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Analog Electronics
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## Operational amplifier as integrator and differentiator

Operational amplifier which is called also called as op-amp has a key role in many electronic applications due to its special characteristics. Name itself indicates that it can perform operations. By using op-amp we can perform different operations like addition, subtraction, multiplication, differentiation and integration. Of these op-amp application as integrator and differentiator is very common.

Before going to see op-amp as integrator and differentiator let us first understand working principle of operational amplifier.

### Operational amplifier working principle:

Let us see the symbol of operational amplifier and its terminals before going to see working of op-amp.
It has two input terminals one is marked negative and other as positive and one output terminal. The input terminal which is marked negative is called inverting input because when we apply an input signal to this inverting input we get a phase shift of 180° in the amplified output signal with respect to the applied input signal. The input terminal which is marked positive is called Non-inverting input because when we apply an input signal to this Non-inverting input there is no phase shift between input signal and amplified output signal.

It has two input power supply terminals +Vs and -Vs. +Vs is connected to positive terminal of battery and -Vs is connected to negative terminal of other battery. If we require a ground we need to provide ground separately as there is no common ground provided in op-amp.

### Open loop operation of op-amp:

In this open loop operation we apply two input signals one at inverting input and other at Non-inverting input as shown in the figure.

So it has differential input(since one is + and other is -) which means difference of two applied input signals and one output is obtained. The gain of open loop operation of op-amp is given by

Gain = output / input

Aol = Vo / V1 - V2

so output voltage is

Vo is output voltage

V1 is voltage at non-inverting terminal

V2 is voltage at inverting terminal

V1 - V2 is differential input voltage

Aol is open loop gain

Open loop gain is very high even a small signal given at input amplifies to large amount but its value will not exceed the supply voltage of op-amp as it obeys law of conservation of energy.

### Closed loop operation of op-amp:

If we introduce a feed back in the circuit it is called closed loop operation. Here a part of output signal is fed back to one of the input terminals. The terminal where feed back is given two signals are present simultaneously one is feed back signal and the other is original applied signal it can be seen in the following diagram.
If we apply feed back signal from output to non-inverting terminal it is called positive feed back which is used in oscillator circuits. And if we apply feed back signal from output to inverting terminal it is called negative feedback here phase shift is present between applied signal and feed back signal. This negative feed back is used for amplifier circuits. Closed loop gain of op-amp is given by

Acl = Vo / V1 - V2

Now output voltage will be

where Vd = V1 - V2.

Acl = closed loop gain.

As we have learnt operation of op-amp now we can clearly understand the application of op-amp as differentiator and integrator.

### Op-amp as integrator:

When an operational amplifier work as integrator we get output as integration of voltage with respect to time. Here we use capacitors at a right place in the circuit which helps to perform integration of applied input voltage. The arrangement can be seen from the following diagram.

The above circuit is called an ideal op-amp integrator circuit.

Here the negative feed back is taken and a capacitor is connected between output terminal and inverting input terminal. Because of negative feedback node X is at virtual ground and if the input voltage is 0 V then no current flows through input resistance Rin then the capacitor will remain uncharged. So we get output as 0 V.

Now if we apply a constant positive DC voltage at the input then  we get a linearly falling voltage at output. If we apply a constant negative DC voltage at input then we get a linearly raising voltage at output. And this rate of change of output voltage will be directly proportional to input voltage.

### Output voltage calculation:

Now we calculate the output voltage of this circuit.

According to virtual ground concept as non-inverting terminal is grounded then node X will also be at ground potential

VX = VY = 0
For an ideal op-amp input impedance is high so input current is very less and the current flowing through resistor R1 also flows through capacitor C.

Input current equation is given by,

I = (Vin – VX) / R1
I = Vin / R1.

output current equation is given by

I = C [d(VX – Vout)/dt] = -Cf[d(Vout)/dt]

Now we equate both current equations as current flowing through resistance R1 (input) and capacitor C (output) are same(since op-amp has high input impedance all current flows through capacitor). We get,

[Vin / R1] = – C [d(Vout)/dt]

Apply integration on both sides, now we get,

The above equation clearly says that output voltage is - 1 / R1 . C times the integration of input voltage. This shows that op-amp acts as an integrator by this circuit arrangement. Here R1.C is called integrator time constant and negative sign shows that there is a phase shift of 180° between input and output voltage. The phase shift is because we have applied signal to inverting terminal.

We use integrator circuit to convert a square wave input to triangular wave output as shown in the following circuit.

As discussed when we apply constant positive dc voltage we get a falling output voltage at linear rate and if we apply constant negative dc voltage we get a rising output voltage at linear rate(if a step signal is integrated we get ramp signal).

An integrator op-amp circuit acts as low pass filter it attenuates high frequency signals.

### OP-amp as differentiator:

A differential op-amp has output voltage which is proportional to rate of change of input voltage. This op-amp can act as differentiator by keeping a capacitor in series with the input voltage source. This can be seen from the diagram given below

The capacitor acts as open circuit for dc input. Here we ground the non-inverting terminal of op-amp and the inverting input terminal is connected to output through a feed back resistor Rf. This makes the circuit to behave as voltage follower. The input current to op-amp is very less due to high input impedance of op-amp we have same current through capacitor and resistor Rf.

### Output voltage calculation:

Now we calculate the output voltage of this circuit.

According to virtual ground concept as non-inverting terminal is grounded then node X will also be at ground potential

VX = VY = 0

Input current is given by,

I = C [d(Vin-Vx)/dt] = C [d(Vin)/dt]

output current is given by,

I = -{(Vout-Vx)/Rf} = -{Vout/Rf}

Now we equate both equations as current through capacitor and resistor are same.

C{d(Vin)/dt} = -Vout/Rf

Vout = -C.Rf {d(Vin)/dt

The above equation clearly says that output voltage is - C.Rf  times the differentiation of input voltage. This shows that op-amp acts as adifferentiator by this circuit arrangement.     Here    Rf.C is called differentiator time constant and negative sign shows that there is a phase shift of 180° between input and output voltage. The phase shift is because we have applied signal to inverting terminal.

Here if we give a square wave input to differentiator the output has to be zero( since differentiation of constant is zero) but we get negative and positive spikes because input signal takes time to change from 0 to Vm. At constant positive DC input we get negative spike at output and at constant negative DC input we get positive spike at output. This can be seen from the following diagram.

If we give a sin wave as input to differentiator we get cos wave as output.

Vout = -C.Rf {d(Vm sin ωt)/dt}

Vout = – Vm. ω. cos ωt consider(C.Rf = 1)

This can be seen in the following diagram

A differential op-amp circuit acts as high pass filter it attenuates low frequency signals.

In this post we have learnt op- amp as integrator and op-amp as differentiator.

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## Barkhausen Criterion of Oscillators

Consider a basic inverting amplifier with an Open loop gain A.The feedback network attenuation factor B is less than unity. As basic amplifier is inverting. it produces a phase shift of 180° between input and output as shown in the below figure.

Now the input Vi applied to the amplifier is to be derived from its output Vo using feedback network.

But the feedback must be positive Le. the voltage derived from output using feedback network must be in phase with V‘. Thus the feedback network must introduce a phase shift of 180° while feeding back the voltage from output to input. This more: positive feedback.

Consider a fictitious voltage Vi applied at the input of the amplifier. Hence we get,

Vo=AVi    -----------------(1)

The feedback factor β decides the feedback to be given to input,

Vf=βVo    -----------------(2)

Substituting equation (1) into equation (2) we get,

Vf=βAVi    -----------------(3)

For the oscillator, we want that feedback should drive the amplifier and hence V, must act as Vi. From equation (3) we can write that, V, is sufficient to act as V, when.

Aβ | = 1

And the phase of Vf is same as Vi i.e. feedback network should introduce 180°phase shift in addition to 180° phase shift introduced by inverting amplifier. This ensures positive feedback. So total phase shift around a loop is 360°.  In this condition, Vf drives the circuit and without external input circuit works as an oscillator.  The two conditions discussed above, required to work the circuit as an oscillator are called Barkhamen Criterion for Oscillation.

The Barkhausen Criterion states that
1.The total phase shift around a loop,as the signal proceeds from input through amplifier, feedback network back to input again. completing a loop, is precisely 0° or 360°.
2. The magnitude of the product of the open loop gain of the amplifier (A) and Magnitude of the feedback factor-β is unity Le. I Aβ I =1.