The voltage gain of a transistor in common em | vedclass

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Question

(A) For a transistor in common emitter configuration, the voltage gain $A_V$ is given by the product of current gain $\beta$ and resistance gain $R_C/

Vedclass · Accepted Answer

$4 	ext{ mA}$

Vedclass · Answer

(A) For a transistor in common emitter configuration, the voltage gain $A_V$ is given by the product of current gain $\beta$ and resistance gain $R_C/R_B$.$A_V = \beta 	imes \frac{R_C}{R_B}$Given: $A_V = 160$, $R_B = 1 	ext{ k}\Omega$, $R_C = 4 	ext{ k}\Omega$, and $\Delta I_B = 100 \mu A$.We know that $\beta = \frac{\Delta I_C}{\Delta I_B}$.Substituting this into the voltage gain formula:$A_V = \left(\frac{\Delta I_C}{\Delta I_B}ight) 	imes \frac{R_C}{R_B}$$160 = \left(\frac{\Delta I_C}{100 	imes 10^{-6} 	ext{ A}}ight) 	imes \left(\frac{4 	ext{ k}\Omega}{1 	ext{ k}\Omega}ight)$$160 = \left(\frac{\Delta I_C}{100 	imes 10^{-6}}ight) 	imes 4$$40 = \frac{\Delta I_C}{100 	imes 10^{-6}}$$\Delta I_C = 40 	imes 100 	imes 10^{-6} 	ext{ A} = 4000 	imes 10^{-6} 	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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