplified !

vi1vi2

vo 100; 0.01d cmA A= =

1

2

1 5 ( ) ;1 5 ( )

i

i

v V mV Sin tv V mV Sin t

ω

ω

= + ×

= − ×

1 2

1 2

10 ( )

12

id in in

in inic

v v v mV Sin tv vv V

ω= − = ×

+= =

1 ( 1) 0o d id cm ic

V Siv A v v

VnA

mtω= +

= +×

6  

vov1v2

A  special  kind  of  difference  amplifier  

1.  Very  High  Differen@al-­‐mode  voltage  gain  2.  Very  High  Common  mode  Rejec@on  ra@o  3.  Very  High  Input  Resistance  4.  Very  Low  output  Resistance  5.  ….  

Operational Amplifier

7  

vov1v2

1.  Infinite  Differen@al-­‐mode  voltage  gain  2.  Infinite  Common  mode  Rejec@on  ra@o  3.  Infinite  Input  Resistance  4.  Zero  output  Resistance  5.  ….  

Ideal Operational Amplifier

8  

Example:  LM  741    

9  

Inside the opamp, there is a complicated circuit containing several transistors and resistors.

10  

+vi-

Ri

v1

R0vo

AOL vi

v2

+-

Simple  equivalent  circuit  model  of  an  opamp  

vov1v2

~105  

~10Ω  

~106  Ω  This assumes very high CMRR

11  

v1v2

vo

+VCC

-VCC

+vi-

Ri

v1

R0vo

AOL vi

v2

+-

vo  

vi  

o ol iv A v= ×VCC  

-­‐VCC  Opamp is said to be saturated

12  

+vi-

Ri

v1

R0vo

AOL vi

v2

+-

vov1v2

vo+ve

+ve

vo

+ve -ve

13  

How  do  we  amplify  this  signal?  

1 ( )sv mV Sin tω=

vS v1v2

vo

+12V

-12V

14  

vS vO

+ 12 V

- 12 V

+vi-

Ri

v1

R0vo

AOL vi

v2

+-

+vi-

Ri

R0vo

AOL vi+-

vS

210 ( )o ol Sv A v Sin tω= × =

But  opamp  voltage  is  limited  to         12V±

15  

510olA =

vS vO

+ 12 V

- 12 V

1 ( )sv mV Sin tω=

Simula&on  Results  

16  

1 ( )sv mV Sin tω= 510olA =

How  do  we  amplify  this  signal  then  ?  

1 ( )sv mV Sin tω=

vS

vS vO

v1v2

vo

+12V

-12V

1.  AYenuate  the  signal  to  0.1mV  and  then  amplify  ?  2.  …….  

17  

If   the   feedback   signal   helps   the   input   voltage   we   have   posi@ve  feedback,  otherwise  nega@ve.  

VO

R1

R2

VS

A  BeIer  Solu&on  

21

oS

v Rv R

=−

Amplifier  has  feedback  

18  

Feedback

vIN vout AV

vIN vout AV

Feedback  network  

Σ  

19  

Negative and Positive feedback

vIN vout AV

Feedback  network  

Σ  

vfb

+

-

Negative feedback

VIN - Vfb

vIN vout AV

Feedback  network  

Σ  

vfb

+

+

Positive feedback

VIN + Vfb

20  

Opamp  circuits  classifica&on  

Opamp  circuits  

Closed  loop  Or  feedback  

Open  loop  

Nega@ve  feedback  

Posi@ve  feedback  

Most  Opamp  Circuits  employ  nega&ve  feedback  

21  

Inver&ng  amplifier  

VO

R1

R2

VS

Ri

vo

AOL vi+-

vS

R0R1

R2

-vi+

+vi-

Ri

vo

+-

vS

R0R1

R2

-AOL vi

+vi-

Ri

R0vo

AOL vi+-

v+

v-

22  

This  is  called  the  Virtual  Ground  property  

+vi-

Ri

vo

+-

vS

R0R1

R2

-AOL vi1 2

(1)s i i i oi

v v v v vR R R− −

= +

Nodal  Analysis  

2(2)OL i o o i

o

A v v v vR R

− − −=

1 2 1 2

1 1 1( ) (3)s oi

i

v v vR R R R R=− + + +

2

2

1 1

(4)1o

i oOLo

R Rv v AR R

+=

−+

As 0OL iA v→∞ →

23  

+vi-

Ri

vo

+-

vS

R0R1

R2

-AOL vi

ii

As 0OL iA v→∞ →

This  implies  that  :     0ii →

No   current   flows   in   or   out   of   either   inver@ng   or   non-­‐inver@ng  terminals  of  an  ideal  opamp    

24  

VO

R1

R2

VS

+ Vi=0 -

Like ‘Ground’ too

vx

i

i

i i

Hence the name Virtual ground

25  For actual ground

Virtual  Ground  Property  

In  an  opamp  with  nega@ve  feedback,  the  voltage  of  the  inver@ng  terminal   is   equal   to   the   voltage   of   the   non-­‐inver@ng   terminal   if  the  gain  of  the  opamp  is  sufficiently  high  

v1v2

vo

+12V

-12V

1 2v v≅

This  property  does  not  hold  under  certain  condi@ons  such  as    q       open  loop,    q       posi@ve  feedback    q     or  if  the  opamp  is  saturated.  

26  

Two  important  property  for  analyzing  ideal  opamp  circuits    under  nega@ve  feedback  

v1v2

vo

+12V

-12V

1 21. v v=

1 22. 0i i= =

At the input side opamp appears to be like a short and an open circuit simultaneously !

27  

i1

i2

Re-­‐analyze  inver@ng  amplifier  with  these  proper@es    

Inver&ng  amplifier  

VO

R1

R2

VS0  

1/s si v R=

2 1

0 o ss

v viR R−

= =

si

21

os

v Rv R

= −28  

Non-­‐Inver&ng  Amplifier  

VO

R2

VS

R1

VS  

1 21. v v=

1 22. 0i i= =1/s si v R=

si

2 1o s s

sv v viR R−

= = 21

1os

v Rv R

= +29  

Examples

VO

1K

10K

10mV VO

1K

10K

10mV21

100oo

S

v R v mVv R

= − ⇒ = −

21

1 110oo

S

v R v mVv R

= + ⇒ =

0

0

1mA

0 110

ov mAK−

=

10ov V=−30  

R1

R2