Question

In a common-base amplifier, the phase difference between the input signal voltage and the output voltage is:
(A) $\dfrac{\pi }{4}$
(B) $\pi $
(C) $0$
(D) $\dfrac{\pi }{2}$

Answer 1

Hint: When we express the voltage biasing and the phase difference, we will use the generally assumed conventions of the input and the output voltages. When we follow these conventions, we will see that only in the common-emitter configuration does the phase difference come out to be ${180^\circ }$.

Complete step by step solution:
In a common base amplifier, the phase difference between the input signal voltage and the output voltage is ${0^\circ }$. The phase difference will be ${0^\circ }$ for both the common base and common collector configurations. Only for common emitter configuration, the phase difference between the input signal voltage and the output voltage will be ${180^\circ }$.
In a bipolar junction transistor, the emitter is heavily doped, with the base being lightly doped and the collector being moderately doped. Also, the collector region is wider than the other two regions combined and the base region is the thinnest.
The various modes of operation are active, cut-off and saturation regions. The fourth mode of operation is the reverse active mode which is generally avoided due to the gain provided being extremely low. In case of forward bias, the emitter-base junction will be forward-biased always.
In case of only the common emitter configuration, the voltage at the output side increases in the magnitude. However, due to the transistor operating in the active region and the collector-base junction being reverse biased, the value of this voltage will be negative. Thus in case of this common emitter configuration only, we will see a ${180^\circ }$ phase shift between the input and the output voltages. Hence the amplifier input and output voltage are in the same phase.
Therefore we can say that the phase difference between the input signal voltage and the output voltage is $0$.
The correct answer is option (C) $0$.

Note:
Simply stating, as we assume the polarities of the voltages at the output and the input sides of the bipolar junction transistor, the phase difference will come out accordingly. The phase differences will thus always be either ${0^\circ }$ or ${180^\circ }$.