In a common emitter configuration with suitable bias, it is given...

In a common emitter configuration with suitable bias, it is given that

{R}_{L}

is the load resistance and

{R}_{BE}

is small-signal dynamic resistance (input side). Then, voltage gain, current gain and power gain are given, respectively, by:

\beta

is current gain,

{I}_{B}

{I}_{C}

and

{I}_{E}

are respectively base, collector, and emitter currents.

Options

A $\beta\dfrac{{R}_{L}}{{R}_{BE}}, \dfrac{\Delta{I}_{C}}{\Delta{I}_{B}}, {\beta}^{2}\dfrac{{R}_{L}}{{R}_{BE}}$
B $\beta\dfrac{{R}_{L}}{{R}_{BE}}, \dfrac{\Delta{I}_{E}}{\Delta{I}_{B}}, {\beta}^{2}\dfrac{{R}_{L}}{{R}_{BE}}$
C ${\beta}^{2}\dfrac{{R}_{L}}{{R}_{BE}}, \dfrac{\Delta{I}_{C}}{\Delta{I}_{B}}, \beta\dfrac{{R}_{L}}{{R}_{BE}}$
D ${\beta}^{2}\dfrac{{R}_{L}}{{R}_{BE}}, \dfrac{\Delta{I}_{C}}{\Delta{I}_{E}}, {\beta}^{2}\dfrac{{R}_{L}}{{R}_{BE}}$

Correct answer

A. $\beta\dfrac{{R}_{L}}{{R}_{BE}}, \dfrac{\Delta{I}_{C}}{\Delta{I}_{B}}, {\beta}^{2}\dfrac{{R}_{L}}{{R}_{BE}}$

Step-by-step solution

For a common emitter configuration, current gain, $\beta = \dfrac{\Delta{I}_{C}}{\Delta{I}_{B}}$ . Voltage gain is the ratio of output voltage to the input voltage. It is given as, ${A}_{V} = \dfrac{\Delta{V}_{CE}}{{R}_{BE}\Delta{I}_{B}} = \beta\dfrac{{R}_{L}}{{R}_{BE}}$ (Since from Ohm's law, ${R}_{L} = \dfrac{∆ {V}_{CE}}{∆ {I}_{B}}$ ). Power gain ${A}_{P} = \beta{A}_{V} = {\beta}^{2}\dfrac{{R}_{L}}{{R}_{BE}}$ .

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