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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