Question

A change of 0.04V takes place between the base and the emitter when an input signal is connected to the CE transistor amplifier. As a result, $20\;{{\mu A}}$ change takes place in the base current, and a change of $2\;{\text{mA}}$ takes place in the collector current. Find the input resistance and A.C current gain.
A) $2\;{\text{k}}\Omega, 100$
B) $1\;{\text{k}}\Omega, 100$
C) $2\;{\text{k}}\Omega, 200$
D) $1\;{\text{k}}\Omega, 200$

Answer 1

Hint: In this problem, first convert the units of the change in current to Ampere then use the formula of the resistance input to obtain the value of input resistance and convert it into kilo ohm. Similarly use the formula of the current gain to obtain the value of the current gain.

Complete step by step answer:
In this question, when an input signal is connected to the Collector-Emitter transistor amplifier, a change in voltage of $0.04\;{\text{V}}$ takes place between the base and the emitter. Due to which the change in the base current will be $20\;{{\mu A}}$ and the change in the collector current will be $2\;{\text{mA}}$. In this problem, we need to calculate the input resistance and the A.C current gain.
First convert the unit of the change in the base current from micro Ampere $\left( {{{\mu A}}} \right)$ to Ampere $\left( A \right)$ as we know that one microAmpere is equal to ${10^{ - 6}}\;{\text{A}}$,
$\Rightarrow \Delta {I_b} = 20\;{{\mu A}} $
$\Rightarrow \Delta {I_b} = 20\;{{\mu A}}\left| {\dfrac{{{{10}^{ - 6}}\;{\text{A}}}}{{1\;{{\mu A}}}}} \right| $
$\Rightarrow \Delta {I_b} = 20 \times {10^{ - 6}}\;{\text{A}} $
Similarly, convert the unit of the change in the collector current from milliAmpere $\left( {{\text{mA}}} \right)$ to Ampere $\left( A \right)$ as we know that one milliAmpere is equal to ${10^{ - 3}}\;{\text{A}}$,
$ \Delta {I_c} = 20\;{\text{mA}}$
$\Rightarrow \Delta {I_c} = 20\;{\text{mA}}\left| {\dfrac{{{{10}^{ - 3}}\;{\text{A}}}}{{1\;{\text{mA}}}}} \right| $
$\Rightarrow \Delta {I_c} = 20 \times {10^{ - 3}}\;{\text{A}} $
Now, calculate the input current by using the resistance formula as given below,
${R_{input}} = \dfrac{{{V_{BE}}}}{{\Delta {I_b}}}$
Here, the input resistance is ${R_{input}}$, the change in the voltage between the base and the emitter is ${V_{BE}}$, and the change in the collector current is $\Delta {I_b}$.
Substitute the given values in the above equation and evaluate as,
${R_{input}} = \dfrac{{{V_{BE}}}}{{\Delta {I_b}}} $
$\Rightarrow {R_{input}} = \dfrac{{0.04}}{{20 \times {{10}^{ - 6}}}} $
$\Rightarrow {R_{input}} = 2 \times {10^3}\;\Omega $
Converting it into kilo-ohms, we get
$\Rightarrow {R_{input}} = 2\;{\text{k}}\Omega$
As we know that, A.C current gain is the ratio of the change in the collector current and the change in the base current, so now we will calculate the A.C current gain as,
$\Rightarrow \beta = \dfrac{{\Delta {I_c}}}{{\Delta {I_b}}} $
$\Rightarrow \beta = \dfrac{{2 \times {{10}^{ - 3}}}}{{20 \times {{10}^{ - 6}}}} $
$\Rightarrow \beta = 100 $

Therefore, the AC current gain is $100$ and the resistance input is $2\;{\text{k}}\Omega $. Hence, the correct option is A.

Note:
Make sure that the units of the change in base current and the change in the collector current are in Ampere. Be careful about the AC current input, as it is the ratio of the change in collector current to the change in the base current instead of the ratio of the change in the base current to the change in the collector current.