TYPICAL DEVICE USING OP-AMPShome.konkuk.ac.kr/~keechan/ch04.pdf · · 2018-03-26We have already...
Transcript of TYPICAL DEVICE USING OP-AMPShome.konkuk.ac.kr/~keechan/ch04.pdf · · 2018-03-26We have already...
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 circuitsusing OpAmps
3. The linear models for OpAmps include dependent sources
TYPICAL DEVICE USING OP-AMPS
1
OP-AMP ASSEMBLED ON PRINTED CIRCUIT BOARD APEX PA03
PIN OUT FOR LM324
DIMENSIONAL DIAGRAM LM 324
LM324 DIP
LMC6294
MAX4240
Unit: inch
2
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
3
CIRCUIT WITH OPERATIONAL AMPLIFIER
DRIVING CIRCUIT
LOAD
OP-AMP
4
TRANSFER PLOTS FOR SOME COMERCIAL OP-AMPS
SATURATIONREGIONLINEAR
REGION
IDENTIFY SATURATION REGIONSOP-AMP IN SATURATION
5
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
6
A
)(0 vvAvR OO
THE IDEAL OP-AMP
iR
ARR iO ,,0IDEAL
i
i
7
THE SCHMITT TRIGGER
CCVRR
R
21
1
CCVRR
R
21
1
CCV
CCV
CCVR
R
2
1
CCVR
R
2
1
CCV
CCV
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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 0.99998MAX4240 0.99995
IDEAL OP-AMP ASSUMPTION YIELDS EXCELLENT APPROXIMATION!
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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 POWER THE SOURCE SUPPLIES NO POWER
THE VOLTAGE FOLLOWER ISOLATES ONECIRCUIT FROM ANOTHER.ESPECIALLY USEFUL IF THE SOURCE HASVERY LITTLE POWER.
CONNECTION WITHOUT BUFFER CONNECTION WITH BUFFER
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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
0i1
2
R
R
V
VG
s
out
FOR COMPARISON, NEXT WE EXAMINE THE SAME CIRCUIT WITHOUT THE ASSUMPTION OF IDEALOP-AMP.
11
REPLACING OP-AMPS BY THEIR LINEAR MODEL
WE USE THIS EXAMPLE TO DEVELOPA PROCEDURE TO DETERMINE OP-AMPCIRCUITS USING THE LINEAR MODELS
12
1. Identify Op Amp nodes
v
v
ov
2. Redraw the circuit cutting outthe Op Amp
v
v
ov
3. Draw components of linear OpAmp (on circuit of step 2)
v
v
oviR
OR
( )A v v
4. Redraw as needed
2R
v
v
13
INVERTING AMPLIFIER: ANALYSIS OF NON IDEAL CASE
NODE ANALYSIS
CONTROLLING VARIABLE IN TERMS OF NODEVOLTAGES
USE LINEAR ALGEBRA
10,10
,108
5
Oi RR
A :AMP-OPTYPICAL 9996994.45,1 21
S
O
v
vkRkR 000.5
S
O
v
vA
14
0i
0 iiRi
vvA
0v
0v
KCL @ INVERTING TERMINAL
000
21
R
v
R
v OS
1
2
R
R
v
v
s
O
GAIN FOR NON-IDEAL CASE
THE IDEAL OP-AMP ASSUMPTION PROVIDES EXCELLENT APPROXIMATION.(UNLESS FORCED OTHERWISE WE WILL ALWAYS USE IT!)
SUMMARY COMPARISON: IDEAL OP-AMP AND NON-IDEAL CASE
NON-IDEAL CASE
REPLACE OP-AMP BY LINEAR MODELSOLVE THE RESULTING CIRCUIT WITHDEPENDENT SOURCES
15
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-).
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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
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LEARNING EXAMPLE: USE IDEAL OP-AMP Ov FIND
1v
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
2211
vvvv
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
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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 o
O 4.810
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“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 1vv
INFINITE GAIN ASSUMPTION
11 vvvv
R2
R1
ivv
0v
20
FIND GAIN AND INPUT RESISTANCE - NON IDEAL OP-AMP
NOW RE-DRAW CIRCUIT TO ENHANCECLARITY. THERE ARE ONLY TWO LOOPS.
DETERMINE EQUIVALENT CIRCUITUSING LINEAR MODEL FOR OP-AMP
v
v
Ov
iR
OR
( )A v v
COMPLETE EQUIVALENTFOR MESH ANALYSIS
Ov
iv v v
MESH 1
MESH 2
CONTROLLNG VARIABLE IN TERMS OF LOOPCURRENTS
)( 21122 iiRiRvO 21
INPUT RESISTANCE
1
1
i
vRin
GAIN
i
O
v
vG
MESH 1
MESH 2
CONTROLLNG VARIABLE IN TERMS OF LOOPCURRENTS
MATHEMATICAL MODEL
)( 21122 iiRiRvO
REPLACE AND PUT IN MATRIX FORM
0)(
)( 1
2
1
211
121 v
i
i
RRRRAR
RRR
Oi
0)(
)( 11
211
121
2
1 v
RRRRAR
RRR
i
i
Oi
THE FORMAL SOLUTION
)())(( 112121 RARRRRRRR iO
)()(
)(
211
121
RRRAR
RRRRAdj
i
O
0)()(
)(1 1
211
121
2
1 v
RRRAR
RRRR
i
i
i
O
THE SOLUTIONS
121
1 vRRR
i O
)( 1
2RAR
i i
???A
1121
1211
22111
))(()(
)(
vRARRR
vRRRR
iRRiRv
iO
O
iRARA 1 inR
1
21
1 R
RR
v
vG O
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A SEMI-IDEAL OP-AMP MODEL
This is an intermediate model, more accurate than the ideal op-ampModel, but simpler than the linear model used so far.
, 0,i O O
R R A A
Non-inverting amplifier and semi-ideal model
0i i !!v v
Replacement Equation
( )O O e O
v A v A v v
v
v
e inv v
2 1
1
(as before)
( ) (replaces )
S
O
O S O
v v
R Rv v
R
A v v v v v
1
! 2
;1
O O
S O
v A R
v A R R
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Si
Ov
R
+-
Sv
Sample Problem
Find the expression for Vo. Indicatewhere and how you are using the IdealOpAmp assumptions.
Set voltages? Svv Use infinite gain assumption Svv v
Use infinite input resistance assumptionand apply KCL to inverting input
0
R
vvi oS
SSo Rivv
0i
v
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Si
Ov
R
Sample Problem
1. Locate nodes
2. Erase Op-Amp
3. Place linear model
4. Redraw if necessary
+-
ov
v
v R
RO
RiiS
A(v+ - v-)
TWO LOOPS. ONE CURRENT SOURCE. USE MESHES.
DRAW THE LINEAR EQUIVALENT CIRCUIT AND WRITE THE LOOP EQUATIONS
1i2i
MESH 1 sii 1
MESH 2 )()()( _22 vvAiRRiiR OSi
CONTROLLING VARIABLE )( 12_ iiRvv i
iR
OR
( )A v v
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SV
OV ANDGAIN FIND
SVv
SVv _
0i
“INVERSE VOLTAGE DIVIDER”OV
SV2R
1R
SO Vk
kkV
1
1100
101S
O
V
VG
VVmVV OS 101.01
LEARNING EXTENSION
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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 X
V V
1 2 4 8X O
V V V V V V !O
V 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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COMPARATOR CIRCUITS
Some REAL OpAmps requirea “pull up resistor.”
ZERO-CROSSING DETECTOR28
LEARNING BY APPLICATION OP-AMP BASED AMMETER
NON-INVERTING AMPLIFIER
1
21R
RG
IRV II
IRR
RGVV IIO
1
21
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1 4 (design eq.)B
A
R
R
LEARNING EXAMPLE DC MOTOR CONTROL - REVISITED
CHOOSE NON-INVERTING AMPLIFIER (WITH POWEROP-AMP PA03)
Constraints: 20M
V VPower dissipationin amplifier 100mW
Simplifying assumptions: , 0i O
R R Significant power lossesOccur only in Ra, Rb
Worst case occurs when Vm=20
2(20 )100
MX
A B
VP mW
R R
4000
A BR R
3B
A
R
R
One solution: 3 , 1B A
R k R k
Standard values at 5%!
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DESIGN EXAMPLE: INSTRUMENTATION AMPLIFIER
DESIGN SPECIFICATIONS
1 2
10OV
GV V
• “HIGH INPUT RESISTENCE”• “LOW POWER DISSIPATION”• OPERATE FROM 2 AA BATTERIES
ANALISIS OF PROPOSED CONFIGURATION
O X YV V V
1 2 1
1
@ : 0XV V V V
AR R
2 1 2
2
@ : 0YV V V V
BR R
1 2 2 1
1 2 1 21 1
O
R R R RV V V V V
R R R R
SIMPLIFY DESIGN BY MAKING 1 2R R
1
1 2
21 ( )
O
RV V V
R DESIGN EQUATION: 1
2 9R R
OV1
V
2V
G
USE LARGE RESISTORS FOR LOW POWER 1 2. ., 100 , 450e g R k R R k
1 2;
A BV V V V Infinite gain
MAX4240
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DESIGN EXAMPLEDESIGN SPECIFICATION 10O
in
V
V
Power loss in resistors should not exceed
100mW 2 2in
V V V when
Design equationS:2
1
1 10O
in
V R
V R
1 2
2
1 2
(20 )100
R R
VP mW
R R
Max Vo is 20V
Solve design equations (by trial and error if necessary)
2 19R R
1 24R R k
1400R
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DESIGN EXAMPLE IMPLEMENT THE OPERATION 1 20.9 0.1
OV V V
DESIGN CONSTRAINTS• AS FEW COMPONENTS AS POSSIBLE• MINIMIZE POWER DISSIPATED• USE RESISTORS NO LARGER THAN 10K
Given the function (weighted sumwith sign change) a basic weighted adder may work
ANALYSIS OF POSSIBLE SOLUTION
1 2
1 21
0
@ : 0O
V
V V VV
R R R
1 2
1 2
O
R RV V V
R R
2 1
2
9
0.1
R R
R
R
DESIGNEQUATIONS
2 1R R R
SOLVE DESIGN EQUATIONS USING TRIAL AND ERROR IF NECESSARY
210 ,5.6 ,...R k k
ANALYZE EACH SOLUTION FOR OTHER CONSTRAINTS AND FACTORS; e.g.DO WE USE ONLY STANDARD COMPONENTS?
33
DESIGN EXAMPLE DESIGN 4-20mA TO 0 – 5V CONVERTER
1. CONVERT CURRENT TO VOLTAGE USING A RESISTOR
I IV RICANNOT GIVE DESIRED RANGE!
2. CHOOSE RESISTOR TO PROVIDE THE 5V CHANGE … AND SHIFT LEVELS DOWN!
5 0312.5
0.020 0.004MAX MIN
MAX MIN
V VR
I I
1.25 6.25I
V MUST SHIFT DOWN BY 1.25V(SUBSTRACT 1.25 V)
2
1 2
I
RV V
R R 1
1 2
( )O SHIFT SHIFT
RV V V V
R R
2
1
O I SHIFT
RV V V
R V V
34
DOES NOT LOAD PHONOGRAPH
1000 OFTION AMPLIFICA ANPROVIDES IT
THATSO DETERMINE 12,RR
LEARNING BY DESIGN
)1)(1(1
2
1 R
R
V
VO
35
LEARNING EXAMPLE
UNITY GAINBUFFER
COMPARATOR CIRCUITS
TT eR 0227.045.57
ONLY ONE LEDIS ON AT ANYGIVEN TIME
36
MATLAB SIMULATION OF TEMPERATURE SENSOR
WE SHOW THE SEQUENCE OF MATLAB INSTRUCTIONS USED TO OBTAIN THE PLOTOF THE VOLTAGE AS FUNCTION OF THE TEMPERATURE
»T=[60:0.1:90]'; %define a column array of temperature values» RT=57.45*exp(-0.0227*T); %model of thermistor» RX=9.32; %computed resistance needed for voltage divider» VT=3*RX./(RX+RT); %voltage divider equation. Notice “./” to create output array» plot(T,VT, ‘mo’); %basic plotting instruction» title('OUTPUT OF TEMPERATURE SENSOR'); %proper graph labeling tools» xlabel('TEMPERATURE(DEG. FARENHEIT)')» ylabel('VOLTS')» legend('VOLTAGE V_T')
37
1vv
0i
1vv
KCL @ v_
EXAMPLE OF TRANSFER CURVE SHOWING SATURATION
THE TRANSFER CURVE
IN LINEAR RANGE
THIS SOURCE CREATES THE OFFSET
OFFSET
OUTPUT CANNOTEXCEED SUPPLY(10V)
38