Q.Point A is held at $-10\,\text{V}$ and point B is earthed (at $0\,\text{V}$). Starting from A, a resistor R is in series with an ideal diode $D_1$ whose arrow (anode to cathode) points from the A/resistor side towards a junction. From that junction the line runs down through a second ideal diode $D_2$ to B; $D_2$'s arrow points upward, from the earthed B side towards the junction (its anode is on the B side, its cathode towards the junction). Assuming the diodes to be ideal, which statement is correct?

Under forward bias the external field opposes the junction's built-in potential. This pushes majority carriers toward the junction, neutralising some of the immobile charge in the depletion region, so the depletion layer width decreases and the barrier potential drops. Consequently the junction resistance is low and a large majority-carrier current flows. Options (A), (B), (D) describe reverse-bias/incorrect behaviour, and (E) is wrong because the forward current is dominated by majority carriers.

Evaluating each option:

• (A) False — an extrinsic (doped) semiconductor has trivalent/pentavalent impurities; intrinsic is pure.
• (B) False — pure Ge or Si is an intrinsic semiconductor.
• (C) False — semiconductor conductivity increases with temperature.
• (D) False — a rectifier does use p-n junction diodes.
• (E) True — in forward bias the external voltage opposes the built-in potential, so the barrier height (and depletion width) decreases.

Assess each pair:

• (A) Cut-in voltage described as the voltage up to which a voltage-independent reverse current exists — wrong; cut-in (threshold) voltage is the forward voltage beyond which the diode conducts strongly. The description actually fits the reverse-saturation region.
• (B) Reverse bias — small leakage current below breakdown. Correct.
• (C) No bias — diffusion and drift balance, no net current. Correct.
• (D) Breakdown voltage — sudden large current (avalanche). Correct.
• (E) Forward bias — moderate ohmic conduction above the barrier voltage. Correct.

The mismatched pair is (A).

In the upper branch the second diode is reverse-biased, so that branch carries no current. All the 5 mA flows through the lower branch, whose two forward-biased ideal diodes drop $0$ V and leave only the $2.4\,\text{k}\Omega$ resistor:

Concept and Intuition

In reverse bias the external field aligns with the built-in junction field, widening the depletion region and raising the potential barrier. This suppresses majority-carrier flow, allowing only a tiny minority-carrier (reverse saturation) current.

Step-by-Step Solution

1. Reverse bias adds to the built-in field.
2. Depletion width increases.
3. Potential barrier height increases.

Common Mistakes

• Confusing with forward bias, which lowers the barrier; and thinking reverse bias creates minority carriers rather than just driving the small existing ones.

ANSWER: E

[!TLDR]

With no external bias to switch it between conducting and blocking, a pn junction diode has no rectifying action and merely presents an ohmic resistance.

Concept

A pn junction develops a depletion region and a built-in potential at equilibrium. Its device functions (rectification, voltage regulation, etc.) all depend on an applied bias that either lowers the barrier (forward) or widens it (reverse). Remove the bias and none of those functions can operate.

Solution

• Rectifier (A): needs an alternating applied voltage so the diode is alternately forward- and reverse-biased; impossible with no bias.
• AC generator (C): a diode cannot generate emf.
• Voltage regulator (D): a Zener diode regulates only when reverse-biased in breakdown.
• Transformer (E): involves mutual induction between coils, unrelated to a diode.
• Resistor (B): with no bias the junction is neither conducting a controlled forward current nor blocking a signal; it simply offers the resistance of its material. This is the only role consistent with "without any voltage biasing."

[!ANSWER]

(B) resistor

The diode equation $I=I_0\left(e^{eV/kT}-1\right)$ gives, for forward bias ($V>0$), a current that rises exponentially with voltage. Hence the I–V characteristic in forward bias is an exponentially increasing curve.

When a p–n junction diode is forward biased (p connected to +, n to −):

• the applied field opposes the built-in field, so the depletion layer narrows;
• the potential barrier decreases;
• majority carriers cross the junction easily, giving a large forward current (milliamperes);
• the junction therefore presents very low resistance.

Options (B), (C), (D) and (E) all describe reverse bias behaviour (wider depletion layer, higher barrier, tiny microampere leakage due to minority carriers). Hence the correct statement is that the junction provides very low resistance.

The built-in potential $V_0$ opposes majority-carrier flow. A forward-bias voltage $V$ is applied so as to oppose $V_0$, reducing the net potential across the junction. Hence the effective barrier height becomes $V_0-V$.

Under reverse bias the p-side is made negative and the n-side positive, so the applied field is in the same direction as the built-in barrier field (not opposite). This widens the depletion layer and raises the potential barrier. Hence the correct statement is that the barrier height increases. (Options A, B, C describe forward-bias/opposite behaviour, and minority carriers do cross, giving the small reverse current.)

In reverse bias, the external field aids the built-in field of the junction. This widens the depletion layer and increases the potential barrier, so majority-carrier current is (nearly) blocked; only a small minority-carrier reverse (leakage) current flows. It does not increase majority carriers or lower the barrier.