3.1 Operational Amplifiers The Ideal Op Amp | unit 3 operational amplifiers

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BEX

Module 03

Operational Amplifiers

3.1 Operational Amplifiers: The Ideal Op Amp

It is commonly known as Op-amp.

It requires an external power source hence is an active device.

It is a versatile device that can be used to amplify ac as well as dc input signals.

It can perform mathematical operations like addition, subtraction, multiplication, integration etc.

Hence was named as Operational Amplifier due to its capability of performing mathematical operations.

When external feedback is provided then the op-amp can be used as ac-dc signal amplifier, oscillator, regulator, active filter etc.

Schematic Symbol

V+ : voltage at inverting input

V- : voltage at non-inverting input

A : voltage gain of op-amp

Characteristics of ideal op-amp

An ideal op-amp is a differential input, single output device.

It has the following characteristics:

Infinite input resistance [Ri =∞]

Zero output resistance [RO = 0]

Infinite voltage gain [AV = ∞]

Infinite bandwidth [BW = ∞]

Infinite Common Mode Rejection Ratio

Infinite slew rate

Zero offset [ ie,V1 = V2 , VO =0]

The above characteristics do not change with the change in temperature.

3.2 Inverting and Non – Inverting configurations

Inverting Amplifier

Fig : inverting op-amp (Ref: 1)

As seen in the above figure, input is applied to only one terminal i.e. inverting input terminal.

The other input terminal is supplied 0V i.e. grounded.

Hence,

V1 = 0V and V2 = Vin

Therefore output voltage Vout is given by,

Vout = A( - Vin )
where, A is the voltage gain of op-amp.

The negative sign implies that the output voltage is 1800 out of phase w.r.to the input voltage.

Hence, the inverting amplifier amplifies the input signal by voltage gain A and inverts it at the output.

Non-Inverting Amplifier

Fig : Non inverting amplifier (Ref: 1)

As seen in the above figure, input is applied to only one terminal i.e. non-inverting input terminal.

The other input terminal is supplied 0V i.e. grounded.

Hence,

V1 = Vin and V2 = 0V

Therefore output voltage Vout is given by,

Vout = AVin

where, A is the voltage gain of op-amp.

Here, the output voltage is more than the input voltage by voltage gain A and are also in phase with each other.

3.3 Op amp application in Integration

Fig: Integrator

It is a circuit which provides output voltage Vo as an integral form of input voltage Vin.

It is obtained by using an inverting amplifier and replacing its feedback resistor Rf with a capacitor C.

Hence,

Ignoring I3 we have, I3 ≈ 0

So, I1 ≈ I2.

We know current across the capacitor is given by,

Now by applying Kirchoffs current law,

However, V = 0 because gain A is very large

Integrating both sides we get

+ Q

Where Q is the integration constant and is proportional to Vo at t = 0 sec .

Therefore, voltage Vo is directly proportional to Vin and inversely proportional to constant RC.

Frequency response of basic integrator circuit is given by,

( for 0 db gain)

3.4 Differentiation and Summing Circuits

Differentiator

A differentiator is a circuit that performs differentiation operation. Hence, the output is a derivative of input.

In an inverting amplifier, if input resistor is replaced by a capacitor then a differentiator circuit is formed.

Fig. : Differentiator

In the above fig. on applying Kirchoff’s law we get,

Since Ib = 0 then,

Ic ≈ If

Since Gain A is very large hence, V1 = 0

So, the output voltage is RC times the negative rate of change of input voltage.

When input is a cosine wave , output is a sine wave i.e. it performs the inverse function of integrator circuit.

Fig.: Input and output waveform of differentiator (Ref. 1)

Frequency response of basic differentiator circuit is given by,

( for 0 db gain)

Summing Amplifier

Fig. : Summing amplifier (Ref: 1)

In the above figure, each terminal is connected with two input sources through resistors. V1 and V2 is connected to the inverting terminal and V4 and V5 is connected to the non-inverting terminal.

Assuming R1=R2=R4=R5=RF=R.

Now, on applying superposition theorem, for finding output voltage due to V1 alone, keep V2=V4=V5=0V.

Hence, now the circuit acts as an inverting amplifier.

Therefore,

Similarly,

Now, for V4, voltages V1=V2=V5=0V then the circuit behaves as a non-inverting amplifier.

Hence,

Similarly,

So, the resultant output voltage by all the 4 input voltages is given by,

Vo = V01 + V02 + V04 +V05

Vo = -V1 – V2 +V4 + V5

The output voltage Vo is equivalent to the sum all input voltages applied at both the terminals.

Numerical 1:

In a summing amplifier, if R = 1kΩ, Va = +3V, Vb = +8V, Vc = +9V, Vd = +5V and supply voltage is ±15V. Find the output voltage Vo.

Solution:

Vo = Sum of all input voltages applied at both the terminals

Vo = Va + Vb + Vc +Vd

Vo = -3 -8 +9 +5

Vo = +3V

Numerical 2:

Find the output voltage for the given circuit diagram if Rf = 5kΩ.

Solution :

We know,

Gain (Av) = =

Hence,

Av1 =

Av2 =

Now, Output voltage Vo = Sum of the two amplified input signals

Vo = Av1 x V1 + Av2 x V2

Vo =(-5 x 3) + ( -2.5x 4) mV

Vo = -25mV

As the above output voltage is negative hence it is an inverting amplifier.

References:

1. Electronic Devices Circuit Theory - by Rober L. Boylestad 11th Edition, Pearson Publication, 2014

2. Microelectronic Circuits by A. S. Sedra and Kenneth C. Smith 7th Edition, Oxford University Press. 2017

3. Digital Design by M. Morris Mano, 5th Edition, Pearson Publication, 2016.

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