The voltage gain of a transistor in common em | vedclass

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Question

(A) Given: Voltage gain $A_V = 160$, Base resistance $R_B = 1 	ext{ k}\Omega$, Collector resistance $R_C = 4 	ext{ k}\Omega$, Change in base current

Vedclass · Accepted Answer

$4 	ext{ mA}$

Vedclass · Answer

(A) Given: Voltage gain $A_V = 160$, Base resistance $R_B = 1 	ext{ k}\Omega$, Collector resistance $R_C = 4 	ext{ k}\Omega$, Change in base current $\Delta I_B = 100 \mu A = 10^{-4} 	ext{ A}$.We know that voltage gain $A_V$ is given by $A_V = \beta 	imes \frac{R_C}{R_B}$, where $\beta = \frac{\Delta I_C}{\Delta I_B}$ is the current gain.Substituting the values: $160 = \left( \frac{\Delta I_C}{100 	imes 10^{-6} 	ext{ A}} ight) 	imes \left( \frac{4 	ext{ k}\Omega}{1 	ext{ k}\Omega} ight)$.$160 = \left( \frac{\Delta I_C}{10^{-4} 	ext{ A}} ight) 	imes 4$.$40 = \frac{\Delta I_C}{10^{-4} 	ext{ A}}$.$\Delta I_C = 40 	imes 10^{-4} 	ext{ A} = 4 	imes 10^{-3} 	ext{ A} = 4 	ext{ mA}$.Thus, the change in output current is $4 	ext{ mA}$.

The voltage gain of a transistor in common emitter configuration is $160$. The resistances in base and collector sides of the circuit are $1 \text{ k}\Omega$ and $4 \text{ k}\Omega$ respectively. If the change in base current is $100 \mu A$, then the change in output current is:

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