Transcript Op-Amp

Universal Collage Of Engineering
And Technology
Subject : Analog Electronics
NAME:
EN.NO:
Jay Bhavsar
 Yagnik Dudharejiya
 JAY Pandya
 Darshan Patel

130460109006
 130460109013
 130460109034
 130460109043

Guidance by : Prof. Kapil Dave
1
 Symbol
of op-amp
 Analyze of op-amp circuit
 Packages of op-amp
 Pin configuration of op-amp
 Applications of op-amp
 Frequency response of op-amp
 Design of op-amp
 Power supplies of op-amp
2
Circuit symbol of an op-amp
•Widely used
•Often requires 2 power supplies + V
•Responds to difference between two signals
3
Characteristics of an ideal op-amp
•Rin = infinity
•Rout = 0
•Avo = infinity (Avo is the open-loop gain, sometimes
A or Av of
the op-amp)
•Bandwidth = infinity (amplifies all frequencies
equally)
4
I+
V+
+
+
IV-
-
Vout = A(V+ - V-)
-
•Usually used with feedback
•Open-loop configuration not used much
5
Vout = A(V+ - V-)
Vout/A = V+ - VLet A
infinity
then,
V+ -V-
0
6
V+ = VI+ = I- = 0
Seems strange, but the input terminals to an
op-amp act as a short and open at the same time
7
•Write node equations at + and - terminals (I+ = I- = 0)
•Set V+ = V-
•Solve for Vout
8

Types of Packages:

Small scale integration(SSI)<10 components

Medium Scale integration of op-amp(MSI)<100 components

Large scale integration (LSI)>100 components

Very large scale integration (VLSI)>1000 components
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I2
I1 = (Vi - V- )/R1
I2 = (V- - Vo)/R2
I1
set I1 = I2,
(Vi - V-)/R1 = (V- Vo)/R2
but V- = V+ = 0
Gain of circuit determined by external components
Vi / R1 = -Vo / R2
Solve for Vo
Vo / Vi = -R2 / R1
10
 TYPES:

The flat pack

Metal can or Transistor Pack

The duel-in-line packages(DIP)
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Very popular circuit
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Current in R1, R2, and R3 add to current in Rf
(V1 - V-)/R1 + (V2 - V-)/R2 + (V3 - V-)/R3 = (V- - Vo)/Rf
Set V- = V+ = 0, V1/R1 + V2/R2 + V3/R3 = - Vo/Rf
solve for Vo, Vo = -Rf(V1/R1 + V2/R2 + V3/R3)
This circuit is called a weighted summer
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Offset Null
1
2
Non-inverting Input 3
Vcc- 4
Inverting Input
A741
8
7
6
5
NC
Vcc+
Output
Offset Null
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 As
a integrator
 As
a differentiator
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I2
I1 = (Vi - V-)/R1
I2 = C d(V- - Vo)/dt
I1
set I1 = I2,
(Vi - V-)/R1 = C d(V- Vo)/dt
Output is the integral of input signal.
CR1 is the time constant
but V- = V+ = 0
Vi/R1 = -C d(Vo)/dt
Solve for Vo
Vo = -(1/CR1)(  Vi dt)
16
17
NON-INVERTING CONFIGURATION
(0 - V-)/R1 = (V- - Vo)/R2
I
I
Vi
But, Vi = V+ = V-,
( - Vi)/R1 = (Vi - Vo)/R2
Solve for Vo,
Vo = Vi(1+R2/R1)
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Vi = V+ = V- = Vo
Vo = Vi
Isolates input
from output
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Write node equations using:
OR
Write node equations using:
V+ = V-
model, let A
I+ = I- = 0
Solve for Vout
infinity
Solve for Vout
Usually easier, can solve most
problems this way.
Works for every op-amp circuit.
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Rin = Vin / I, from definition
VI
V+
V
Rin = Vin / 0
Rin = infinity
21
Rin = Vin / I, from definition
I
I = (Vin - Vout)/R
V-
I = [Vin - A (V+ - V-)] / R
V+

Vout =
A(V+ - V-) But V+ = 0
I = [Vin - A( -Vin)] / R
Rin = VinR / [Vin (1+A)]
As A approaches infinity,
Rin = 0
22
Inverting configuration
Non-inverting configuration
Vi
Rin = 0 at
this point
Vo /Vi = - R2/R1
Vo/Vi = 1+R2/R1
Rin = R1
Rin = infinity
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DIFFERENCE AMPLIFIER
Use superposition,
set V1 = 0, solve for Vo
(non-inverting amp)
set V2 = 0, solve for Vo
(inverting amp)
Fig. A difference amplifier.
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DIFFERENCE AMPLIFIER
Vo1 = -(R2/R1)V1
Vo2 = (1 + R2/R1) [R4/(R3+R4)] V2
Add the two results
Vo = -(R2/R1)V1 + (1 + R2/R1)
[R4/(R3+R4)] V2
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Vo = -(R2/R1)V1 + (1 + R2/R1) [R4/(R3+R4)]V2
For Vo = V2 - V1
Set R2 = R1 = R, and set R3 = R4 = R
For Vo = 3V2 - 2V1
Set R1 = R, R2 = 2R, then
3[R4/(R3+R4)] = 3
Set R3 = 0
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When measuring Rin at one input,
ground all other inputs.
Rin at V1 = R1, same as inverting
amp
Rin at V2 = R3 + R4
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Add buffer amplifiers to the inputs
Rin = infinity at both V1 and V2
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where w0 = 1/RC
(a) Magnitude response of (single time constant) STC
networks of the low-pass type.
29
OPEN-LOOP FREQUENCY RESPONSE OF OP-AMP
Open-loop gain
at low
frequencies
Break frequency(bandwidth),
occurs where Ao drops 3dB
below maximum
Unity gain frequency, occurs
where Ao = 1 (A = 0dB)
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•Open-loop op-amp
•Inverting and non-inverting amplifiers
•Low-pass filter
•High-pass filter
31
FREQUENCY RESPONSE OF OPEN-LOOP OP-AMP
Open-loop op-amp: ft = Ao fb
where Ao is gain of op-amp
32
Inverting or noninverting amplifier:
ft = |A| fb, where A = gain of circuit
- 20 dB/dec
A = - R2 / R1,
inverting
A = 1 + R2/R1,
non-inverting
|A|
fb
ft
33
Z2
Z1
Vi
A = - Z2 / Z1
1  R2
R2
1
R1
A   sC
 
1  R2 R1 1  sCR2
sC
•At large frequencies A becomes zero.
•Passes only low frequencies.
34
Low-pass filter: C acts as a short at high
frequencies, gain drops to zero at high frequencies,
ft |A| fb. fb = 1/2pR2C
Due to external
C
Due to
Vi
- 20
fb
35
Z2
Z1
C
VVi
A = - Z2 / Z1
A 
R2
R1  1
sC
•At large frequencies A becomes - R2 / R1.
•Passes only high frequencies.
36
High-pass filter: C acts as an open at low
frequencies, gain is zero at low frequencies,
C
Vi
Due to external capacitor
fL = 1/2pR1CDue to op-amp
bandwidth
37
C
Design the circuit to obtain:
High-frequency Rin = 1KW
High-frequency gain = 40dB
lower 3 dB frequency = 100Hz
•Rin = R1 + 1/sC. At high frequencies, s becomes large, Rin  R1.
Let R1 = 1KW
•A = - R2 / (R1 + 1/sC). At high frequencies, s becomes large, A  R2 / R1 .
A = 40dB = 100,
100 = R2 / 1KW,
R2 = 100KW.
•fL = 1/2pR1C
C = 1/2p R1 fL,
1/2p(1KW)100 = 1.59mF
C=
38
20 DB/DECADE
(DUE TO CAPACITOR)
-20 DB/DECADE
(DUE TO OP-AMP)
FL = 100HZ
39
OUTPUT OF HIGH-PASS FILTER
IN EXAMPLE
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C2
C1
BANDPASS FILTER
• Both C2 and C1 act as shorts at high frequencies.
• C2 limits high-frequency gain
• C1 limits low-frequency gain
• The gain at midrange frequencies = - R2 / R1
•fL = 1/2pR1 C1
•fH = 1/2pR2 C2
-20 dB/decade
(due to C2)
bandwidth
fL
fH
41
Saturation:
Input must be small enough so the output remains less than the
supply voltage.
 Slew rate:
Maximum slope of output voltage. Response time of op-amps are
described by a slew rate rather than a delay.
42
 PSRR:(Power
supply rejection ratio)
 CMRR:(Common
mode rejection ratio)
43

Definition:
The change in an op-amp input offset voltage (Vios)
caused by variation in the supply voltage and it is called as
“power supply rejection ratio”.

It is also called a “SVRR”, AND “PSS”.

Equation:
PSRR:Vios/v
44
 Definition:
It is the ratio of common mode gain and
differential gain.
 Equation : Vcm=(V1+V2)/2

Vo=AdVd+AcmVcm
 Where Ad=Differential gain and

Vcm= Common mode gain
 CMRR=(Acm/Ad)
 Final equation :
Vo=[V1-V2+Vcm/CMRR]Ad
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www.google.com
 1. Op-Amp and Linear integrated Circuit technologyRamakant A Gayakwad, PHI Publication
 2. Digital Fundamentals by Morris and Mano, PHI
Publication
 3. Micro Electronics Circuits by SEDAR/SMITH.Oxford
Pub F.COUGHLIN, FREDERICK F. DRISCOLL

4. Operational Amplifier and Linear integrated Circuits By
K.LAL kishore.
 5. Fundamentals of Logic Design by Charles H. Roth
Thomson
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Thank you 
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