Question

The phase difference between the input and output voltage of CE amplifier circuit is ?

Answer 1

Firstly, we have to find the relation between the collector voltage, collector-emitter reverse biased voltage, load resistance and collector current. From the help of this formula, we will eventually find the difference in input and output voltage.

Complete step by step answer:

The circuit details of the amplifier circuit are given in the figure.

Emitter-base (input) circuit is forward biased with battery ${V_{BB}}$ of voltage ${V_{EB}}$, and the collector-emitter (output) circuit is reverse biased with battery ${V_{CC}}$ of voltage ${V_{CE}}$. ${R_C}$ is a load resistance connected in a collector circuit.

When no A.C. voltage is applied to the input circuit but emitter-base circuit is closed, let us consider that ${I_e}$, ${I_b}$, ${I_c}$ be the emitter current, base current and collector current respectively.Then according to Kirchhoff’s Law we get,

${I_e} = {I_b} + {I_c}$

Due to collector current ${I_c}$, voltage drops across ${R_c} = {I_c}{R_c}$. If ${V_c}$ is collector voltage then,

${V_{CE}} = {V_c} + {I_c}{R_c}$

$\Rightarrow {V_c} = {V_{CE}} - {I_c}{R_c} - - - - \left( 1 \right)$

When input voltage is fed to the emitter base circuit, it will change the emitter voltage and hence the emitter current, which in turn will change the collector current. Due to it, the collector voltage ${V_c}$ will vary with the equation $\left( 1 \right)$. These variations in collector voltage appear as amplified output.

Now, we get the phasor relationship as,

(i) When the positive half cycle of input a.c. signal voltage comes, it supports the forward biasing of the emitter-base circuit. Due to this, the emitter current increases and consequently the collector current increases. As a result of it, the collector voltage ${V_c}$ decreases from equation $\left( 1 \right)$. Since the collector is connected to the positive terminal of the ${V_{CE}}$ battery, the decrease in collector voltage means the collector will become less positive, which means negative with respect to initial value. This indicates that during the positive half cycle of input a.c. signal voltage, the output signal voltage at the collector varies through a negative half cycle.

(ii) When negative half cycle of input a.c. signal voltage comes, it opposes the forward biasing of the emitter-base circuit, due to it the emitter current decreases and hence collector current decreases. Consequently the collector voltage ${V_c}$, increases from equation $\left( 1 \right)$. Thus, the collector becomes more positive. This indicates that during the negative half cycle of input a.c. signal voltage, the output signal voltage varies through the positive half cycle. Thus, in a common emitter amplifier circuit, the input signal voltage and the output collector voltage are in opposite phase which implies ${180^ \circ }$ out of phase, also shown in figure.

Note: It must be noted that the resistance in the input circuit is low as it is forward biased while in the output circuit is high as it is in reverse biased. The low a.c. input signal is applied across the base-emitter circuit and the amplified a.c. signal is obtained as the change in collector voltage.