In an n-p-n common emitter (CE) transistor the collector current changes from 5 $\mathrm{mA}$ to $16 \mathrm{~mA}$ for the change in base current from $100~ \mu \mathrm{A}$ and $200 ~\mu \mathrm{A}$, respectively. The current gain of transistor is __________.
Solution
The current gain of a transistor in common emitter configuration is given by:<br/><br/>
$\beta = \frac{I_C}{I_B}$<br/><br/>
where $I_C$ is the collector current and $I_B$ is the base current.
<br/><br/>
In this case, the collector current changes from $5 \mathrm{~mA}$ to $16 \mathrm{~mA}$ for the change in base current from $100~ \mu \mathrm{A}$ to $200 ~\mu \mathrm{A}$. Therefore, we have:<br/><br/>
$\Delta I_C = 16 \mathrm{~mA} - 5 \mathrm{~mA} = 11 \mathrm{~mA}$<br/><br/>
$\Delta I_B = 200~ \mu \mathrm{A} - 100~ \mu \mathrm{A} = 100~ \mu \mathrm{A}$
<br/><br/>
Therefore, the current gain of the transistor is:<br/><br/>
$$\beta = \frac{\Delta I_C}{\Delta I_B} = \frac{11 \mathrm{~mA}}{100~ \mu \mathrm{A}} = 110$$
<br/><br/>
Therefore, the current gain of the transistor is 110.
About this question
Subject: Physics · Chapter: Electronic Devices · Topic: Transistors
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