Current $I$ through a given $p-n$ junction when a voltage $V$ is applied across it is given by $I = I_0 \left( e^{\frac{V}{2 V_T}} - 1 \right)$, where $I_0$ and $V_T$ are constants. If $r_d(I)$ is the dynamic resistance of the junction, then $r_d(1000 I_0) = \alpha r_d(10 I_0)$, where $\alpha$ is approximately equal to

$P-N$ junction is said to be forward biased,when

When a silicon $PN$ junction is in forward biased condition with a series resistance,it has a knee voltage of $0.6\, V$. The current flowing through it is $5\, mA$ when the $PN$ junction is connected to a $2.6\, V$ battery. The value of the series resistance is.....$\Omega$.

If the forward voltage in a semiconductor diode is doubled,the width of the depletion layer will

Three identical p-n junction diodes $D_1$, $D_2$ and $D_3$ are connected across a battery as shown in the figure. If the widths of the depletion regions of $D_1$, $D_2$ and $D_3$ are $W_1$, $W_2$ and $W_3$, respectively, then the correct option is:

Consider a circuit consisting of a capacitor $(20 \mu\text{F})$, a resistor $(100 \Omega)$, and two identical diodes connected in anti-parallel as shown in the figure. The resistance of each diode under forward biasing condition is $10 \Omega$. The time constant of the circuit is $\alpha \times 10^{-3} \text{ s}$. The value of $\alpha$ is . . . . . . .