Applications of Operational Amplifiers: Differential Amplifier

Differential Amplifier:

The basic differential amplifier is shown in fig. 1.

Fig. 1

Since there are two inputs superposition theorem can be used to find the output voltage. When V_b= 0, then the circuit becomes inverting amplifier, hence the output due to V_a only is

V_o(a) = -(R_f / R₁) V_a

Similarly when, V_a = 0, the configuration is a inverting amplifier having a voltage divided network at the noninverting input

Example - 1

Find v_out and i_out for the circuit shown in fig. 2. The input voltage is sinusoidal with amplitude of 0.5 V.

Fig. 2

Solution:

We begin by writing the KCL equations at both the + and – terminals of the op-amp.

For the negative terminal,

Therefore,

15 v_- = v_out

For the positive terminal,

This yields two equations in three unknowns, v_out, v₊ and v_-. The third equation is the relationship between v₊ and v_- for the ideal OPAMP,

v₊ = v_-

Solving these equations, we find

v_out = 10 v_in = 5 sinωt V

Since 2 kΩ resistor forms the load of the op-amp, then the current i_out is given by

Example - 2

For the different amplifier shown in fig. 3, verify that

Fig. 3

Solution:

Since the differential input voltage of OPAMP is negligible, therefore,

v₁= v_x
and v₂ = v_y

The input impedance of OPAMP is very large and, therefore, the input current of OPAMP is negligible.

Thus 
And 
From equation (E-1)

or       

From equation (E-2)

or       

The OPAMP3 is working as differential amplifier, therefore,