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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 vo = wava + wbvb, 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 va = 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 vb. Since no current flows at terminal 2, it follows that
wb = - R2 / R1
Next, by grounding the left terminal of R1, let us make vb = 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.
D flipflop
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