For a transistor amplifier in common emitter configuration with a load impedance of $1 \ k\Omega$ ($h_{fe} = 50$ and $h_{oe} = 25 \ \mu A/V$),the current gain is:

$A$ transistor is operating in the active mode. $v_1$ is the potential barrier across the base-emitter junction and $v_2$ is the potential barrier across the collector-base junction. $b_1$ is the width of the depletion layer of the base-emitter junction and $b_2$ is the width of the collector-base junction.

In a common emitter transistor amplifier,the load resistance is $6.5 \text{ k}\Omega$ and the input resistance is $1.3 \text{ k}\Omega$. If the current gain is $78$,the voltage gain is:

$A$ common emitter amplifier is designed with an $NPN$ transistor $(\alpha = 0.99)$. The input impedance is $1 \ k\Omega$ and the load resistance is $10 \ k\Omega$. The voltage gain will be:

In a common base amplifier circuit,calculate the change in base current if the change in emitter current is $2\, mA$ and $\alpha = 0.98$.

The typical transfer characteristic of a transistor in $CE$ configuration is shown in the figure. $A$ load resistor of $2\,k\Omega$ is connected in the collector branch of the circuit used. The input resistance of the transistor is $0.50\,k\Omega$. The voltage gain of the transistor is: