Download Chapter 4 - McMaster University > ECE

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
LM324 DIP
LMC6294
MAX4240
OP-AMP ASSEMBLED ON PRINTED CIRCUIT BOARD
APEX PA03
DIMENSIONAL DIAGRAM LM 324
PIN OUT FOR LM324
CIRCUIT SYMBOL FOR AN OP-AMP SHOWING POWER SUPPLIES
LINEAR MODEL
INPUT RESISTANCE
TYPICAL VALUES
Ri : 105   1012 
RO : 1  50
A : 10  10
5
7
GAIN
OUTPUT RESISTANCE
CIRCUIT WITH OPERATIONAL AMPLIFIER
LOAD
OP-AMP
DRIVING CIRCUIT
TRANSFER PLOTS FOR SOME COMERCIAL OP-AMPS
LINEAR
REGION
IDENTIFY SATURATION REGIONS
SATURATION
REGION
OP-AMP IN SATURATION
CIRCUIT AND MODEL FOR UNITY GAIN
BUFFER
WHY UNIT GAIN BUFFER?
PERFORMANCE OF REAL OP-AMPS
Op-Amp BUFFER GAIN
LM324
0.99999
LMC6492
0.9998
MAX4240
0.99995
KVL :  Vs  Ri I  RO I  AOVin  0
KVL : - Vout  RO I  AOVin  0
CONTROLLING VARIABLE : Vin  Ri I
SOLVING
1
BUFFER Vout 
GAIN
Ri
Vs 1 
RO  AO Ri
V
AO    out  1
VS
THE IDEAL OP-AMP
IDEAL  RO  0, Ri  , A  
i
i
RO  0  vO  A(v  v )
Ri   
A
THE UNITY GAIN BUFFER – IDEAL OP-AMP ASSUMPTION
vOUT  v S
v  v s

vOUT
v  v
vOUT  v  
vOUT

1
vS
USING LINEAR (NON-IDEAL) OP-AMP MODEL WE OBTAINED
Vout

Vs
1
Ri
1
RO  AO Ri
PERFORMANCE OF REAL OP-AMPS
Op-Amp BUFFER GAIN
LM324
0.99999
LMC6492
0.9998
MAX4240
0.99995
IDEAL OP-AMP ASSUMPTION YIELDS EXCELLENT APPROXIMATION!
WHY USE THE VOLTAGE FOLLOWER OR UNITY GAIN BUFFER?
vO  v S
v  v s
v  v
vO  v
CONNECTION WITHOUT BUFFER
THE VOLTAGE FOLLOWER ACTS AS
BUFFER AMPLIFIER
THE VOLTAGE FOLLOWER ISOLATES ONE
CIRCUIT FROM ANOTHER
ESPECIALLY USEFUL IF THE SOURCE HAS
VERY LITTLE POWER
CONNECTION WITH BUFFER
vO  v S
THE SOURCE SUPPLIES NO POWER
THE SOURCE SUPPLIES POWER
LEARNING EXAMPLE
DETERMINE THE GAIN G 
APPLY KCL @ vVs  0 Vout  0

0
R1
R2
G
Vout
R
 2
Vs
R1
Ao    v  v v  0
Ri    i  i  0
v  0
i  0
v  0
Vout
Vs
LEARNING EXAMPLE: DIFFERENTIAL AMPLIFIER
THINK NODES!
OUTPUT CURRENT IS NOT KNOWN
THE OP-AMP IS DEFINED BY ITS 3 NODES. HENCE IT NEEDS 3 EQUATIONS
KCL AT V_ AND V+ YIELD TWO EQUATIONS
(INFINITE INPUT RESISTANCE IMPLIES THAT i-, i+ ARE KNOWN)
DON’T USE KCL AT OUTPUT NODE. GET THIRD EQUATION FROM INFINITE
GAIN ASSUMPTION (v+ = v-)
LEARNING EXAMPLE: DIFFERENTIAL AMPLIFIER
IDEAL OP-AMP CONDITIONS
NODES @ INVERTING TERMINAL
R4
R4
v2  v 
v2
R3  R4
R3  R4

 R 
R
R 
R 
vO  1  2 v  2 v1  2  1  1 v  v1 
R1 
R1
R1   R2 


i  0  v  
NODES @ NON INVERTING TERMINAL
R4  R2 , R3  R1  vO 
R2
(v2  v1 )
R1
LEARNING EXAMPLE: USE IDEAL OP-AMP
v 1
FIND vO
vo1
v 1
v 1  v m 1
v 2  v m 2
FINISH WITH INPUT NODE EQUATIONS…
USE INFINTE GAIN ASSUMPTION
vm1
v1  v1
v 2
vo 2
 v m 1  v1
v 2  v 2
vm 2  v2
USE REMAINING NODE EQUATIONS
v 2
vm 2
6 NODE EQUATIONS + 2 IDEAL OP-AMP
v 1  v 1
v 2  v 2
@ vm1 :
v1  v01 v1  v 2 v1  vo 2


00
R2
RG
R1
v 2  v o 2 v 2  v1 v 2
@ vm 2 :


00
R1
RG
R2
ONLY UNKWONS ARE OUTPUT NODE VOLTAGES
SOLVE FOR REQUIRED VARIABLE v o  v o1
LEARNING EXTENSION
FIND IO . ASSUME IDEAL OP - AMP
v  12V
AO    v  12V
v  12V
Ri    i  0
12  Vo 12

 0  Vo  84V
12k
2k
V
 IO  o  8.4mA
10k
KCL@ v :
LEARNING EXTENSION
v0
NONINVERTING AMPLIFIER - IDEAL OP-AMP
v
v_
vo
R2
v  vi
i  0
R1
“inverse voltage divider”
SET VOLTAGE
v  vi
v  vi  v  vi
INFINITE GAIN ASSUMPTION
INFINITE INPUT RESISTANCE
R1
R1  R2
vi 
v0  v0 
vi
R1  R2
R1
LEARNING EXAMPLE
UNDER IDEAL CONDITIONS BOTH CIRCUITS SATISFY
VO  8V1  4V2
If 1V  V1  2V , 2V  V2  3V
dc supplies are  10V
DETERMINE IF BOTH IMPLEMENTATIONS PRODUCE THE
FULL RANGE FOR THE OUTPUT
VX  2V1  V2
1V  V1  2V , 2V  V2  3V  1V  VX  2V VX OK!
VO  4VX
1V  VX  2V  4V  VO  8V VO OK !
VY  8V1
1V  V1  2V  16V  VY  8V
EXCEEDS SUPPLY VALUE.
THIS OP-AMP SATURATES!
POOR IMPLEMENTATION