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Operational Amplifier

VCC

-VEE

VCC and VCC are the dc power supply

connections. Usually VCC -VEE .

Normally they are not shown in schematics.

Common Mode and Differential Voltages

vicm Common Mode Voltage 0.5(v1 v2) Thus

Common Mode Voltage is the average of voltages at

the non-inverting and inverting inputs vid

Differential Voltage (v1- v2) Thus Differential

Voltage is the difference of voltages at the

non-inverting and inverting inputs

Zout

Zin

Equivalent Circuit of an Op-Amp

- Ideally
- Zin ?
- Zo 0
- AOL ?
- No parasitic capacitance or inductance
- (bandwidth ?)
- Zero gain for common mode voltages.

Using Op-Amps

- For Stability Op-Amps use negative feedback. This

can be - achieved by feeding back the output to the

inverting input. Some - Examples Inverting, Non-inverting Amplifiers,

Integrators. - This does not mean positive feedback is not

useful. Comparators, - Schmitt Triggers are some examples where positive

feedback is - used.

Inverting Amplifiers

Analysis of Inverting Amplifiers

Since Zin ?, we can safely

assume that current entering the inverting

terminal is 0. This means the inverting input is

at zero potential but no current can flow through

it. Thus we say that the inverting input is at

virtual ground

Analysis of Inverting Amplifiers (2)

Thus we can say that

LP81

and

LP82

Applying KCL at the inverting input i1- i2 0

LP83

or

LP84

LP85

or

Analysis of Inverting Amplifiers (3)

Input impedance of inverting amplifier

Zin

LP86

Output impedance of inverting amplifier

Since output voltage expression (LP85) is

independent of load resistance, Zo 0

Inverting Adder

Inverting Adder(2)

Applying KCL at the inverting input

LP87

LP88

LP89

Let R1R2R3Rn R0 R

LP810

LP811

?

LP812

?

Example of inverting Adder

If v1 2 V, v2 -3 V, R1 100k ?, R2 33 k?, R0

47 k? in the above circuit calculate v0.

Example of inverting Adder(2)

Applying KCL at the inverting input

?

or

or

Output Impedance of an Inverting Amplifiers with

Finite AOL, rin, and r OL

Worked out on Greenboard

Non-Inverting Amplifiers

Non-Inverting Amplifiers(2)

Since AOL ?, for proper

amplification the voltage at the inverting input

(v-) must be equal to the voltage at the

non-inverting input (v). Otherwise, vo (v-

v-)AOL will be saturated.

? v- vin

LP813 Since Zin

?, ii 0

LP814

LP815

Equating LP813 and LP815

LP816

Voltage Follower

A voltage follower is a special case of a

non-inverting amplifier. In this the output is

directly connected to the inverting input.

Thus, v0 vin

LP817 This amplifier has very high

input impedance Zin Thus it is used as an input

buffer in many applications.

Differential Amplifier

- Uses concepts of both inverting and non-inverting

amplifier - Used as rear-end of Instrumentation amplifier

(Voltage followers - form the front-end)
- Needed in applications with high input impedance

and CMRR - (CMRR Common Mode Rejection Ratio)

Ad Differential gain Acm Common mode gain

Differential Amplifier Analysis

Since for linear operation input voltages at the

inverting and non-inverting input are equal

Rearranging

Choosing R2 R4 and R1 R3

- The configuration rejects any common mode signal

existing - between v1 and v2.

A Voltage to Current converter

Show that when Q1 and Q2 have high ? and working

in active region

where IL is the load current.

Solution to the Voltage to Current Converter

Circuit

U1 works like a voltage follower and thus its

inverting input is at vin. Thus the collector

voltage of Q1 ? Vcc-vin. U2 also works like a

voltage follower and thus its inverting input is

as at Vcc-vin.This implies that the emitter

current of Q2 ? (Vcc-( Vcc-vin))/R vin/R ?

IL

Positive Feedback

QuestionWhy wont the above circuit work as an

amplifier?

Positive Feedback(2)

Answer Because of positive feedback

To start with, v- v vin and

LP818

Let us say due to some disturbance v vin ?v

LP819

Then

Subsequently,

LP820

Positive Feedback(3)

Thus due to positive feedback, the disturbance

gets amplified and drives the op-amp to either

positive or negative saturation ( the output

goes near VCC or VEE).

Positive Feedback(4)

Example In the above circuit VCC 15V and

-VEE -15V. R1 R2 10 k?,Vin 5V. Initially v0

10V and v 5V. What happens to v0 if due

to some disturbance v becomes i) 5.01 V, ii)

4.99 V. The output Saturation voltage is ? 13V.

AOL 105.

Positive Feedback(4)

- In this case ?v v - v- 5.01-5 0.01V.
- Hence AOL ?v 1050.01 103. Since the output

can increase only - as far as 13V, v0 saturates to 13V.
- In this case ?v v - v- 4.99-5 -0.01V.
- Hence AOL ?v -1050.01 -103. Since the output

can increase only - as far as -13V, v0 saturates to -13V.

Negative feedback ,on the other hand, opposes any

disturbance and stabilizes amplifiers.

Op-Amp Integrator

Op-Amp Integrator (2)

Since the inverting input is at virtual ground

LP821

LP822

Applying KCL at the inverting input

i1i2 0

LP823

LP824

LP825

Example on Op-Amp Integrator

V

A rectangular waveform is applied to the input of

an integrator. If R 10 k? and C 0.1 ?F

sketch vo. Assume zero initial value for vo.

Example on Op-Amp Integrator (2)

RC 1040.110-6 10-3 sec

Thus at t 0.5 ms, vo -2(0.5-0)10-3/10-3

0 -1.0V

Example on Op-Amp Integrator (2)

Thus at t 1.5 ms, vo 2(1.5-0.5)10-3)/10-3

-1.0V 1V . etc.

Op-Amp Differentiator

Op-Amp Differentiator (2)

Since the inverting input is at virtual ground

LP826

LP827

Applying KCL at the inverting input

i1i2 0

LP828

LP829

LP830

Differentiators are avoided in practice as they

amplify noise

Imperfections and Limitations of Op-Amps

- Bias and Offset Currents
- Due to inherent manufacturing imperfections,

undesirable base or - gate currents (dc) enter the inverting and the

non-inverting inputs of the - op-amps. They are in the order of nA for BJT

based op-amps and pA - for FET based op-amps. If IB- is the magnitude of

this current entering - the inverting input and IB the current entering

the non-inverting input

Then bias current, IB

and offset current, Ioff

Normally,

- Offset voltage
- The output of an op-amp will usually not be zero

with zero input - voltage. This is because of input offset voltage.

Imperfections and Limitations of Op-Amps(2)

- Slew-rate (SR) limitations
- The magnitude of the rate of change of the

output voltage is limited. This - is called the slew-rate limitation

Example for a 741, SR 0.5 V/?s

Imperfections and Limitations of Op-Amps(3)

- Output voltage swing
- For a specified supply voltage, the output

voltage range is normally 2-3 - volts less. For example, when a 741 is supplied

with ? 15V, the - guaranteed output range is ? 12V.

Imperfections and Limitations of Op-Amps(4)

- Output current limit
- An Op-Amp can supply a specified maximum value of

current. For a 741 - it is ? 40 mA.

Imperfections and Limitations of Op-Amps(5)

- Full Power Bandwidth
- Frequency range (0-fFP) for which the Op-Amp can

produce an - undistorted sinusoidal output with its peak

amplitude equal to the guaranteed maximum output.

Relationship between SR and fFP

Let

for undistorted output

Clearly

At the upper limit

Imperfections and Limitations of Op-Amps(6)

or

or

Example for a LF 351, Vom 10V and SR 12V/

?s. What is the fFP for this amplifier?

fFP

Imperfections and Limitations of Op-Amps(7)

- Gain-bandwidth product
- Because of internal capacitances used in the

op-amps, there exists an upper limit on gain with

respect to the frequency of input signal. To be

more precise the gain-bandwidth product is

constant above a certain frequency of operation.

Under closed loop operation we normally trade-off

gain with bandwidth.(? maximum bandwidth at - unity gain.
- Example If for the following op-amp we can get

an open loop gain - of 100 dB (105) upto 40Hz, find its maximum band

width for a - closed loop gain of 40 (100) dB. (Ans 40 kHz).

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