Question

In a common emitter transistor amplifier, the output resistance is \[500{\text{ k}}\Omega \] and the current gain \[\beta = 49\]. If the power gain of the amplifier is \[5 \times {10^6}\], the input resistance is
A. \[325{\text{ k}}\Omega \]
B. \[165{\text{ k}}\Omega \]
C. \[{\text{198 k}}\Omega \]
D. \[240{\text{ k}}\Omega \]

Answer 1

Hint:In this question, we need to find the input resistance. For this, we will use the formula of power gain for a common emitter transistor amplifier. After simplification, we will get the final result.

Formula used:
The formula for power gain for a common emitter transistor amplifier is given below.
Power gain \[ = {\beta ^2} \times \dfrac{{{R_o}}}{{{R_i}}}\]
Here, \[\beta \] is the current gain, \[{R_o}\] is the output resistance and \[{R_i}\] is the input resistance.

Complete step by step solution:
We know that the power gain for a common emitter transistor amplifier is Power gain \[ = {\beta ^2} \times \dfrac{{{R_o}}}{{{R_i}}}\]
But the current gain \[\left( \beta \right)\]is 49.
Also, the output resistance is \[\left( {{R_o}} \right) = 500{\text{ k}}\Omega \]
But \[1{\text{ k}}\Omega = 1000{\text{ }}\Omega = {10^3}{\text{ }}\Omega \]
So, \[\left( {{R_o}} \right) = 500 \times 1000 = 500 \times {10^3}{\text{ }}\Omega \]
Also, power gain is \[5 \times {10^6}\]
So, we get
\[5 \times {10^6} = {\left( {49} \right)^2} \times \dfrac{{500 \times {{10}^3}{\text{ }}}}{{{R_i}}}\]
\[5 \times {10^6}\left( {{R_i}} \right) = {\left( {49} \right)^2} \times 500 \times {10^3}\]

By simplifying, we get
\[\left( {{R_i}} \right) = \dfrac{{{{\left( {49} \right)}^2} \times 500 \times {{10}^3}}}{{5 \times {{10}^6}}}\]
\[\Rightarrow \left( {{R_i}} \right) = {\left( {49} \right)^2} \times 100 \times {10^{3 - 6}}\]
\[\Rightarrow \left( {{R_i}} \right) = {\left( {49} \right)^2} \times {10^2} \times {10^{ - 3}}\]
\[\Rightarrow \left( {{R_i}} \right) = {\left( {49} \right)^2} \times {10^{2 - 3}}\]

By simplifying further, we get
\[\left( {{R_i}} \right) = 2401 \times {10^{2 - 3}}\]
\[\Rightarrow \left( {{R_i}} \right) = 2401 \times {10^{ - 1}}\]
\[\Rightarrow \left( {{R_i}} \right) = \dfrac{{2401}}{{10}}\]
This gives, \[\left( {{R_i}} \right) = 240.1{\text{ }}\Omega \]
That is \[\left( {{R_i}} \right) \approx 240{\text{ }}\Omega \]
Hence, the value of input resistance is approximately \[240{\text{ }}\Omega \].

Therefore, the correct option is (D).

Additional information: We know that an amplifier is a type of electronic circuit often used to boost the strength of a poor input signal in terms of voltage, current, or power. So, the common emitter amplifier is a voltage amplifier that consists of three basic single-stage bipolar junction transistors. This amplifier's input is captured from the base terminal, its output is gathered from the collector terminal, and both terminals share the emitter terminal.

Note: Many students generally make mistakes in writing the formula of power gain of an amplifier. They generally write \[{\beta ^2} \times \dfrac{{{R_i}}}{{{R_o}}}\] instead of \[{\beta ^2} \times \dfrac{{{R_o}}}{{{R_i}}}\]. Also, while doing calculations, they may get confused with the power of 10.