An NPN BJT with $alpha$ = 0.90 is used in a common-base amplifier configuration and is biased such that $I_c$ = 5 mA with $R_c$ = 1.8 k$Omega$. It is connected to an ac input signal with a signal resistance of 1 k$Omega$ and is feeding a resistive load of 1 k$Omega$. Assume that $V_T$ = 25 mV and determine the overall voltage gain, $G_v$. $G_v$ = 0.618 $G_v$ = 0.554 $G_v$ = 0.576 $G_v$ = 0.631
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Calculate the transconductance,Gt: Gt=O.9O*Ic/Rc=5mA/1.8k=0.6 Show more…
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