Q. Illustrate working of Differential Amplifier?

Figure shows a weighted differencing amplifier, where "weighted" refers to the fact that the output voltage has the form v_o = w_av_a + w_bv_b, in which the weighting coefficients wa and wb are the voltage gains seen by the inputs. Since the network is linear, superposition can be applied for analysis. First, by grounding the input terminal of R₃, let us make v_a = 0. Except for the presence of R₃ in parallel with R₄, between terminal 2 and ground, the circuit is that of an inverting amplifier of gain -R₂/R₁ to the input v_b. Since no current flows at terminal 2, it follows that

w_b = - R₂ / R₁

Next, by grounding the left terminal of R₁, let us make v_b = 0. With respect to the voltage on terminal 2, note that the circuit is that of a noninverting amplifier with gain 1 + R₂/R₁. The voltage divider, consisting of R₃ and R₄, corresponds to a gain of R₄/(R₃ + R₄) from the input to terminal 2. Thus the overall gain becomes

which corresponds to the response of a differential amplifier. Usually resistor values are so chosen that R₃ = R₁ and R₄ = R₂ for some practical reasons. Improved versions of differential amplifiers are available commercially.