Question

An NPN transistor is used in common emitter configuration as an amplifier with 1KOhm of load resistance. A signal voltage of 10 mV is applied across the base-emitter. This produces $3mA$ change in the collector current and $15\mu A$ change in the base current of the amplifier. The input resistance and voltage gain are:
A) $0.33k\mathrm{\Omega },1.5$
B) $0.67k\mathrm{\Omega },200$
C) $0.33k\mathrm{\Omega },300$
D) $0.67k\mathrm{\Omega },300$

Answer 1

Hint:Voltage gain is defined as the ratio of the output voltage to the input voltage. The input resistance is the ratio of change in input voltage to change in input current.

Complete step by step answer:
So we are given a NPN transistor in common emitter mode, the input resistance can be calculated if we know the change in input current when an input voltage is applied. So the input resistance is,
${\text{R}}_{\text{i}}={\displaystyle \frac{\mathrm{\Delta }{\text{V}}_{\text{i}}}{\mathrm{\Delta }{\text{I}}_{\text{i}}}}$
$R_i={\displaystyle \frac{10\times {10}^{-3}}{15\times {10}^{-6}}}$
$\therefore {\text{R}}_{\text{i}}=0.67K\mathrm{\Omega }$
So the input resistance is ${\text{R}}_{\text{i}}=0.67K\mathrm{\Omega }$.
The voltage gain is defined as the ratio of change in the output voltage to the change in input voltage. We know that $\mathrm{\Delta }{\text{V}}_{\text{i}}=\text{10mV}$, in order to calculate the change in output voltage we multiply the change in change in output current with the load resistance. So we can write,
$\mathrm{\Delta }{\text{V}}_{\text{o}}=\mathrm{\Delta }{\text{I}}_{\text{o}}{\text{R}}_{\text{L}}$
Substituting the values of change in output voltage and load resistance, we get,
$\mathrm{\Delta }{\text{V}}_{\text{o}}=(3\times {10}^{-3})\left(1000\right)$
$\therefore \mathrm{\Delta }{\text{V}}_{\text{o}}=\text{3V}$
So the voltage gain is given by,
${\text{A}}_{\text{v}}={\displaystyle \frac{\mathrm{\Delta }{\text{V}}_{\text{o}}}{\mathrm{\Delta }{\text{V}}_{\text{i}}}}$
${\text{A}}_{\text{v}}={\displaystyle \frac{3}{10\times {10}^{-3}}}$
$\therefore {\text{A}}_{\text{v}}=300$
So the voltage gain for the transistor is ${\text{A}}_{\text{v}}=300$.
So the answer to the question is option (D)- $0.67k\mathrm{\Omega },300$

Note:
The current gain of a transistor is defined as the ratio of output current to the input current. It is denoted by the Greek letter $\beta$ for common emitter configuration and $\alpha$ for common base configuration.
The current gain in the common base configuration is defined by,
$\alpha ={\displaystyle \frac{\mathrm{\Delta }{\text{I}}_{\text{c}}}{\mathrm{\Delta }{\text{I}}_{\text{e}}}}$
Where,
$\mathrm{\Delta }{\text{I}}_{\text{c}}$ is the change in collector current.
$\mathrm{\Delta }{\text{I}}_e$ is the change in emitter current.
The current gain in the common emitter configuration is defined by,
$\beta ={\displaystyle \frac{\mathrm{\Delta }{\text{I}}_{\text{c}}}{\mathrm{\Delta }{\text{I}}_{\text{b}}}}$
Where,
$\mathrm{\Delta }{\text{I}}_{\text{c}}$ is the change in collector current.
$\mathrm{\Delta }{\text{I}}_{\text{b}}$ is the change in emitter current.