Operational amplifier

Operational amplifier:

Operational amplifier is direct coupled differential amplifier having two inputs with single output. Thus operational amplifier is basically a differential amplifier with additional circuitry added with it to increase the performance of it by giving some specific feature of it. The operational amplifier performs the mathematical operation such as summation subtraction, multiplication, differentiation, integration etc.

Basic symbol for operational amplifier is shown below,

Where ‘-‘and ‘+’ are two terminal and A is the open loop gain.

Figure

The input terminal of operational amplifier are denoted by ‘-‘ and ‘+’ sign as in figure with input voltage \(v_i -\) and \(v_i+\) representing the inverting input and non-inverting input. If only \(v_i\)- is given by grounding \(v_i^+\) then output signal is in opposite phase to that of input signal. So that the \(v_i^-\) is called as inverting input. Whereas when only \(v_i^+\) is given by grounding \(v_i^-\) the output signal is in same phase to that of input signal. So, \(v_i^+\) is called as non-inverting input.

The operational amplifier having the following characteristics;

1. The very high voltage gain.
2. The input impedance is very high (\(r_{in}\) very high)
3. The output impedance is very low(\(r_{out}\)very low)
4. The large bandwidth (first bandwidth)
5. The very high frequency resonance etc.

The operational amplifier is said to be ideal operational amplifier as it has following characteristics;

1. It has infinite voltage gain
2. It has infinite input impedance
3. It has zero output impedance
4. It has infinite bandwidth
5. It has infinite CMRR

A)Inverting operational amplifier

The operational amplifier with input signal is given to inverting input terminal (-) by grounding its non-inverting input is called as inverting operational amplifier.

The inverting operational amplifier with \(V_{in}\) as input signal is given to the inverting input terminal through resistor \(R_1\). The feedback resistor \(R_f\) is connected between \(v_i^-\) and \(v_\circ\) which provides the feedback path. The open loop voltage gain of operational amplifier is A. \(i_1,i_f\) and \(i^-\) are the current flowing through \(R_1,R_f\) and input terminal of operational amplifier. Since the amplifier is in inverting mode so gain of amplifier is –A.

$$-A=\frac{v_\circ}{v_i^-}$$

$$v_i-=\frac{-v}{A}\dotsm(1)$$Now from the ohm’s law, current flowing through resistor \(R_1\) is given by,

$$v_{in}=i_1 R_1+v_i^-$$

$$i_1=\frac{v_{in}-v_i}{R_1}\dotsm(2)$$

Also current flowing through \(R_f\) is given by,

$$v_i^-= i_f R_f+v_\circ$$

$$i_f=\frac{v_i^--v_\circ}{R_f}\dotsm(3)$$Lei \(i^-\) be the current entering to the amplifier through inverting input terminal. Hence, from Kirchhoff’s current law,

$$i_1=i_f+i^-\dotsm(4)$$

For ideal operational input impedance is very large so that current entering through input terminal is zero (that is neglected).

$$ i.e. i_1=i_f\dotsm(5)$$

$$\Rightarrow \frac{v_{in}-v_i^-}{R_1}=\frac{v_{i^-}-v_\circ}{R_f}$$

$$\frac{v_{in}}{R_1}-\frac{v_i^-}{R_1}=\frac{v_i^-}{R_f}-\frac{v_\circ}{R_f}\dotsm(6)$$

Since, ideal operational amplifier has infinite gain i.e. |A|=\(\infty\)

Then from(1),

$$v_i^-=\frac{-v_\circ}{A}$$

$$\Rightarrow v_i^-=0$$

Then from (6)

$$\frac{v_{in}}{R_1}=\frac{-v_\circ}{R_f}(7)$$

$$\frac {v_{in}}{v_\circ}=\frac{-R_1}{R_f}\dotsm(7)$$This gives the overall gain or close loop gain of inverting operational amplifier,

$$\frac{v_circ}{v_{in}}=A’=A_{cl}=\frac{-R_f}{R_1}$$

$$A_{cl}=\frac{-R_f}{R_1}\dotsm(8)$$Hence, the overall gain of inverting operational amplifier is independent to its open loop gain (A) but dependent only upon the ratio of resistor. The ‘-‘ ve sign is due to the phase reversal output of this amplifer.

1. B) Non-inverting amplifier:

The operational amplifier in which input signal is given to no-inverting input terminal whereas input terminal is grounded is called as non-inverting operational amplifier.

Let us consider the non-inverting operational amplifier with open loop gain A in which \(v_{in}\) is given through ‘+’ve input and feedback resistor \(R_f\) is connected between \(v_\circ\) and \(v_i^-\). The inverting input ‘-‘ is grounded through load resistor\(R_1\). Now, let \(i_f,i_1\)and \(i^-\)be the current flowing through \(R_f,R_1\) and inverting input.

Now current flowing through \(R_f\),

$$i.e.i_f={v_\circ-v_i^-}{R_f}$$and that through \(R_1\)

$$i.e. i_1=\frac{v_i^-}{R_1}$$Also from Kirchhoff’s law,

$$i_f=i_1+i^-$$

Since the input impedance of ideal amplifier is finite. So,\(i^-=0,i_f=i_1\)

$$\frac{v_\circ}{R_f}-\frac{v_i^-}{R_f}=\frac{v_i^-}{R_1}\dotsm(1)$$Since, the gain of ideal operational amplifier is infinite i.e.|A|\(\Rightarrow \infty\)

The gain of operational amplifier is given by,

$$A=\frac{v_\circ}{v_{i^+}-v_i^-}$$

$$\Rightarrow v_{i^+}-v_{i^-}=0$$

$$\therefore v_{i^+}=v_i=v_{in}\dotsm(2)$$Using (2) in (1) we get

$$\frac{v_\circ}{R_f}-\frac{v_{in}}{R_f}=\frac{v_{in}}{R_1}$$

$$\frac{v_\circ}{v_{in}}=1+\frac{R_f}{R_1}$$

$$A’=\frac{v_\circ}{v_{in}}=\biggl(1+\frac{R_f}{R_1}\biggr)$$

This is the overall gain of non-inverting operational amplifier which is also called as close loop gain. And is independent of A but depends upon resistor.

References:

(1)Theraja, B.L. Basic Electronics. N.p.: S.Chand, n.d. Print.

(2)C.L.Arora. Refresher Course in Physics. Vol. II and III. N.p.: S.Chand, 2006. Print.

(3)Malvino. Electronic Principles. N.p.: Tata McGraw-Hill, n.d. Print.

(4)N.Nelkon and P.Parker. Advanced Level Physics. 5th ed. N.p.: Arnold Heinemann, n.d. Print.

(5)Priti Bhakta Adhikari,Diya Nidhi Chaatkuli, Ishowr Prasad Koirala. A Textbook of Physics (2nd Year). N.p.: Sukunda Pustak Bhawan, 2070. Prin