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CIRCUITS WITH 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
COMMERCIAL PACKAGING OF TYPICAL OP-AMP
LMC 6294 DIP
OP-AMP ASSEMBLED ON PRINTED CIRCUIT BOARD
DIMENSIONAL DIAGRAM LM 324
PIN OUT FOR LM324
CIRCUIT SYMBOL
FOR AN OP-AMP
LINEAR MODEL
OUTPUT RESISTANCE
INPUT RESISTANCE
TYPICAL VALUES
Ri : 105   1012 
RO : 1  50
A : 105  107
GAIN
CIRCUIT WITH OPERATIONAL AMPLIFIER
LOAD
OP-AMP
DRIVING CIRCUIT
COMMERCIAL OP-AMPS AND THEIR MODEL VALUES
MANUFACTURER
National
National
Maxim
PART No
LM324
LMC6492
MAX4240
A
100,000
50,000
20,000
Ri[MOhm]
1
10
45
Ro[Ohm]
20
150
160
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 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
NEXT WE EXAMINE THE SAME CIRCUIT
WITHOUT THE ASSUMPTION OF IDEAL
OP-AMP
REPLACING OP-AMPS BY THEIR LINEAR MODEL
CONNECT THE EXTERNAL COMPONENTS
LABEL THE NODES FOR TRACKING
b - d
b - a
DRAW THE LINEAR EQUIVALENT FOR OP-AMP
REDRAW CIRCUIT FOR INCREASED CLARITY
R2
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
IDEAL OP-AMP ASSUMPTIONS
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
IDEAL OP-AMP CONDITIONS
KCL @ 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  
KCL @ NON INVERTING TERMINAL
R4  R2 , R3  R1  vO 
R2
(v2  v1 )
R1
LEARNING EXAMPLE: USE IDEAL OP-AMP
FIND vO
Which voltages are set?
v1  v1, v2  v2
What voltages are also known due to
infinite gain assumption?
Use now the infinite resistance assumption
i  0
v1
KCL@v1
v1
v2
KCL@v2
va
v2
v2
CAUTION: There could be
currents flowing OUT of
the OpAmps
THE
CIRCUIT REDUCES TO THIS vO , va
Solve for vo
to be determined
LEARNING EXTENSION
v0
NONINVERTING AMPLIFIER - IDEAL OP-AMP
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
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 :
FIND GAIN AND INPUT RESISTANCE - NON IDEAL OP-AMP
v1


v

Ri

v
RO
vO
A( v   v  )
ADD THE INPUT VOLTAGE SOURCE
R2
DETERMINE EQUIVALENT CIRCUIT
USING LINEAR MODEL FOR OP-AMP

v

Ri

v
THE OP-AMP
RO
vO
A( v   v  )
v1

v

Ri
v

R1
RO
vO
A( v   v  )
AND THE EXTERNAL RESISTORS
NOW RE-DRAW CIRCUIT TO ENHANCE
CLARITY. THERE ARE ONLY TWO LOOPS
NON-INVERTING OP-AMP. NON IDEAL OP-AMP
COMPLETE EQUIVALENT
FOR MESH ANALYSIS

vO

R2
MESH 1
MESH 2

v1

v


Ri
v
R1
RO
vO
A( v   v  )
CONTROLLNG VARIABLE IN TERMS OF LOOP
CURRENTS
vO   R2i2  R1 (i1  i2 )
COMPLETE EQUIVALENT CIRCUIT
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
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. Place remaining components

v


R
i2
i1
vO
iS
RO
Ri
iS
+
-

1. Locate nodes
vo
A(v + - v -)
vo
v
v
2. Place the nodes in
linear circuit model
RO TWO LOOPS. ONE CURRENT SOURCE. USE MESHES
3. Place linear model
Ri
MESH 1 i1
+
-
MESH 2
 is
Ri (i2  iS )  ( R  RO )i2  A(v  v_ )
CONTROLLING VARIABLE
v
v  v_  Ri (i2  i1 )
LEARNING EXTENSION
FIND GAIN AND VO
v  VS
v _  VS
i  0
VS 
“INVERSE VOLTAGE DIVIDER”
100k  1k
VO 
VS
1k
V
G  O  101
VS
VS  1mV  VO  0.101V
VO
VS
R2
R1
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 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')
EXAMPLE OF TRANSFER CURVE SHOWING SATURATION
v  v1
i  0
v  v1
THIS SOURCE CREATES THE OFFSET
THE TRANSFER CURVE
OUTPUT CANNOT
EXCEED SUPPLY
(10V)
KCL @ v_
IN LINEAR RANGE
OFFSET