operational amplifiers why do we study them at this point??? 1. opamps are very useful electronic...
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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
TYPICAL DEVICE USING OP-AMPS
OP-AMP ASSEMBLED ON PRINTED CIRCUIT BOARD APEX PA03
PIN OUT FOR LM324
DIMENSIONAL DIAGRAM LM 324
LM324 DIP
LMC6294
MAX4240
CIRCUIT SYMBOL FOR AN OP-AMP SHOWING POWER SUPPLIES
LINEAR MODEL OUTPUT RESISTANCEINPUT RESISTANCE
GAIN75
125
1010:
501:
1010:
A
R
R
O
i
TYPICAL VALUES
TRANSFER PLOTS FOR SOME COMERCIAL OP-AMPS
SATURATIONREGIONLINEAR
REGION
IDENTIFY SATURATION REGIONSOP-AMP IN SATURATION
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
THE UNITY GAIN BUFFER – IDEAL OP-AMP ASSUMPTION
svv
vv
OUTv v
OUT Sv v
OUTv
1OUT
S
v
v
USING LINEAR (NON-IDEAL) OP-AMP MODEL WE OBTAINED
1
1
out
is
O O i
VRV
R A R
PERFORMANCE OF REAL OP-AMPS
Op-Amp BUFFER GAINLM324 0.99999LMC6492 0.9998MAX4240 0.99995
IDEAL OP-AMP ASSUMPTION YIELDS EXCELLENT APPROXIMATION!
WHY USE THE VOLTAGE FOLLOWER OR UNITY GAIN BUFFER?
svv
vv
vvO
SO vv
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
CONNECTION WITHOUT BUFFER CONNECTION WITH BUFFER
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
LEARNING EXAMPLE: DIFFERENTIAL AMPLIFIER
THE OP-AMP IS DEFINED BY ITS 3 NODES. HENCE IT NEEDS 3 EQUATIONS
THINK NODES!
KCL AT V_ AND V+ YIELD TWO EQUATIONS(INFINITE INPUT RESISTANCE IMPLIES THAT i-, i+ ARE KNOWN)
OUTPUT CURRENT IS NOT KNOWN
DON’T USE KCL AT OUTPUT NODE. GET THIRD EQUATION FROM INFINITEGAIN ASSUMPTION (v+ = v-)
LEARNING EXAMPLE: DIFFERENTIAL AMPLIFIER
NODES @ INVERTING TERMINAL
NODES @ 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 FIND1v
1v
2v
2v
1ov
2ov
1mv
2mv
6 NODE EQUATIONS + 2 IDEAL OP-AMP
22
11
vv
vv
22
11
m
m
vv
vv
FINISH WITH INPUT NODE EQUATIONS…USE INFINTE GAIN ASSUMPTION
2211vvvv
2211vvvv
mm
USE REMAINING NODE EQUATIONS
00:@1
2121
2
011
1
R
vv
R
vv
R
vvv o
G
m
00:@2
212
1
22
2
R
v
R
vv
R
vvv
G
o
m
ONLY UNKWONS ARE OUTPUT NODE VOLTAGES
1 VARIABLE REQUIREDFOR SOLVE
oovv
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 4.810
“inverse voltage divider”
ii vR
RRvv
RR
Rv
1
2100
21
1
0i
INFINITE INPUT RESISTANCE
NONINVERTING AMPLIFIER - IDEAL OP-AMP
LEARNING EXTENSION
v
_vov
SET VOLTAGE iv v
INFINITE GAIN ASSUMPTION
i iv v v v
R 2
R 1
ivv
0v
LEARNING EXAMPLE UNDER IDEAL CONDITIONS BOTH CIRCUITS SATISFY
1 28 4
OV V V
1 2If 1 2 , 2 3
dc supplis are 10
V V V V V V
V
DETERMINE IF BOTH IMPLEMENTATIONS PRODUCE THEFULL RANGE FOR THE OUTPUT
1 22
XV V V
1 21 2 , 2 3 1 2
XV V V V V V V V V OK!
XV
4O XV V
1 2 4 8X O
V V V V V V !OV OK
18
YV V
11 2 16 8
YV V V V V V
EXCEEDS SUPPLY VALUE.THIS OP-AMP SATURATES!
POOR IMPLEMENTATION
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