Operational Amplifier Part 1 Mukesh N Tekwani tekwani@ email.com

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.

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 V 1 Non-Inverting i/p V 2

Circuit Symbol

Internal Block Diagram

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

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!

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

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

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 V o given by: V o = A (V 2 – V 1 ) A is called as the gain of the amplifier. V 1 is the voltage applied at the inverting input, V 2 is the voltage applied at the non-inverting input,

The op-amp can be regarded as a device which generates an voltage V o given by:

V o = A (V 2 – V 1 )

A is called as the gain of the amplifier.

V 1 is the voltage applied at the inverting input,

V 2 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 10 5 . But the gain decreases 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 10 5 .

But the gain decreases with frequency.

Variation of Output voltage with V 1 V o = A (V 2 – V 1 )  When V 2 = 0, V o = -AV 1 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

V o = A (V 2 – V 1 )

 When V 2 = 0, V o = -AV 1

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 V 2 V o = A (V 2 – V 1 )  When V 1 = 0, V o = AV 2 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

V o = A (V 2 – V 1 )

 When V 1 = 0, V o = AV 2

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 V o = A (V 2 – V 1 ) If V 2 > V 1 , V o is positive If V 2 < V 1 , V o is negative If V 2 = V 1 , V o is zero

V o = A (V 2 – V 1 )

If V 2 > V 1 , V o is positive

If V 2 < V 1 , V o is negative

If V 2 = V 1 , V o 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 V 1 and V 2 is zero: V 2 – V 1 = 0

Infinite input resistance means the current into the inverting input is zero:

i - = 0

Infinite gain means the difference between V 1 and V 2 is zero:

V 2 – V 1 = 0

The Basic Inverting Amplifier R 2 V in – + + – V out R 1 + – I 1 I 2 Resistor used to control amplification

How to Calculate the Gain For an Inverting amplifier: Gain = -R 2 / R 1 Example : if R 2 is 100 kilo-ohm and R 1 is 10 kilo-ohm, Gain = -100 / 10 = -10 If the input voltage is 0.5V then the output voltage would be V in x Gain: V out = 0.5V X -10 = -5V

For an Inverting amplifier:

Gain = -R 2 / R 1

Example : if R 2 is 100 kilo-ohm and R 1 is 10 kilo-ohm, Gain = -100 / 10 = -10 If the input voltage is 0.5V then the output voltage would be V in x Gain: V out = 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. R 2 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 .

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.

R 2 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.

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: i 1 + i 2 + i - =0 – R 1 R 2 i 1 i - i 2

Apply KCL at the inverting input:

i 1 + i 2 + i - =0

KCL

Solve for V o Amplifier gain: Thus, Gain of an op-amp depends only on the two resistances and not on the op-amp characteristics

Amplifier gain:

Assumptions made in deriving gain equation Each input draws zero current from the signal source. Typically, i/p current is 1  A That is, input impedances are infinite The i/ps are both at the same potential if the op-amp is not saturated.

Each input draws zero current from the signal source.

Typically, i/p current is 1  A

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 V o -V s +V s saturation saturation Vin B A

Transfer Characteristics of a Non-inverting Amplifier V o -V s +V s saturation saturation V 2 – V 1 V 2 > V 1 V 2 < V 1 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. V s or -V s

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. V s or -V s

Transfer Characteristics of an OpAmp V s -V s V o Value V 0 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 = 10 5 (Remember A = V o / V in ) So, max i/p voltage is V in = V o / A  V in = + 9 / 10 5 = + 90  V This is the maximum input voltage swing. A smaller value of A would allow greater input.

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 = 10 5 (Remember A = V o / V in )

So, max i/p voltage is V in = V o / A

 V in = + 9 / 10 5 = + 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 &quot;flattened&quot; at the top and bottom.

Problem 1: In this circuit, we want a gain of ten. If R 1 is 5 K ohm, what is the value you need to use for R 0 ? Give your answer in ohms. 50,000 ohm

In this circuit, we want a gain of ten. If R 1 is 5 K ohm, what is the value you need to use for R 0 ? Give your answer in ohms.

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

In this circuit, you have it set up for a gain of -10. The input voltage is 0.24v. What is the output voltage?

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

In this circuit, Ro and R1 values are shown. The input signal is also shown. Sketch the o/p signal.

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.

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

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 R 2 V in P Q – + + – V out R 1 + – I 1 I 2 V Q V P

Virtual earth In the previous figure, V Q = 0 and  V P = 0 P is called a virtual earth or ground point even though it is not connected to the ground.

In the previous figure, V Q = 0 and  V P = 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 – + V i V o R f R i

Non-inverting Amplifier The output (Vo) is in phase with the input. R f and R i form a voltage divider circuit. A fraction of o/p voltage (Vo) developed across R f 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 + R f R i There is no virtual earth at the non-inverting i/p terminal.

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

R f and R i form a voltage divider circuit.

A fraction of o/p voltage (Vo) developed across R f 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 + R f

R i

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

Voltage Follower – + V i V o

Voltage Follower This is a special case of the non-inverting amplifier. In case of non-inverting amplifier, gain A = 1 + R f R i If we set R f = 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 V i without drawing any current.

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

In case of non-inverting amplifier, gain

A = 1 + R f

R i

If we set R f = 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 V i 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

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

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

Thank You