The base of an $n p n$ BJT $T 1$ has a linear doping profile $N_B(x)$ as shown below. The base of another $n p n$ BJT $T 2$ has a uniform doping $N_B$ of $10^{17} \mathrm{~cm}^{-3}$. All other parameters are identical for both the devices. Assuming that the hole density profile is the same as that of doping, the common - emitter current gain of $T 2$ is

In the voltage regulator shown below, $V_I$ is the unregulated input at 15 V . Assume $V_{B E}=0.7 \mathrm{~V}$ and the base current is negligible for both the BJTs. If the regulated output $V_0$ is 9 V , the value of $R_2$ is $\_\_\_\_$ $\Omega$.

The components in the circuit given below are ideal. If $R=2 \mathrm{k} \Omega$ and $C=1 \mu \mathrm{~F}$, the -3 dB cut-off frequency of the circuit in Hz is

For the BJT in the amplifier shown below, $V_{B E}=0.7 \mathrm{~V}, \frac{k T}{q}=26 \mathrm{mV}$. Assume that the BJT output resistance ( $r_0$ ) is very high and the base current is negligible. The capacitors are also assumed to be short circuited at signal frequencies. The input $V_i$ is direct coupled. The low frequency voltage gain $\frac{V_0}{V_i}$ of the amplifier is