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(B) The small signal current gain (current amplification factor) $\beta_{ac}$ is defined as the ratio of the change in collector current to the change

Vedclass · Accepted Answer

$400$

Vedclass · Answer

(B) The small signal current gain (current amplification factor) $\beta_{ac}$ is defined as the ratio of the change in collector current to the change in base current at a constant collector-emitter voltage $(V_{CE})$.Given:Change in collector current,$\Delta I_C = 6\,mA - 4\,mA = 2\,mA = 2 	imes 10^{-3}\,A$Change in base current,$\Delta I_B = 25\,\mu A - 20\,\mu A = 5\,\mu A = 5 	imes 10^{-6}\,A$The formula for current gain is:$\beta_{ac} = \frac{\Delta I_C}{\Delta I_B}$Substituting the values:$\beta_{ac} = \frac{2 	imes 10^{-3}}{5 	imes 10^{-6}}$$\beta_{ac} = \frac{2}{5} 	imes 10^3$$\beta_{ac} = 0.4 	imes 1000 = 400$Thus,the current amplification factor is $400$.

For a constant collector-emitter voltage of $8\,V$,the collector current of a transistor changed from $4\,mA$ to $6\,mA$,whereas the base current changed from $20\,\mu A$ to $25\,\mu A$. If the transistor is in the active state,the small signal current gain (current amplification factor) will be:

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