Operational Amplifier Part 1

Operational Amplifier Part 1Mukesh N [email protected]

Mukesh N. Tekwani 2May 3, 2023

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 impedanceno current flows into inputs

Infinite voltage gaina voltage difference at the two inputs is

magnified to a very large extent in practice, voltage gain ~ 200000means difference between + terminal and

terminal is amplified by 200,000!


Op Amp Characteristics Explained

Infinite bandwidth In practice, bandwidth limited to few MHz

rangeslew 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


The Basic Inverting AmplifierR2

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 greaterAmplification is linear – i.e. distortion of o/p is lessGain 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


KCL0i

111 R

vRvvi inin

222 R

vRvvi outout


Solve for Vo

Amplifier gain:

21 Rv

Rv outin

1

2RRA

vvAin

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

+Vssaturation

saturation

VinB

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 earthR2

VinP

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


Education

Operational Amplifier Part 1