Question

In a common emitter configuration with suitable bias, it is given than 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:
A. $\beta \dfrac{{{R_L}}}{{{R_{BE}}}},\dfrac{{\Delta {I_E}}}{{\Delta {I_B}}},{\beta ^2}\dfrac{{{R_L}}}{{{R_{BE}}}}$
B. ${\beta ^2}\dfrac{{{R_L}}}{{{R_{BE}}}},\dfrac{{\Delta {I_C}}}{{\Delta {I_B}}},\beta \dfrac{{{R_L}}}{{{R_{BE}}}}$
C. ${\beta ^2}\dfrac{{{R_L}}}{{{R_{BE}}}},\dfrac{{\Delta {I_C}}}{{\Delta {I_E}}},{\beta ^2}\dfrac{{{R_L}}}{{{R_{BE}}}}$
D. $\beta \dfrac{{{R_L}}}{{{R_{BE}}}},\dfrac{{\Delta {I_C}}}{{\Delta {I_B}}},{\beta ^2}\dfrac{{{R_L}}}{{{R_{BE}}}}$

Answer 1

Hint: You need to calculate the voltage and current gain using their formulas, and for the power gain, you need to use the voltage and current gain. In common emitter transistors, the emitter is common between the input circuit and the input circuit.

Complete step by step answer:
The base is at the input side, and the collector is on the output side. And any gain can be calculated by dividing the output value to the input value. The input circuit consists of the base and emitter while the output circuit consists of emitter and collector.
 Here the current gain is given, that means it is the io of input current and output current. In common emitter transistors, the input current is ${I_B}$, and the output current is ${I_C}$.
The current gain is the ratio of collector current to the base current, and it is given by the formula $\beta = \dfrac{{{I_C}}}{{{I_B}}}$
The voltage gain is the ratio of collector-emitter voltage to the base-emitter voltage, and it is given by the formula
${\rm{Voltage}}\;{\rm{Gain}} = \dfrac{{{V_{CE}}}}{{{V_{BE}}}}$
So,${\rm{Voltage}}\;{\rm{Gain}} = \beta \dfrac{{{R_L}}}{{{R_{BE}}}}$
The power of any electrical circuit is given by voltage multiplied by the current. Similarly, The Power Gain is also calculated by multiplying the voltage gain and current gain. It is now given by
$\rm{Power\; gain} = \rm{voltage\; gain} \times \rm{current\;gain} $
So , $Power\;gain = {\beta ^2}\dfrac{{{R_L}}}{{{R_{BE}}}}$

So, the correct answer is “Option B”.

Note:
You may go wrong in calculating the power gain here because power gain is based on voltage gain and current gain. We know that any gain can be calculated by dividing output by input. But power gain can’t be calculated by output power to input power.