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 $\_\_\_\_$ $\Omega$.

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

For the input v_in(t)as shown in the figure, the transistor switches between the cut-off and saturation regions of operation, when T is large. Assume collector-to-emitter voltage saturation V_CE(sat) = 0.2V and base-to-emitter voltage V_BE = 0.7V. The minimum value of the common-base current gain$$\left( \alpha \right)$$ of the transistor for the switching should be _________.