Download IDEAL OP-AMP ASSUMPTION USING LINEAR

Introduction to 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
Vout
Vs
v  0
i  0
v  0
FOR COMPARISON, NEXT WE EXAMINE THE SAME
CIRCUIT WITHOUT THE ASSUMPTION OF IDEAL
OP-AMP
REPLACING OP-AMPS BY THEIR LINEAR MODEL
WE USE THIS EXAMPLE TO DEVELOP
A PROCEDURE TO DETERMINE OP-AMP
CIRCUITS USING THE LINEAR MODELS
1.
Identify Op Amp nodes
v
vo
3. Draw components of linear OpAmp
(on circuit of step 2)
v
vo
Ri
v
v
RO


2. Redraw the circuit cutting out
the Op Amp
A(v  v )
4. Redraw as needed
v
R2
v
vo
v
v
INVERTING AMPLIFIER: ANALYSIS OF NON IDEAL CASE
USE LINEAR ALGEBRA
NODE ANALYSIS
CONTROLLING VARIABLE IN TERMS OF NODE
VOLTAGES
TYPICAL OP - AMP : A  105 ,
Ri  108 , RO  10
R1  1k, R2  5k 
vO
v
 4.9996994 A    O  5.000
vS
vS
SUMMARY COMPARISON: IDEAL OP-AMP AND NON-IDEAL CASE
v  0
i  0
v  0
Ri    i  i  0
A    v   v
KCL @ INVERTING TERMINAL
NON-IDEAL CASE
REPLACE OP-AMP BY LINEAR MODEL
SOLVE THE RESULTING CIRCUIT WITH
DEPENDENT SOURCES
GAIN FOR NON-IDEAL CASE
0  v S 0  vO
v
R

0  O  2
R1
R2
vs
R1
THE IDEAL OP-AMP ASSUMPTION PROVIDES EXCELLENT APPROXIMATION.
(UNLESS FORCED OTHERWISE WE WILL ALWAYS USE IT!)
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 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  v1
v  v1  v  v1
INFINITE GAIN ASSUMPTION
INFINITE INPUT RESISTANCE
R1
R1  R2
vi 
v0  v0 
vi
R1  R2
R1
FIND GAIN AND INPUT RESISTANCE - NON IDEAL OP-AMP
v
vO
vi  v  v
COMPLETE EQUIVALENT
FOR MESH ANALYSIS
Ri
RO
v



vO
A(v  v )

DETERMINE EQUIVALENT CIRCUIT
USING LINEAR MODEL FOR OP-AMP
MESH 1
NOW RE-DRAW CIRCUIT TO ENHANCE
CLARITY. THERE ARE ONLY TWO LOOPS
MESH 2
CONTROLLNG VARIABLE IN TERMS OF LOOP
CURRENTS
vO   R2i2  R1 (i1  i2 )
THE SOLUTIONS
MATHEMATICAL MODEL
R1  v1 
 i1  1 ( R1  R2  RO )

i     ( AR  R ) ( R  R )  0 
 2

i
1
1
2  
MESH 1
MESH 2
i1 
R1  R2  RO
v1

i2 
 ( ARi  R1 )

CONTROLLNG VARIABLE IN TERMS OF LOOP
CURRENTS
vO   R2i2  R1 (i1  i2 )
INPUT RESISTANCE
Rin 
v1
i1
GAIN
G
vO
vi
REPLACE AND PUT IN MATRIX FORM
 R1
( R1  R2 )
  i1  v1 
 AR  R ( R  R  R ) i    0 
 
 i
1
1
2
O  2 
THE FORMAL SOLUTION
1
 R1
 v1 
 i1  ( R1  R2 )
i    AR  R ( R  R  R )  0 
 2  i
1
1
2
O   
  ( R1  R2  RO )( R1  R2 )  R1( ARi  R1)
R1 
( R  R2  RO )
Adj   1

  ( ARi  R1 ) ( R1  R2 )
vO  R1i1  ( R1  R2 )i2

R1 ( R1  R2  RO )
( R  R2 )( ARi  R1 )
v1  1
v1


A   ???
A      AR1Ri
Rin  
G
vO
R  R2
 1
v1
R1
A SEMI-IDEAL OP-AMP MODEL
This is an intermediate model, more accurate than the ideal op-amp
model but simpler than the linear model used so far
Ri   , RO  0, A  AO  i  i  0 v   v  !!
v
ve  vin
v
Replacement Equation
vO  AO ve  AO (v  v )
Non-inverting amplifier and semi-ideal model
v  v S
vO 
R2  R1
v (as before)
R1

vO
AO
R1

; 
v S 1  AO 
R!  R2
actual gain-ideal gain  1
GE 
1 A 
ideal gain
AO (v S  v )  vO (replaces v   v  )
O
Sample Problem
R
Set voltages?
i  0
v


Use infinite gain assumption

v
vO
iS
+
-
vS
v  vS

Find the expression for Vo. Indicate
where and how you are using the Ideal
OpAmp assumptions
v  vS
Use infinite input resistance assumption
and apply KCL to inverting input
vo  v
iS 
0
R
vo  vS  Ri S
Sample Problem
DRAW THE LINEAR EQUIVALENT CIRCUIT AND WRITE THE LOOP EQUATIONS
R
4. Redraw if necessary



Ri
iS
v
R
RO


vO
A(v  v )
i2
i1
iS
vo
RO
Ri
+
-

A(v + - v -)
1. Locate nodes
v
2. Erase Op-Amp
3. Place linear model
TWO LOOPS. ONE CURRENT SOURCE. USE MESHES
MESH 1 i1
 is
Ri (i2  iS )  ( R  RO )i2  A(v  v_ )
CONTROLLING VARIABLE v  v _  Ri (i2  i1 )
MESH 2
LEARNING EXTENSION
FIND GAIN AND VO
v  VS
v _  VS
VS 
i  0
“INVERSE VOLTAGE DIVIDER”
100k  1k
VO 
VS
1k
V
G  O  101
VS
VS  1mV  VO  0.101V
VO
VS
R2
R1
LEARNING EXAMPLE
UNDER IDEAL CONDITIONS BOTH CIRCUITS SATISFY
VO  8V1  4V2
If 1V  V1  2V , 2V  V2  3V
dc supplis are  10V
DETERMINE IF BOTH IMPLEMENTATIONS PRODUCE THE
FULL RANGE FOR THE OUTPU
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
COMPARATOR CIRCUITS
Some REAL OpAmps require
a “pull up resistor.”
ZERO-CROSSING DETECTOR
LEARNING BY APPLICATION
OP-AMP BASED AMMETER
NON-INVERTING AMPLIFIER
G 1
R2
R1
VI  RI I
 R 
VO  GVI  1  2  RI I
 R1 
LEARNING EXAMPLE
DC MOTOR CONTROL - REVISITED
CHOOSE NON-INVERTING
AMPLIFIER (WITH POWER
OP-AMP PA03)
1
Constraints: VM
RB
 4 (design eq.)
RA
 20V
Power dissipation
in amplifier  100mW
Significant power losses
Simplifying assumptions: Ri   , RO  0  Occur only in Ra, Rb
Worst case occurs when Vm=20
(20V )2
PMX 
 100mW  RA  RB  4000
RA  RB
RB
3
RA
One solution: RB  3k  , RA  1k 
Standard values at 5%!
DESIGN EXAMPLE: INSTRUMENTATION AMPLIFIER
V1
G
V2
DESIGN SPECIFICATIONS
G
VO
VO
 10
V1  V2
• “HIGH INPUT RESISTENCE”
• “LOW POWER DISSIPATION”
• OPERATE FROM 2 AA BATTERIES
MAX4240
VO  VX  VY
ANALISIS OF PROPOSED CONFIGURATION
VA  V1 ; VB  V2
Infinite gain
V1  V2 V1  VX

0
R
R1
V  V1 V2  VY
@B: 2

0
R
R2
@ A:
R 
R 
R
R


VO  V1  1  1   V2  1  2   V1 2  V2 1
R
R
R
R


SIMPLIFY DESIGN BY MAKING R1  R2  V   1  2 R1  (V  V )


O
2
R  1

DESIGN EQUATION: 2 R1  9 R
USE LARGE RESISTORS FOR LOW POWER e . g ., R  100k  , R1  R2  450k 
DESIGN EXAMPLE
DESIGN SPECIFICATION
VO
 10
Vin
Power loss in resistors should not exceed
100mW when 2V  Vin  2V
Design
Max Vo is 20V
VO
R2
 1
 10
equationS:
Vin
R1
PR  R
1
2
(20V )2

 100mW
R1  R2
 R2  9 R1
 R1  R2  4k 
Solve design equations (by trial and error if necessary)
R1  400
IMPLEMENT THE OPERATION VO  0.9V1  0.1V2
DESIGN EXAMPLE
DESIGN CONSTRAINTS
• AS FEW COMPONENTS AS POSSIBLE
• MINIMIZE POWER DISSIPATED
• USE RESISTORS NO LARGER THAN 10K
Given the function (weighted sum
with sign change) a basic weighted
adder may work
ANALYSIS OF POSSIBLE SOLUTION
V  0
R
R
 VO   V1  V2
V V
V
@V :  

0
R1
R2
O

R
1
R1 1
2
R2
R2  9 R1
DESIGN
EQUATIONS
R
 0.1
R2
R2  R1  R
SOLVE DESIGN EQUATIONS USING TRIAL AND ERROR IF NECESSARY
R2  10k ,5.6k ,...
ANALYZE EACH SOLUTION FOR OTHER CONSTRAINTS AND FACTORS; e.g.
DO WE USE ONLY STANDARD COMPONENTS?

DESIGN EXAMPLE
DESIGN 4-20mA TO 0 – 5V CONVERTER
1. CONVERT CURRENT TO VOLTAGE USING A RESISTOR
I
VI  RI
CANNOT GIVE DESIRED RANGE!
2. CHOOSE RESISTOR TO PROVIDE THE 5V CHANGE
… AND SHIFT LEVELS DOWN!
R
VMAX  VMIN
50

 312.5
I MAX  I MIN 0.020  0.004
1.25  VI  6.25
V 
MUST SHIFT DOWN BY 1.25V
(SUBSTRACT 1.25 V)
R2
VI
R1  R2
R1
V 
(VO  VSHIFT )  VSHIFT
R1  R2
V  V  VO  R2 VI  VSHIFT 
R1
LEARNING BY DESIGN
DOES NOT LOAD PHONOGRAPH
DETERMINE R2 , R1 SO THAT
IT PROVIDES AN AMPLIFICATION OF 1000
VO
R
 (1)(1  2 )
V1
R1
LEARNING EXAMPLE
RT  57.45e 0.0227T
UNITY GAIN
BUFFER
ONLY ONE LED
IS ON AT ANY
GIVEN TIME
COMPARATOR CIRCUITS
MATLAB SIMULATION OF TEMPERATURE SENSOR
WE SHOW THE SEQUENCE OF MATLAB INSTRUCTIONS USED TO OBTAIN THE PLOT
OF 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')