A transistor is used as an amplifier in CB mo | vedclass

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Question

(A) In common base $(CB)$ mode,the current gain $\alpha = 0.98$,load resistance $R_L = 5 \, k\Omega = 5000 \, \Omega$,and input resistance $R_{in} = 7

Vedclass · Accepted Answer

$70, 68.6$

Vedclass · Answer

(A) In common base $(CB)$ mode,the current gain $\alpha = 0.98$,load resistance $R_L = 5 \, k\Omega = 5000 \, \Omega$,and input resistance $R_{in} = 70 \, \Omega$.The voltage gain $(A_v)$ is given by the formula: $A_v = \alpha 	imes \frac{R_L}{R_{in}}$.Substituting the values: $A_v = 0.98 	imes \frac{5000}{70} = 0.98 	imes 71.428 \approx 70$.The power gain $(A_p)$ is given by the product of current gain and voltage gain: $A_p = \alpha 	imes A_v$.Substituting the values: $A_p = 0.98 	imes 70 = 68.6$.Thus,the voltage gain is $70$ and the power gain is $68.6$.

$A$ transistor is used as an amplifier in $CB$ mode with a load resistance of $5 \, k\Omega$. The current gain of the amplifier is $0.98$ and the input resistance is $70 \, \Omega$. The voltage gain and power gain respectively are:

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