The power gain and voltage gain of a transist | vedclass

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(A) Given: Power gain $(A_p)$ = $1800$, Voltage gain $(A_v)$ = $60$, Change in emitter current $(\Delta I_e)$ = $0.62 	ext{ mA}$.We know that Power g

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$0.60$

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(A) Given: Power gain $(A_p)$ = $1800$, Voltage gain $(A_v)$ = $60$, Change in emitter current $(\Delta I_e)$ = $0.62 	ext{ mA}$.We know that Power gain $(A_p)$ = Current gain $(\beta)$ $	imes$ Voltage gain $(A_v)$.Therefore, $\beta = A_p / A_v = 1800 / 60 = 30$.We also know that the current gain in common emitter configuration is $\beta = \Delta I_c / \Delta I_b$, and $\Delta I_e = \Delta I_b + \Delta I_c$.Substituting $\Delta I_b = \Delta I_c / \beta$, we get $\Delta I_e = \Delta I_c / \beta + \Delta I_c = \Delta I_c (1 + 1/\beta) = \Delta I_c ((\beta + 1) / \beta)$.Rearranging for $\Delta I_c$: $\Delta I_c = \Delta I_e 	imes (\beta / (\beta + 1))$.Substituting the values: $\Delta I_c = 0.62 	imes (30 / 31) = 0.62 	imes (0.9677) \approx 0.60 	ext{ mA}$.

The power gain and voltage gain of a transistor connected in common emitter configuration are $1800$ and $60$ respectively. If the change in the emitter current is $0.62 \text{ mA}$, then the change in the collector current is: (in $\text{ mA}$)

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