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(C) The $AC$ current gain $\beta_{ac}$ is defined as the ratio of the change in collector current $\Delta I_{C}$ to the change in base current $\Delta

Vedclass · Accepted Answer

$150$

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(C) The $AC$ current gain $\beta_{ac}$ is defined as the ratio of the change in collector current $\Delta I_{C}$ to the change in base current $\Delta I_{B}$ at a constant collector-emitter voltage $V_{CE}$.From the graph,at $V_{CE} = 10\, V$,the collector current $I_{C} = 4.0\, mA$ corresponds to a base current $I_{B} = 30\, \mu A$.To find $\beta_{ac}$ around this operating point,we consider the adjacent curves for $I_{B} = 20\, \mu A$ and $I_{B} = 40\, \mu A$.At $V_{CE} = 10\, V$:For $I_{B} = 20\, \mu A$,$I_{C} = 3.0\, mA$.For $I_{B} = 40\, \mu A$,$I_{C} = 6.0\, mA$.Therefore,the change in base current is $\Delta I_{B} = 40\, \mu A - 20\, \mu A = 20\, \mu A = 20 	imes 10^{-6}\, A$.The change in collector current is $\Delta I_{C} = 6.0\, mA - 3.0\, mA = 3.0\, mA = 3.0 	imes 10^{-3}\, A$.The $AC$ current gain is $\beta_{ac} = \frac{\Delta I_{C}}{\Delta I_{B}} = \frac{3.0 	imes 10^{-3}}{20 	imes 10^{-6}} = \frac{3000}{20} = 150$.

The output characteristics of a transistor are shown in the figure. When $V_{CE} = 10\, V$ and $I_{C} = 4.0\, mA$,then the value of $\beta_{ac}$ is:

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