An $n-p-n$ transistor with current gain $\beta=100$ in common emitter configuration is shown in the figure. The output voltage of the amplifier will be $.....V$.

In a transistor,the base is very lightly doped as compared to the emitter because by doing so:

In an $n-p-n$ transistor,$200$ electrons enter the emitter in $10^{-8} \ s$. If $1 \%$ of electrons are lost in the base,then the current that enters the emitter and the current amplification factor are respectively $\left[e=1.6 \times 10^{-19} \ C\right]$

For the given Common Emitter $(CE)$ transistor circuit,the base current is $I_B = 0.04 \text{ mA}$. If the current gain $\beta = 100$ and the collector-emitter voltage $V_{CE} = 12 \text{ V}$,find the value of the collector resistance $R_C$ in $\text{k}\Omega$.

In the study of a transistor as an amplifier,if $\alpha = \frac{I_{C}}{I_{E}}$ and $\beta = \frac{I_{C}}{I_{B}}$,where $I_{C}$,$I_{B}$,and $I_{E}$ are the collector,base,and emitter currents respectively,then:

$A$ 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: