Question
When the base current in a transistor is changed from $30\mu\text{A}$ to $80\mu\text{A},$ the collector current is changed from 1.0mA to 3.5mA. Find the current gain $\beta.$
✓
Answer
$\delta\text{I}_\text{b}=80\mu\text{A}-30\mu\text{A}=50\mu\text{A}=50\times10^{-6}\text{A}$
$\delta\text{I}_\text{c}=3.5\text{mA}-1\text{mA}$ $=-2.5\text{mA}=2.5\times10^{-3}\text{A}$
$\beta=\Big(\frac{\delta\text{I}_\text{c}}{\delta\text{I}_\text{b}}\Big)\text{V}_\text{ce}$ = constant
$\Rightarrow\frac{2.5\times10^{-3}}{50\times10^{-6}}=\frac{2500}{50}=50$
Current gain = 50.
$\delta\text{I}_\text{c}=3.5\text{mA}-1\text{mA}$ $=-2.5\text{mA}=2.5\times10^{-3}\text{A}$
$\beta=\Big(\frac{\delta\text{I}_\text{c}}{\delta\text{I}_\text{b}}\Big)\text{V}_\text{ce}$ = constant
$\Rightarrow\frac{2.5\times10^{-3}}{50\times10^{-6}}=\frac{2500}{50}=50$
Current gain = 50.
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