Operationalamplifier

Operational

Amplifier

Part 1

Mukesh N Tekwani

[email protected]

Mukesh N. Tekwani 2 May 5, 2014

Operational Amplifier

Originally an op-amp was an electronic

circuit that could carry out mathematical

operations of addition, subtraction,

differentiation and integration.

Hence the word “operational”

Op-amp is used to amplify DC and AC

signals.


Operational Amplifier Symbol

Circuit Symbol

-

+

+ve supply

-ve supply

output

Inverting i/p

V1

Non-Inverting

i/p V2


Internal Block Diagram

Vi

Ri

AVi

Ro

Vo

+

_

+

_

+

_

Vp

Vn

ip

in

+

_


Characteristics of Ideal Op-Amp

Infinite input impedance (about 2 Mohm)

Low output impedance (about 200 ohm)

Very large voltage gain at low frequency

Thus, small changes in voltages can be amplified by using an op-amp

Infinite bandwidth (all frequencies are amplified by same factor

No slew rate – no delay between change in i/p and changes in o/p


Op Amp Characteristics Explained

Infinite input impedance

no current flows into inputs

Infinite voltage gain

a voltage difference at the two inputs is magnified to a very large extent

in practice, voltage gain ~ 200000

means difference between + terminal and terminal is amplified by 200,000!


Op Amp Characteristics Explained

Infinite bandwidth

In practice, bandwidth limited to few MHz

range

slew rate limited to 0.5–20 V/s


Op Amp Slew Rate Explained The o/p of an op amp does not change

instantaneously.

The rate of change of o/p of an op amp is

limited (about 0.5 V/ sec)

So, if we want to change the o/p voltage from 0

to 10 V, it would take 20 s


Op Amp Slew Rate Explained


Operational Amplifier Without Feedback

The op-amp can be regarded as a device

which generates an voltage Vo given by:

Vo = A (V2 – V1)

A is called as the gain of the amplifier.

V1 is the voltage applied at the inverting input,

V2 is the voltage applied at the non-inverting input,


Variation of Gain with Frequency

The value of gain A depends on the

frequency of the i/p signal and is very high at

low frequencies.

At DC, (f = 0 Hz), gain A is about 105.

But the gain decreases with frequency.


Variation of Output voltage with V1

Vo = A (V2 – V1)

When V2 = 0, Vo = -AV1

So, the output voltage is out of phase with

the input voltage applied to the inverting

input.

That is why it is called the “inverting” input


Variation of Output voltage with V2

Vo = A (V2 – V1)

When V1 = 0, Vo = AV2

So, the output voltage is in phase with the

input voltage applied to the non-inverting

input.

That is why it is called the “non-inverting”

input


Variation of Output with Input Voltages

Vo = A (V2 – V1)

If V2 > V1, Vo is positive

If V2 < V1, Vo is negative

If V2 = V1, Vo is zero


Consequences of Ideal characteristics

Infinite input resistance means the current

into the inverting input is zero:

i- = 0

Infinite gain means the difference between V1

and V2 is zero:

V2 – V1 = 0

Mukesh N. Tekwani 16 May 5, 2014 16

The Basic Inverting Amplifier

R2

Vin

–

+ +

–

Vout

R1

+

–

I1

I2

Resistor used to control amplification


How to Calculate the Gain

For an Inverting amplifier:

Gain = -R2 / R1

Example : if R2 is 100 kilo-ohm and R1 is 10 kilo-ohm, Gain = -100 / 10 = -10 If the input voltage is 0.5V then the output voltage would be Vin x Gain: Vout = 0.5V X -10 = -5V


Inverting Amplifier

The i/p voltage to be amplified is fed to the inverting i/p

A fraction of the o/p signal is fed back to the op-amp

through the inverting i/p.

R2 is the feedback resistance in this circuit

Since we have used the inverting i/p, the o/p is out of

phase with the i/p signal.

This process is called negative feedback.


Inverting Amplifier

It is called negative feedback because the overall gain

of the amplifier reduces.

So why use negative feedback if gain is reduced?

The gain is constant over a wide range of input

frequencies and input voltages.

Stability is greater

Amplification is linear – i.e. distortion of o/p is less

Gain is independent of the characteristics of op

amp.


Solving the Amplifier Circuit

Apply KCL at the inverting input:

i1 + i2 + i-=0

–

R1

R2

i1 i-

i2


KCL

0i

111

R

v

R

vvi inin

22

2R

v

R

vvi outout


Solve for Vo

Amplifier gain:

21 R

v

R

v outin

1

2

R

RA

v

vA

in

out

Thus, Gain of an

op-amp

depends only

on the two

resistances and

not on the op-

amp

characteristics


Assumptions made in deriving gain equation

Each input draws zero current from the signal

source.

Typically, i/p current is 1A

That is, input impedances are infinite

The i/ps are both at the same potential if the op-

amp is not saturated.


Transfer Characteristics of Inverting Amplifier

Vo

-Vs

+Vs saturation

saturation

Vin B

A


Transfer Characteristics of a Non-inverting Amplifier

Vo

-Vs

+Vs

saturation

saturation

V2 – V1

V2 > V1

V2 < V1

B

A


Transfer Characteristics of an Op-Amp

The output (Vo) is directly proportional to the

input only within the range AOB. In this region,

the op-amp behaves linearly. There is very little

distortion of the amplifier output.

If the inputs are outside this linear range, then

saturation occurs. That is output is close to the

maximum value it can have i.e. Vs or -Vs


Transfer Characteristics of an OpAmp

Vs

-Vs

Vo

Value V0 might have for an ac i/p if

opamp did not saturate


Transfer Characteristics of an OpAmp

Consider an opamp connected to +9 V supply.

The o/p voltage can never exceed these values.

max value of o/p voltage can be +9V or -9V

Let A = 105 (Remember A = Vo / Vin )

So, max i/p voltage is Vin = Vo / A

Vin = +9 / 105 = + 90 V

This is the maximum input voltage swing.

A smaller value of A would allow greater input.


Saturation Effect in Op Amp Suppose gain is -10. Assume the input is a signal of amplitude of 1.4v. We would

expect the output of the amplifier to be a signal of amplitude of 14V because the

amplitude of the input is 1.4v and the gain is -10. But, if you take saturation into

account, you will get a signal that is "flattened" at the top and bottom.


Problem 1:

In this circuit, we want a gain of ten. If R1 is 5 K

ohm, what is the value you need to use for

R0? Give your answer in ohms.

50,000 ohm


Problem 2: In this circuit, you have it set up for a gain of -10. The input

voltage is 0.24v. What is the output voltage?

Gain = - Vo / Vi

Vo = Gain x Vi

Vo = (-10) x 0.24

Vo = -2.4 V


Problem 3: In this circuit, Ro and R1 values are shown. The input signal

is also shown. Sketch the o/p signal.

10 K ohm

2.7 K ohm


Problem 3:


Problem 3: • Gain A = Ro / R1

So, A = - 10 K / 2.7 K = -3.7

Amplitude of i/p signal is 4 V

So max o/p voltage is Vo = A x Vin

Vo = 3.7 x 4 = 14.8 V

But power supply is only +9V

So 9V is the max o/p the amplifier can provide.


Problem 3: • Amplifier is saturated

• It will remain saturated as long as size of i/p

voltage is greater than 9V / 3.7 = 2.4 V

• That is why we observe that the o/p gets

clipped as soon as the i/p rises above 2.4 V


Concept of virtual earth

R2

Vin P

Q

–

+ +

–

Vout

R1

+

–

I1

I2

VQ

VP


Virtual earth

In the previous figure, VQ = 0 and VP = 0

P is called a virtual earth or ground point even

though it is not connected to the ground.


Non-inverting Amplifier


Non-inverting op amp

–

+

Vi

Vo

Rf

Ri


Non-inverting Amplifier

The output (Vo) is in phase with the input.

Rf and Ri form a voltage divider circuit.

A fraction of o/p voltage (Vo) developed across Rf is fed back to the inverting i/p

This fraction is called feedback factor and is given by

= Ri / (Ri + Rf) Gain of this amplifier is:

A = 1 + Rf

Ri

There is no virtual earth at the non-inverting i/p terminal.


Voltage Follower

–

+

Vi

Vo


Voltage Follower

This is a special case of the non-inverting amplifier.

In case of non-inverting amplifier, gain

A = 1 + Rf

Ri

If we set Rf = 0, A = 1 (unity gain)

This is called voltage follower because the o/p voltage is locked to the i/p voltage (both are same)

Advantage: op amp has very high i/p impedance so it can measure Vi without drawing any current.


Characteristics of Voltage Follower

This is a special case of the non-inverting

amplifier.

Gain A = 1

The o/p voltage “follows” the i/p voltage

Op amp has very high i/p impedance and very

low i/p impedance


Voltage Follower used for measuring charge

Test Plate


Voltage Follower used for measuring charge

This circuit uses a capacitor to make a charge-measuring device.

If a charged object touches the test plate, it will transfer charge to the capacitor.

The p.d. between the plates of the capacitor rises

If the capacitor is connected directly to a voltmeter, this charge will drain away through the meter and incorrect reading would be obtained.

Op-amp has very high i/p impedance and so practically no charge is removed from the capacitor and yet measured by the voltmeter


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Operationalamplifier