Question

A npn transistor operates as a common emitter amplifier, with a power gain of \[60\,{\text{dB}}\] . The input circuit resistance is \[100\,\Omega \] and the output load resistance is \[10\,{\text{k}}\Omega \] . The common emitter current gain \[\beta \] is:
A. \[60\]
B. \[{10^4}\]
C. \[6 \times {10^2}\]
D. \[{10^2}\]

Answer 1

Hint: First of all, we will convert the power gain, which is in decibels. After that, we will use the expression for the power gain which is the product of voltage gain and current gain. We will replace voltage gain by current gain along with some necessary modifications. We will manipulate accordingly and find the result.

Formula used:
The power gain can be converted which is in decibels can be converted using the formula:
\[{A_{{{\text{p}}_{{\text{dB}}}}}} = 10 \times {\log _{10}}{A_{\text{p}}}\] …… (1)
Where,
\[{A_{{{\text{p}}_{{\text{dB}}}}}}\] indicates power gain in decibels.
The formula which gives the power gain and relates voltage gain and current gain is shown below:
\[{A_{\text{p}}} = {A_{\text{v}}} \times \beta \] …… (2)
\[{A_{\text{p}}}\] indicates power gain.
\[{A_{\text{v}}}\] indicates voltage gain.
\[\beta \] indicates current gain.
Again, the formula which gives the relation between the voltage gain and the current gain is as follows:
\[{A_{\text{v}}} = \beta \times \dfrac{{{R_{{\text{out}}}}}}{{{R_{{\text{in}}}}}}\] …… (3)

Complete step by step answer:
In the given question, we are supplied with the following data:
There is an npn transistor which operates as a common amplifier.There is power gain of \[60\,{\text{dB}}\] .The inside circuit resistance of a common emitter amplifier is \[100\,\Omega \] .The output load resistance is \[10\,{\text{k}}\Omega \] .We are asked to find the current gain \[\beta \] .

Let us begin to solve the problem. First, we will find a way to convert the power gain which is in decibels. For this we will substitute the required values in the equation (1) and we get:
${A_{{{\text{p}}_{{\text{dB}}}}}} = 10 \times {\log _{10}}{A_{\text{p}}} \\
\Rightarrow 60 = 10 \times {\log _{10}}{A_{\text{p}}} \\
\Rightarrow 6 = {\log _{10}}{A_{\text{p}}} \\
\Rightarrow {A_{\text{p}}} = {10^6} \\$
Now, we substitute the equation (3) in equation (2) and we get:
${A_{\text{p}}} = {A_{\text{v}}} \times \beta \\
\Rightarrow {A_{\text{p}}} = \beta \times \dfrac{{{R_{{\text{out}}}}}}{{{R_{{\text{in}}}}}} \times \beta \\
\Rightarrow {A_{\text{p}}} = {\beta ^2} \times \dfrac{{{R_{{\text{out}}}}}}{{{R_{{\text{in}}}}}} \\
\Rightarrow \beta = \sqrt {\dfrac{{{A_{\text{p}}} \times {R_{{\text{in}}}}}}{{{R_{{\text{out}}}}}}} \\$
Now, we substitute the required values in the above equation found:
$\beta = \sqrt {\dfrac{{{A_{\text{p}}} \times {R_{{\text{in}}}}}}{{{R_{{\text{out}}}}}}} \\
\Rightarrow \beta = \sqrt {\dfrac{{{{10}^6} \times 100}}{{10 \times {{10}^3}}}} \\
\Rightarrow \beta = \sqrt {{{10}^4}} \\
\therefore \beta = 100$
Hence, the required current gain is found to be \[100\] .

The correct option is D.

Note:While solving the problem, we should remember that the power gain which is in decibels must be converted. Most of the students tend to make mistakes by not converting the power gain and they get irrelevant results.