universal collage of engineering and technology subject : analog electronics
TRANSCRIPT
Universal Collage Of Engineering And Technology
Subject : Analog Electronics
NAME:EN.NO:
Jay Bhavsar Yagnik Dudharejiya JAY Pandya Darshan Patel
130460109006 130460109013 130460109034 130460109043
2Guidance by : Prof. Kapil Dave
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
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Circuit symbol of an op-amp
•Widely used•Often requires 2 power supplies + V•Responds to difference between two signals
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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)
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V+
V-
Vout = A(V+ - V-)+
-
+
-
•Usually used with feedback•Open-loop configuration not used much
I-
I+
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Vout = A(V+ - V-)
Vout/A = V+ - V-
Let A infinity
then,
V+ -V- 0
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V+ = V-
I+ = I- = 0
Seems strange, but the input terminals to an op-amp act as a short and open at the same time
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•Write node equations at + and - terminals (I+ = I- = 0)
•Set V+ = V-
•Solve for Vout
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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I1 = (Vi - V- )/R1
I2 = (V- - Vo)/R2
set I1 = I2,
(Vi - V-)/R1 = (V- - Vo)/R2
but V- = V+ = 0
Vi / R1 = -Vo / R2
Solve for Vo
Vo / Vi = -R2 / R1
Gain of circuit determined by external components
I1
I2
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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A741Offset Null
Inverting Input
Non-inverting Input
Vcc-
Vcc+
Output
Offset Null
NC
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As a integrator
As a differentiator
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I1 = (Vi - V-)/R1
I2 = C d(V- - Vo)/dt
set I1 = I2,
(Vi - V-)/R1 = C d(V- - Vo)/dt
but V- = V+ = 0
Vi/R1 = -C d(Vo)/dt
Solve for Vo
Vo = -(1/CR1)( Vi dt)
Output is the integral of input signal.
CR1 is the time constant
I1
I2
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NON-INVERTING CONFIGURATION
(0 - V-)/R1 = (V- - Vo)/R2
But, Vi = V+ = V-,
( - Vi)/R1 = (Vi - Vo)/R2
Solve for Vo,
Vo = Vi(1+R2/R1)
Vi
I
I
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Vi = V+ = V- = Vo
Vo = Vi
Isolates inputfrom output
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Write node equations using:
V+ = V-
I+ = I- = 0
Solve for Vout
Usually easier, can solve mostproblems this way.
Write node equations using:
model, let A infinity
Solve for Vout
Works for every op-amp circuit.
OR
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V
V-
V+
Rin = Vin / I, from definition
Rin = Vin / 0
Rin = infinity
I
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Vout = A(V+ - V-)
V-
V+
Rin = Vin / I, from definition
I = (Vin - Vout)/R
I = [Vin - A (V+ - V-)] / R
But V+ = 0
I = [Vin - A( -Vin)] / R
Rin = VinR / [Vin (1+A)]
As A approaches infinity,
Rin = 0
I
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Vi
Inverting configuration Non-inverting configuration
Vo/Vi = 1+R2/R1
Rin = infinity
Vo /Vi = - R2/R1
Rin = R1
Rin = 0 atthis point
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Fig. A difference amplifier.
DIFFERENCE AMPLIFIER
Use superposition,
set V1 = 0, solve for Vo (non-inverting amp)
set V2 = 0, solve for Vo (inverting amp)
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Add the two results
Vo = -(R2/R1)V1 + (1 + R2/R1) [R4/(R3+R4)] V2
DIFFERENCE AMPLIFIER
Vo1 = -(R2/R1)V1 Vo2 = (1 + R2/R1) [R4/(R3+R4)] V2
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For Vo = V2 - V1Set R2 = R1 = R, and set R3 = R4 = R
For Vo = 3V2 - 2V1Set R1 = R, R2 = 2R, then 3[R4/(R3+R4)] = 3Set R3 = 0
Vo = -(R2/R1)V1 + (1 + R2/R1) [R4/(R3+R4)]V2
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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 inputsRin = infinity at both V1 and V2
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(a) Magnitude response of (single time constant) STC networks of the low-pass type.
where w0 = 1/RC
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OPEN-LOOP FREQUENCY RESPONSE OF OP-AMP
Unity gain frequency, occurswhere Ao = 1 (A = 0dB)
Break frequency(bandwidth), occurs where Ao drops 3dB below maximum
Open-loop gainat low frequencies
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•Open-loop op-amp
•Inverting and non-inverting amplifiers
•Low-pass filter
•High-pass filter
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FREQUENCY RESPONSE OF OPEN-LOOP OP-AMP
Open-loop op-amp: ft = Ao fbwhere Ao is gain of op-amp
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Inverting or noninverting amplifier:ft = |A| fb, where A = gain of circuit
A = - R2 / R1, inverting
A = 1 + R2/R1, non-invertingfb ft
|A|
- 20 dB/dec
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Vi
A = - Z2 / Z1
Z2
Z1 sCR2 1R
R
R
1
RsC1
RsC1
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A
•At large frequencies A becomes zero.
•Passes only low frequencies.
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Vi
C
Low-pass filter: C acts as a short at highfrequencies, gain drops to zero at high frequencies,ft |A| fb. fb = 1/2pR2C
Due to external Due to
fb
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C
VVi
A = - Z2 / Z1
Z2
Z1
•At large frequencies A becomes - R2 / R1.
•Passes only high frequencies.
sC1 R
R1
2
A
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C
Vi
High-pass filter: C acts as an open at lowfrequencies, gain is zero at low frequencies,
fL = 1/2pR1C
Due to external capacitor Due to op-amp
bandwidth
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C Design the circuit to obtain:
High-frequency Rin = 1KWHigh-frequency gain = 40dBlower 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, C = 1/2p(1KW)100 = 1.59mF
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FL = 100HZ
20 DB/DECADE(DUE TO CAPACITOR)
-20 DB/DECADE(DUE TO OP-AMP)
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OUTPUT OF HIGH-PASS FILTER IN EXAMPLE
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BANDPASS FILTERC2
C1 • 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
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Saturation:Input must be small enough so the output remains less than thesupply voltage.
Slew rate:Maximum slope of output voltage. Response time of op-amps aredescribed by a slew rate rather than a delay.
PSRR:(Power supply rejection ratio)
CMRR:(Common mode rejection ratio)
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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
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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 technology-
Ramakant 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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