PN Junction and Diode Questions in English

(B) The width of the depletion region $(w)$ in a $p-n$ junction is inversely proportional to the doping concentration $(N)$ on that side, expressed as $w \propto 1/N$.
Given that the $p$-side has a lower doping concentration $(10^{15} \text{ atoms/cm}^3)$ compared to the $n$-side $(10^{18} \text{ atoms/cm}^3)$, the depletion region will extend further into the $p$-side.
Therefore, the depletion region width is greater on the $p$-side than on the $n$-side.

(C) The time constant of an $RC$ circuit is given by $\tau = R_{eq}C$.
In the given circuit, there is a resistor $R = 100 \Omega$ in series with a parallel combination of two identical diodes.
Each diode has a resistance of $10 \Omega$ in the forward bias condition.
The equivalent resistance of the two diodes in parallel is $R_d = \frac{10 \times 10}{10 + 10} = 5 \Omega$.
The total resistance of the circuit is $R_{total} = R + R_d = 100 \Omega + 5 \Omega = 105 \Omega$.
The capacitance is $C = 20 \mu\text{F} = 20 \times 10^{-6} \text{ F}$.
Thus, the time constant $\tau = R_{total} \times C = 105 \Omega \times 20 \times 10^{-6} \text{ F} = 2100 \times 10^{-6} \text{ s} = 2.1 \times 10^{-3} \text{ s}$.
Comparing this with $\alpha \times 10^{-3} \text{ s}$, we get $\alpha = 2.1$.

(C) Statement $A$ is correct: In a $p-n$ junction, forward bias reduces the potential barrier, allowing current to rise sharply after the threshold voltage.
Statement $B$ is incorrect: The current in forward bias is called forward current. The term 'reverse saturation current' refers to the very small, nearly constant current that flows in a $p-n$ junction diode when it is in reverse bias.
Therefore, Statement $A$ is true and Statement $B$ is false.

(B) The circuit consists of four parallel branches connected to a $10 \text{V}$ $DC$ source.
Each branch contains a resistor and a diode.
Analyzing the polarity of the diodes with respect to the $10 \text{V}$ battery:
$1$. The top branch ($4 \Omega$ resistor) has the diode in forward-biased condition.
$2$. The second branch ($3 \Omega$ resistor) has the diode in reverse-biased condition (it acts as an open circuit).
$3$. The third branch ($2 \Omega$ resistor) has the diode in forward-biased condition.
$4$. The bottom branch ($5 \Omega$ resistor) has the diode in reverse-biased condition (it acts as an open circuit).
Only the branches with $4 \Omega$ and $2 \Omega$ resistors are active.
These two resistors are in parallel, so the equivalent resistance $R_{\text{eq}}$ is:
$R_{\text{eq}} = \frac{4 \times 2}{4 + 2} = \frac{8}{6} = \frac{4}{3} \Omega$.
The total current $I$ drawn from the battery is:
$I = \frac{V}{R_{\text{eq}}} = \frac{10}{4/3} = \frac{30}{4} = 7.5 \text{A} = \frac{15}{2} \text{A}$.
Thus, the correct option is $B$.

(C) $1$. Analyze the biasing of each diode:
- Diode $D_1$ is forward-biased because its p-side is connected to the positive terminal of the battery.
- Diode $D_2$ is in an open circuit (due to the gap in the wire), so no current flows through it. It acts as an unbiased diode.
- Diode $D_3$ is reverse-biased because its n-side is connected to the positive terminal of the battery.
$2$. Relate bias to depletion width:
- Forward bias reduces the depletion width $(W_f < W_0)$.
- Unbiased state has a standard depletion width $(W_0)$.
- Reverse bias increases the depletion width $(W_r > W_0)$.
$3$. Compare the widths:
- For $D_1$ (forward-biased): $W_1$ is minimum.
- For $D_2$ (unbiased): $W_2$ is intermediate.
- For $D_3$ (reverse-biased): $W_3$ is maximum.
Therefore, $W_3 > W_2 > W_1$.

(C) In a $p-n$ junction diode, the depletion region width increases with an increase in reverse bias voltage.
When the reverse bias voltage is increased to a critical value known as the breakdown voltage, the electric field across the junction becomes very strong.
This strong electric field causes a large number of charge carriers to be generated (either through Zener breakdown or Avalanche breakdown), leading to a sudden increase in the reverse current.
Therefore, breakdown occurs when the reverse bias is increased to this critical value.

(C) An ideal diode in forward bias acts as a short circuit (zero resistance), and in reverse bias, it acts as an open circuit (infinite resistance).
Case $1$: Diode $D$ is forward biased.
The upper branch has a resistance of $30 \ \Omega$ in series with the diode (which acts as $0 \ \Omega$). The lower branch has a resistance of $30 \ \Omega$. These two branches are in parallel.
$R_1 = (30 \ \Omega \parallel 30 \ \Omega) = \frac{30 \times 30}{30 + 30} = \frac{900}{60} = 15 \ \Omega$.
Case $2$: Diode $D$ is reverse biased.
The upper branch acts as an open circuit (infinite resistance). Thus, only the lower branch with $30 \ \Omega$ resistance is connected between $A$ and $B$.
$R_2 = 30 \ \Omega$.
Ratio $R_1/R_2 = 15/30 = 1/2$.

(B) $1$. Analyze the circuit: The diodes $D_1$ and $D_2$ are forward-biased, while $D_3$ is reverse-biased.
$2$. Since $D_3$ is reverse-biased, it acts as an open circuit (infinite resistance), so no current flows through the branch containing $D_3$.
$3$. The circuit simplifies to two parallel branches connected in series with the $100 \text{ } \Omega$ resistor.
$4$. Branch $1$ (containing $D_1$): Total resistance $R_1 = R_{D1} + 150 \text{ } \Omega = 50 \text{ } \Omega + 150 \text{ } \Omega = 200 \text{ } \Omega$.
$5$. Branch $2$ (containing $D_2$): Total resistance $R_2 = R_{D2} + 50 \text{ } \Omega = 50 \text{ } \Omega + 50 \text{ } \Omega = 100 \text{ } \Omega$.
$6$. The equivalent resistance of the parallel combination of $R_1$ and $R_2$ is $R_p = \frac{R_1 \times R_2}{R_1 + R_2} = \frac{200 \times 100}{200 + 100} = \frac{20000}{300} = \frac{200}{3} \text{ } \Omega$.
$7$. The total resistance of the circuit is $R_{total} = R_p + 100 \text{ } \Omega = \frac{200}{3} + 100 = \frac{500}{3} \text{ } \Omega$.
$8$. The total current $I$ flowing from the battery is $I = \frac{V}{R_{total}} = \frac{5}{500/3} = \frac{15}{500} = 0.03 \text{ A} = 30 \text{ mA}$.

(D) diode is forward biased if the potential at the $p$-side (anode) is higher than the potential at the $n$-side (cathode). Otherwise, it is reverse biased.
In Fig. $(1)$:
The $p$-side of diode $D_1$ is at $-5 \ V$ and the $n$-side is connected to $-3 \ V$ through resistor $R_1$. Since $-5 \ V < -3 \ V$, the potential at the $p$-side is lower than the potential at the $n$-side. Therefore, diode $D_1$ is reverse biased.
In Fig. $(2)$:
The $p$-side of diode $D_2$ is at $0 \ V$ and the $n$-side is connected to $-4 \ V$ through resistor $R_2$. Since $0 \ V > -4 \ V$, the potential at the $p$-side is higher than the potential at the $n$-side. Therefore, diode $D_2$ is forward biased.
Thus, Fig. $(1)$ is reverse biased and Fig. $(2)$ is forward biased.

(B) The given circuit consists of a $5 \text{ V}$ $DC$ source, a silicon diode, and a resistor of $200 \text{ } \Omega$ connected in series.
Since the diode is forward-biased, it will conduct current when the applied voltage exceeds its knee voltage.
The effective voltage across the resistor $(V_R)$ is given by:
$V_R = V_{\text{source}} - V_{\text{knee}}$
$V_R = 5 \text{ V} - 0.7 \text{ V} = 4.3 \text{ V}$
Using Ohm's law, the current $(I)$ flowing through the circuit is:
$I = \frac{V_R}{R}$
$I = \frac{4.3 \text{ V}}{200 \text{ } \Omega} = 0.0215 \text{ A}$
To convert the current into milliamperes (mA):
$I = 0.0215 \times 1000 \text{ mA} = 21.5 \text{ mA}$
Therefore, the reading in the milliammeter is $21.5 \text{ mA}$.

(A) In a $p-n$ junction, the depletion region is formed by the diffusion of charge carriers across the junction.
When the junction is forward biased, the external electric field opposes the internal electric field, which pushes the charge carriers towards the junction, thereby decreasing the width of the depletion region.
When the junction is reverse biased, the external electric field supports the internal electric field, which pulls the charge carriers away from the junction, thereby increasing the width of the depletion region.
Therefore, the depletion region increases if the junction is reverse biased.

(B) $1$. Germanium $(Ge)$ is a group $14$ element. Indium $(In)$ is a group $13$ element. Doping $Ge$ with $In$ creates a p-type semiconductor. Thus, $X$ is p-type.
$2$. Arsenic $(As)$ is a group $15$ element. Doping $Ge$ with $As$ creates an n-type semiconductor. Thus, $Y$ is n-type.
$3$. The junction formed is a p-n junction.
$4$. In the circuit diagram, the p-side $(X)$ is connected to the negative terminal of the battery and the n-side $(Y)$ is connected to the positive terminal of the battery.
$5$. When the p-side is connected to the negative terminal and the n-side is connected to the positive terminal, the p-n junction is reverse biased.

(A) When a $p-n$ junction diode is forward biased, the positive terminal of the external battery is connected to the $p$-type region and the negative terminal to the $n$-type region.
This external electric field opposes the internal electric field of the depletion region.
As a result, the majority charge carriers are pushed towards the junction, which reduces the width of the depletion layer.
Consequently, the potential barrier height decreases, allowing current to flow through the diode.

(B) diode is reverse biased when the potential at the $p$-terminal (anode) is lower than the potential at the $n$-terminal (cathode).
$(1)$ $p$-terminal = $5 \text{ V}$, $n$-terminal = $0 \text{ V}$ (ground). Since $5 \text{ V} > 0 \text{ V}$, it is forward biased.
$(2)$ $p$-terminal = $-20 \text{ V}$, $n$-terminal = $-10 \text{ V}$. Since $-20 \text{ V} < -10 \text{ V}$, it is reverse biased.
$(3)$ $p$-terminal = $15 \text{ V}$, $n$-terminal = $10 \text{ V}$. Since $15 \text{ V} > 10 \text{ V}$, it is forward biased.
$(4)$ $p$-terminal = $20 \text{ V}$, $n$-terminal = $-5 \text{ V}$. Since $20 \text{ V} > -5 \text{ V}$, it is forward biased.
Thus, the diode is reverse biased in Figure $(2)$.

(D) $1$. Arsenic $(As)$ is a pentavalent impurity. Doping germanium with arsenic creates an $n$-type semiconductor. Thus, $X$ is $n$-type.
$2$. Indium $(In)$ is a trivalent impurity. Doping germanium with indium creates a $p$-type semiconductor. Thus, $Y$ is $p$-type.
$3$. In the circuit diagram, the positive terminal of the battery is connected to $X$ ($n$-type) and the negative terminal is connected to $Y$ ($p$-type).
$4$. When the $n$-side is connected to the positive terminal and the $p$-side is connected to the negative terminal, the $p-n$ junction is reverse biased.
$5$. Therefore, $X$ is $n$-type, $Y$ is $p$-type, and the junction is reverse biased.