The barrier potential of a silicon diode is approximately, (a) 0.7 V (b) 0.3 V (c) 2.0 V (d) 2.2 V

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Q. The barrier potential of a silicon diode is approximately,

(a) 0.7 V

(b) 0.3 V

(d) 2.2 V

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Concept understanding — P-N Junction Formation

P-N Junction Formation: From Intuition to Precision

Imagine two rooms connected by a door. One room is filled with people who have extra energy (they want to give it away), and the other room is filled with people who are missing energy (they want to take it). The moment you open the door, what happens? People rush from the high-energy room to the low-energy room until both rooms reach a balance. That rush, and the final balanced state, is the essence of a P-N junction.

In a semiconductor, the "people" are charge carriers: electrons (negative charge) and holes (the absence of an electron, which behaves like a positive charge). A P-type semiconductor has an excess of holes (positive carriers), and an N-type semiconductor has an excess of electrons (negative carriers). When you bring them together, they don't just sit still — they interact.

The Intuitive Picture

Take a P-type crystal and an N-type crystal. At the instant they touch, there is a huge concentration difference: lots of holes on the P-side, lots of electrons on the N-side. Nature hates steep gradients, so carriers begin to diffuse — they move from where they are abundant to where they are scarce.

Electrons from the N-side cross into the P-side.
Holes from the P-side cross into the N-side.

But here's the catch: when an electron from the N-side meets a hole on the P-side, they recombine — the electron fills the hole, and both disappear as free carriers. This recombination doesn't happen everywhere; it happens in a narrow region near the interface, called the depletion region (or space-charge region).

Why "depletion"? Because in that region, free electrons and free holes have been used up. All that remains are the fixed, immovable ions: positive donor ions on the N-side (which lost their electron) and negative acceptor ions on the P-side (which gained an electron). These fixed charges create an electric field that points from the N-side (positive ions) to the P-side (negative ions).

This electric field is crucial. It acts like a bouncer: it pushes electrons back toward the N-side and holes back toward the P-side. This drift motion opposes the initial diffusion. Eventually, the diffusion current (driven by concentration difference) exactly balances the drift current (driven by the electric field). The system reaches thermal equilibrium — no net current flows.

The Precise Statement

\text{A P-N junction is formed by bringing P-type and N-type semiconductors into intimate contact. At equilibrium, a depletion region of fixed ions creates a built-in electric field that prevents further net diffusion of carriers.}

More formally:

Diffusion: Majority carriers (electrons from N-side, holes from P-side) diffuse across the junction due to the concentration gradient.
Recombination: These carriers recombine near the interface, leaving behind fixed ionized impurities (donors on N-side, acceptors on P-side).
Depletion region: A region devoid of free carriers, containing only fixed charges, forms at the junction.
Built-in electric field: The fixed charges create an electric field ( $\vec{E}$ ) pointing from N to P.
Equilibrium: The drift current due to $\vec{E}$ exactly cancels the diffusion current. The net current is zero.

The width of the depletion region ( $W$ ) depends on the doping concentrations. For a one-sided junction (heavily doped on one side), the depletion region extends mostly into the lightly doped side.

Important

The P-N junction is not a physical gap — it is a metallurgical interface. The depletion region is a few micrometers wide, and the built-in potential ( $V_{bi}$ ) is typically 0.6–0.8 V for silicon at room temperature.

Why This Matters

This tiny region — the depletion region with its built-in field — is the heart of every diode, transistor, solar cell, and LED. It controls current flow: apply a forward bias (P positive, N negative), and the barrier shrinks, allowing current to flow. Apply a reverse bias (P negative, N positive), and the barrier grows, blocking current. That one-way street for current is the foundation of modern electronics.

Note

The built-in potential $V_{bi}$ is given by:

$V_{bi} = \frac{kT}{q} \ln\left(\frac{N_A N_D}{n_i^2}\right)$

where $N_A$ and $N_D$ are acceptor and donor concentrations, $n_i$ is the intrinsic carrier concentration, $k$ is Boltzmann's constant, $T$ is temperature, and $q$ is the electron charge. You don't need to memorize this now — just know that it arises from the balance of diffusion and drift.

The formation of the p-n junction and its depletion region is one of the most examined topics in the NCERT Class 12 Physics Semiconductor Electronics chapter, and "p-n junction formation and depletion region explained" is a heavily searched query for CBSE board and JEE Main revision. Because every diode and transistor question builds on this idea, it is consistently featured in "semiconductor electronics important questions" for competitive exams.

Why this formula?

Why the P-N Junction Forms: The Physics Behind the Barrier

A p-n junction isn't just two pieces of semiconductor stuck together. The key to understanding it is this: nature hates sharp gradients in carrier concentration. When you bring p-type (excess holes) and n-type (excess electrons) material into contact, carriers immediately begin to diffuse across the junction — holes from p to n, electrons from n to p.

This diffusion is the engine that drives everything else.

Step 1: Diffusion Creates a Depletion Region

As holes leave the p-side, they leave behind fixed, negatively charged acceptor ions ( $A^-$ ). As electrons leave the n-side, they leave behind fixed, positively charged donor ions ( $D^+$ ). These ions are immobile — they're locked in the crystal lattice.

The region near the junction that gets stripped of mobile carriers is called the depletion region (or space-charge region). It contains only fixed ions, creating an electric field that points from the n-side (positive ions) toward the p-side (negative ions).

Watch out

Do not confuse "depletion" with "no charge." The depletion region is highly charged — it's just that the charge is from fixed ions, not mobile carriers.

Step 2: The Electric Field Opposes Diffusion

The built-in electric field $E$ exerts a force on any mobile carrier that tries to cross:

Holes (positive) feel a force pushing them back toward the p-side.
Electrons (negative) feel a force pushing them back toward the n-side.

This field grows stronger as more carriers diffuse and more ions are uncovered. Eventually, the field becomes strong enough that the drift current (carriers swept by the field) exactly balances the diffusion current (carriers moving due to concentration gradient). At this point, the net current is zero — thermal equilibrium is reached.

Step 3: The Built-in Potential Barrier

Because the electric field exists over a distance, there is a potential difference across the depletion region. This is the built-in potential $V_0$ (also called $V_{bi}$ ). It represents the energy barrier that a majority carrier must overcome to cross to the other side.

V_0 = \frac{kT}{q} \ln\left(\frac{N_A N_D}{n_i^2}\right)

Where:

$k$ = Boltzmann constant
$T$ = absolute temperature
$q$ = electron charge magnitude
$N_A$ = acceptor doping concentration (p-side)
$N_D$ = donor doping concentration (n-side)
$n_i$ = intrinsic carrier concentration

Why This Formula Holds: The Derivation

The derivation comes from equating the Fermi levels on both sides. In equilibrium, the Fermi level must be constant throughout the entire structure.

On the p-side, the Fermi level $E_F$ lies close to the valence band. The position relative to the intrinsic Fermi level $E_i$ is:

E_F - E_i = kT \ln\left(\frac{N_A}{n_i}\right) \quad \text{(for p-type)}

On the n-side, the Fermi level lies close to the conduction band:

E_F - E_i = -kT \ln\left(\frac{N_D}{n_i}\right) \quad \text{(for n-type)}

The difference in $E_i$ between the two sides (which is the same as the difference in $E_F$ between the two sides before contact) must be accommodated by the built-in potential. The total band bending $qV_0$ equals this difference:

qV_0 = \left[ kT \ln\left(\frac{N_A}{n_i}\right) \right] - \left[ -kT \ln\left(\frac{N_D}{n_i}\right) \right]

qV_0 = kT \left[ \ln\left(\frac{N_A}{n_i}\right) + \ln\left(\frac{N_D}{n_i}\right) \right]

qV_0 = kT \ln\left(\frac{N_A N_D}{n_i^2}\right)

Dividing both sides by $q$ gives the formula for $V_0$ .

Important

The built-in potential depends only on doping concentrations and temperature — not on the junction area or geometry. It is typically 0.6–0.8 V for silicon and 0.2–0.3 V for germanium at room temperature.

Step 4: The Depletion Width

The electric field and potential drop are not uniform — they vary with position. Solving Poisson's equation in the depletion region (with the charge density being $+qN_D$ on the n-side and $-qN_A$ on the p-side) gives the width of the depletion region:

W = \sqrt{\frac{2\epsilon_s}{q} \left(\frac{1}{N_A} + \frac{1}{N_D}\right) V_0}

Where $\epsilon_s$ is the permittivity of the semiconductor.

The derivation is straightforward: Poisson's equation $\frac{d^2V}{dx^2} = -\frac{\rho}{\epsilon_s}$ is integrated twice with the boundary conditions that the electric field is zero at the edges of the depletion region and continuous at the metallurgical junction. The result shows that the depletion region extends farther into the lightly doped side — a fact that becomes critical for device design.

The Key Insight

The p-n junction is a self-limiting structure. Diffusion tries to equalise carrier concentrations, but the resulting electric field prevents complete mixing. The equilibrium is a balance between two opposing forces — diffusion and drift — and the built-in potential is the quantitative measure of that balance.

Step 1. The barrier (junction) potential is set by the material the diode is made from, not by any external circuit values.

Step 2. For silicon at $25^\circ$ C this equilibrium potential is approximately 0.7 V (for germanium it is about 0.3 V, ruling out option (b)).

✓Final answer

(a) 0.7 V

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Q.The barrier potential of a silicon diode is approximately,

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