In a common emitter amplifier circuit using an $$n$$-$$p$$-$$n$$ transistor, the phase difference between the input and the output voltages will be:

We have a common-emitter (CE) amplifier that employs an $$n$$-$$p$$-$$n$$ bipolar junction transistor. In this configuration the input signal $$v_i$$ is applied between the base and the emitter, while the output signal $$v_o$$ is taken between the collector and the emitter through an external collector resistor $$R_C$$.

To see how the phase relationship arises, we recall the small-signal relation for a CE stage. Using the small-signal hybrid-π model, the incremental collector current $$i_c$$ is

$$i_c \;=\; \beta\,i_b$$

where $$i_b$$ is the incremental base current and $$\beta$$ is the transistor’s current gain. The corresponding incremental collector voltage $$v_c$$ measured with respect to the emitter is obtained from Ohm’s law across the collector resistor $$R_C$$:

$$v_c \;=\; V_{CC} \;-\; i_c R_C.$$

For the small-signal change we write

$$\Delta v_c \;=\; -\,\Delta i_c\,R_C.$$

Substituting $$\Delta i_c = \beta \,\Delta i_b$$, we get

$$\Delta v_c \;=\; -\,\beta\,R_C\,\Delta i_b.$$

The minus sign is crucial: it shows that an increase in base current (which is proportional to the increase in input voltage $$\Delta v_i$$) produces a decrease in the collector voltage $$\Delta v_c$$. In other words, when $$v_i$$ goes up, $$v_o$$ comes down, and vice-versa.

A reversal of the direction of change corresponds to a phase difference of $$180^\circ$$ between the two sinusoidal signals. More formally, if we write the input and output as sinusoidal functions,

$$v_i(t) = V_{i\text{(max)}}\sin(\omega t),$$

then with the negative sign the output becomes

$$v_o(t) = -\,A\,V_{i\text{(max)}}\sin(\omega t) = A\,V_{i\text{(max)}}\sin(\omega t + \pi),$$

because $$-\sin(\omega t) = \sin(\omega t + \pi)$$. The angle $$\pi$$ radians is exactly $$180^\circ$$, confirming the phase inversion.

Therefore the phase difference between the input and the output voltages in a common-emitter amplifier is $$180^\circ$$.

Hence, the correct answer is Option A.