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? (a) $D_1$ is forward biased and $D_2$ is reverse biased and hence current flows from A to B. (b) $D_2$ is forward biased and $D_1$ is reverse biased and hence no current flows from B to A and vice versa. (c) $D_1$ and $D_2$ are both forward biased and hence current flows from A to B. (d) $D_1$ and $D_2$ are both reverse biased and hence no current flows from A to B and vice versa.

> [!TLDR] > B (earthed, $0\,\text{V}$) is at a higher potential than A ($-10\,\text{V}$), so the circuit tries to drive current from B to A. On that path $D_1$ is reverse biased and blocks it. Since $D_1$ and $D_2$ are in series, no current flows either way. ### Concept A is fixed at $-10\,\text{V}$ and B is earthed at $0\,\text{V}$, so $V_B > V_A$. Conventional current would flow from the higher potential (B) to the lower (A), i.e. along B $\to D_2 \to$ R $\to$ A. ### Test each diode on that path - $D_2$ has its anode on the B (earth) side, so B $\to D_2$ is anode $\to$ cathode = **forward biased** (it would conduct). - $D_1$ has its cathode facing the junction and anode on the A/resistor side, so travelling from the junction back to R is cathode $\to$ anode = **reverse biased** (it blocks). Because the two diodes are in series and $D_1$ is reverse biased, the branch is open — no current flows from B to A. Flow from A to B is impossible as well, since $V_A [!ANSWER] > **(B)** $D_2$ is forward biased and $D_1$ is reverse biased, so no current flows from B to A (or vice versa).

Point A is held at -10V and point B is earthed (at 0V). Starting from A, a resistor R is in series with an ide

NCERT Exemplar · Q3

-10\,\text{V}

(a)

D_1

is forward biased and

D_2

is reverse biased and hence current flows from A to B.

(b)

D_2

is forward biased and

D_1

is reverse biased and hence no current flows from B to A and vice versa.

(c)

D_1

and

D_2

are both forward biased and hence current flows from A to B.

(d)

D_1

and

D_2

are both reverse biased and hence no current flows from A to B and vice versa.

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✓ Free question

Concept understanding — P N Junction Biasing

P-N Junction Biasing: The First Meeting

Imagine you have two rooms separated by a door. One room is full of people who want to leave (electrons in the N-side), and the other room is full of empty chairs (holes in the P-side). The door is initially locked — that's the depletion region, a zone with no free charges. Now, what happens if you push on the door from one side, or pull it from the other? That's exactly what biasing does to a P-N junction.

The Unbiased Junction (The Starting Point)

When you bring P-type and N-type semiconductors together, something interesting happens at the boundary. Electrons from the N-side (which has extra electrons) rush toward the P-side (which has extra holes). Holes from the P-side rush toward the N-side. They meet and recombine — an electron falls into a hole, and both disappear as free charges.

This leaves behind a region near the junction with no free charges — just fixed positive ions on the N-side (where electrons left) and fixed negative ions on the P-side (where holes left). This is the depletion region. It acts like a tiny battery: the N-side is positive relative to the P-side, creating a built-in electric field that stops further diffusion. The voltage across this region is about 0.7 V for silicon, 0.3 V for germanium.

Note

The depletion region is not a physical gap — it's a region depleted of mobile charge carriers. The crystal is still continuous.

Forward Bias: Pushing the Door Open

Connect the positive terminal of a battery to the P-side and the negative terminal to the N-side. This is forward bias.

What happens? The external battery's positive terminal repels holes in the P-side toward the junction. The negative terminal repels electrons in the N-side toward the junction. Both carriers are pushed toward each other — they overcome the built-in electric field. The depletion region shrinks.

When the applied voltage exceeds about 0.7 V (for silicon), the depletion region becomes so thin that carriers can cross freely. A large current flows. The junction is now "on" — like a switch closed.

Watch out

Never apply forward bias without a current-limiting resistor. The junction has very low resistance once forward-biased, and the current can destroy it instantly.

Reverse Bias: Pulling the Door Shut

Now reverse the battery: positive to N-side, negative to P-side. This is reverse bias.

The positive terminal attracts electrons from the N-side away from the junction. The negative terminal attracts holes from the P-side away from the junction. Both carriers are pulled apart. The depletion region widens.

The built-in electric field is now reinforced by the external field. Only a tiny current flows — the reverse saturation current, caused by thermally generated electron-hole pairs in the depletion region. This current is typically in the nanoampere range and is almost independent of the applied voltage (until breakdown).

The junction is "off" — like a switch open.

Important

In reverse bias, the depletion region acts as an insulator. The junction blocks current flow except for a negligible leakage current.

The Precise Statement

I = I_S \left( e^{\frac{qV}{nkT}} - 1 \right)

This is the Shockley diode equation. Here:

$I$ = diode current
$I_S$ = reverse saturation current (very small, typically $10^{-12}$ to $10^{-15}$ A for silicon)
$q$ = electron charge ( $1.6 \times 10^{-19}$ C)
$V$ = applied voltage (positive for forward bias, negative for reverse bias)
$n$ = ideality factor (1 for ideal, 1–2 for real diodes)
$k$ = Boltzmann constant ( $1.38 \times 10^{-23}$ J/K)
$T$ = absolute temperature (K)

For forward bias ( $V > 0$ ), the exponential term dominates — current grows rapidly. For reverse bias ( $V < 0$ ), the exponential term becomes negligible, and $I \approx -I_S$ — a tiny constant current.

Summary Table

Condition	Bias	Depletion Region	Current
No external voltage	Unbiased	Moderate width	Zero net current
P positive, N negative	Forward bias	Shrinks	Large (exponential)
P negative, N positive	Reverse bias	Widens	Tiny (saturation)

Why This Matters

Every diode, LED, solar cell, and transistor relies on this principle. A solar cell is just a P-N junction under forward bias from light. A transistor uses two junctions back-to-back. The ability to control current flow with a voltage — to switch between "on" and "off" — is the foundation of all modern electronics.

Tip

Remember the mnemonic: Positive to P-side = Forward bias (current flows). Negative to P-side = Reverse bias (current blocked). The arrow in the diode symbol points from P to N — the direction of conventional current when forward-biased.

Forward and reverse biasing of the p-n junction, along with the Shockley diode equation, is a core numerical and conceptual topic in the NCERT Class 12 Physics Semiconductor Electronics chapter, frequently searched as "p-n junction biasing important questions" by CBSE board and JEE Main aspirants. This concept is also essential groundwork for understanding rectifiers and transistor circuits covered later in the same syllabus.

Why this formula?

Why a PN Junction Biases the Way It Does

A PN junction is not a resistor. Its current-voltage behaviour is fundamentally asymmetric — and that asymmetry comes directly from the physics of the depletion region and the energy barrier it creates.

The Unbiased Junction: A Built-in Barrier

When P-type and N-type semiconductors are joined, holes from the P-side diffuse into the N-side, and electrons from the N-side diffuse into the P-side. This diffusion leaves behind fixed, charged ions: negative acceptor ions on the P-side, positive donor ions on the N-side. These ions create an electric field that opposes further diffusion.

The result is a depletion region — a zone with no free carriers — and a built-in potential $V_0$ across it. This potential acts as an energy barrier that prevents net current flow at equilibrium.

Important

At equilibrium (zero bias), the net current is zero. The diffusion current (due to concentration gradient) is exactly balanced by the drift current (due to the built-in electric field).

Forward Bias: Lowering the Barrier

Apply a positive voltage $V$ to the P-side relative to the N-side. This external voltage opposes the built-in potential. The net barrier becomes:

V_{\text{net}} = V_0 - V

The depletion width shrinks. More importantly, the energy barrier for majority carriers (holes from P-side, electrons from N-side) is reduced. A larger number of carriers now have enough energy to cross the junction.

The current that flows is diffusion current — carriers injected across the junction become minority carriers on the other side, where they recombine. The relationship is exponential because the number of carriers with energy above the barrier follows a Boltzmann distribution.

I = I_0 \left( e^{qV / kT} - 1 \right)

Here $I_0$ is the reverse saturation current (very small), $q$ is the electron charge, $k$ is Boltzmann's constant, and $T$ is absolute temperature.

Why the exponential? The fraction of carriers with enough energy to surmount a barrier of height $q(V_0 - V)$ is proportional to $e^{-q(V_0 - V)/kT}$ . At equilibrium ( $V=0$ ), this gives a current that exactly cancels the drift current. When $V>0$ , the barrier drops, and the net current becomes proportional to $e^{qV/kT}$ .

Reverse Bias: Raising the Barrier

Apply a negative voltage to the P-side relative to the N-side. Now the external voltage adds to the built-in potential:

V_{\text{net}} = V_0 + V_R

The barrier becomes higher. The depletion region widens. Diffusion of majority carriers is almost completely suppressed. The only current that flows is a tiny reverse saturation current $I_0$ , carried by minority carriers (electrons from the P-side, holes from the N-side) that are swept across by the electric field.

This current is essentially constant with voltage because the number of minority carriers is fixed by thermal generation — it does not depend on the barrier height once the barrier is large enough to block majority carriers.

Watch out

The reverse current is not zero — it is very small (nanoamps to microamps for silicon) but present. It doubles roughly every 10°C rise in temperature because thermal generation of minority carriers increases.

The Complete Picture: The Diode Equation

The single equation that captures both forward and reverse behaviour is:

I = I_0 \left( e^{qV / nkT} - 1 \right)

where $n$ is the ideality factor (typically 1 for ideal diodes, 1–2 for real diodes).

Forward bias ( $V > 0$ ): The exponential term dominates, current grows rapidly.
Reverse bias ( $V < 0$ ): The exponential term vanishes, $I \approx -I_0$ (a small constant).
At $V=0$ : $I = 0$ — the equation correctly gives zero net current.

Tip

For quick calculations at room temperature (300 K), remember $\frac{kT}{q} \approx 0.026$ V. So $e^{V/0.026}$ gives the factor by which current increases for every 26 mV of forward bias — a handy rule of thumb.

Why Not Ohm's Law?

A PN junction does not obey Ohm's law because the number of carriers available to conduct current is not constant — it depends exponentially on the applied voltage. The junction is a non-linear device: its resistance changes dramatically with bias direction and magnitude.

In forward bias, the resistance is low and decreases as voltage increases. In reverse bias, the resistance is extremely high (megohms) until breakdown occurs.

This asymmetry — the ability to conduct in one direction and block in the other — is the fundamental reason the PN junction is the building block of almost all semiconductor devices.

B (earthed, $0\,\text{V}$ ) is at a higher potential than A ( $-10\,\text{V}$ ), so the circuit tries to drive current from B to A. On that path $D_1$ is reverse biased and blocks it. Since $D_1$ and $D_2$ are in series, no current flows either way.

Concept

A is fixed at $-10\,\text{V}$ and B is earthed at $0\,\text{V}$ , so $V_B > V_A$ . Conventional current would flow from the higher potential (B) to the lower (A), i.e. along B $\to D_2 \to$ R $\to$ A.

Test each diode on that path

$D_2$ has its anode on the B (earth) side, so B $\to D_2$ is anode $\to$ cathode = forward biased (it would conduct).
$D_1$ has its cathode facing the junction and anode on the A/resistor side, so travelling from the junction back to R is cathode $\to$ anode = reverse biased (it blocks).

Because the two diodes are in series and $D_1$ is reverse biased, the branch is open — no current flows from B to A. Flow from A to B is impossible as well, since $V_A < V_B$ .

Why the other options fail

(A), (C): require current from A to B, but $V_A < V_B$ , and $D_1$ blocks that direction anyway.
(D): $D_2$ is actually forward biased, not reverse.

✓Final answer

D_2

Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

Higher Secondary (+2 Stage) Examination 2025Set ANNUAL1 markMCQ
$Q. When a p-n junction diode is reverse biased, then —$
(a) the potential barrier decreases and the depletion region widens
(b) the potential barrier increases and the depletion region narrows
(c) the potential barrier decreases and the depletion region narrows
(d) the potential barrier increases and the depletion region widens
›Reveal solutionSolution

Reverse bias pulls majority carriers further away from the junction, so both the depletion width and the barrier potential increase.

In reverse bias, the p-side is connected to the negative terminal and the n-side to the positive terminal of the external battery. This external field points in the same direction as the internal junction field, so it pulls holes (in the p-region) and electrons (in the n-region) further away from the junction on either side. This widens the depletion region and increases the potential barrier across the junction -- which is why only a tiny reverse saturation current can flow.

✓Final answer
$(d) the potential barrier increases and the depletion region widens.$
Higher Secondary (+2 Stage) Examination 2024Set ANNUAL1 markMCQ
$Q. In which of the following connections is the diode reverse-biased?$
(a) +5V — diode (anode left, cathode right) — resistor — +3V
(b) −4V — diode (anode left, cathode right) — resistor — −3V
(c) 0V — diode (anode left, cathode right) — resistor — −2V
(d) +2V — diode (anode left, cathode right) — resistor — −2V
›Reveal solutionSolution
A diode is reverse-biased whenever the terminal on its anode (p) side is at a LOWER potential than the terminal on its cathode (n) side; since no current flows in reverse bias, there is no potential drop across the series resistor, so the resistor's far-end voltage equals the cathode's potential.

In each option, the anode of the diode faces the left-hand terminal, and the resistor carries no voltage drop if the diode blocks conduction (reverse bias), so the cathode is effectively at the same potential as the right-hand terminal. The diode conducts (forward bias) only if the anode-side terminal is at a higher potential than the cathode-side terminal; it blocks conduction (reverse bias) if the anode-side terminal is at a LOWER potential.

Check each option (anode-terminal vs cathode-terminal):

(a) +5V vs +3V: anode side higher → forward biased.

(b) −4V vs −3V: anode side (−4V) is LOWER than cathode side (−3V) → reverse biased.

(c) 0V vs −2V: anode side higher → forward biased.

(d) +2V vs −2V: anode side higher → forward biased.

Only option (b) has the anode side at a lower potential than the cathode side.

✓Final answer
$(b) -4V — diode — resistor — -3V (reverse-biased).$
Higher Secondary (+2 Stage) Examination 2023Set ANNUAL1 markMCQ
$Q. Which of the following represents a forward-biased diode?$
(a) 0V —(diode + resistor)— -2V
(b) -4V —(diode + resistor)— -3V
(c) +3V —(diode + resistor)— +5V
(d) -2V —(diode + resistor)— +2V
›Reveal solutionSolution
A p-n junction diode conducts (is forward biased) when its anode (the tail of the arrow symbol) is at a higher potential than its cathode (the arrowhead side); check each option for which end is higher.

The diode symbol's arrow points from anode (p-side) to cathode (n-side), which is also the direction conventional current flows when the diode is forward biased. Forward bias requires: potential at the arrow's TAIL (left terminal here) > potential at the arrow's HEAD (right terminal).

Checking each option (left terminal vs right terminal):

0V vs −2V: 0 > −2 ✓ forward biased

−4V vs −3V: −4 < −3 ✗ reverse biased

+3V vs +5V: 3 < 5 ✗ reverse biased

−2V vs +2V: −2 < 2 ✗ reverse biased Only option (a) has the left (anode) terminal at a higher potential than the right (cathode) terminal, so only in (a) does the diode conduct.

✓Final answer
$(a) 0V —(diode+resistor)— -2V.$

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Concept

Test each diode on that path

Why the other options fail

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