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Why Depletion Region Width Changes with Bias (and Temperature)

Physics · Semiconductor Electronics : Materials, Devices And Simple Circuits · NEET

Forward bias narrows the depletion region and reverse bias widens it. In forward bias the applied voltage opposes the built-in field, so fewer immobile ions are needed and the region shrinks. In reverse bias the applied voltage adds to the built-in field, so more ions are exposed and the region grows. Memory hook: "Forward = Feed carriers in = thin wall. Reverse = Rip carriers out = thick wall."

At a glance

Depletion widthNarrows (thinner)Widens (thicker)
Effective barrierV0 minus V (lower)V0 plus V (higher)
Applied field vs built-in fieldOpposes built-in fieldAdds to built-in field
CurrentLarge (mA), rises fastVery small (microamp)
Depletion Width vs BiasForward biaspnthinnarrowsNo biaspnmidequilibriumReverse biaspnwidewidens
The yellow band is the depletion region. Forward bias makes it thin, an unbiased junction keeps its equilibrium width, and reverse bias makes it wide.

Your doubts, answered

Does forward bias increase or decrease the depletion width?

Forward bias DECREASES (narrows) the depletion width. The battery pushes majority carriers toward the junction. These carriers move into the depletion region and neutralise some of the exposed immobile ions, so the charged region becomes thinner. At the same time the potential barrier (built-in voltage) drops from V0 to about (V0 minus V). Thinner wall plus lower barrier means current can flow easily.

Why does reverse bias increase the depletion width?

In reverse bias the positive terminal connects to the n-side and the negative terminal to the p-side. This pulls majority carriers AWAY from the junction. As they leave, more immobile donor and acceptor ions are exposed on both sides, so the space-charge region gets wider. The barrier rises to about (V0 plus V). A wider region and higher barrier is why almost no current flows in reverse bias.

What happens to the potential barrier when bias changes?

The built-in barrier V0 (about 0.3 V for Ge, about 0.7 V for Si) is not fixed once you apply a voltage. Forward bias lowers the effective barrier to (V0 minus V). Reverse bias raises it to (V0 plus V). Barrier height and depletion width always move together: lower barrier goes with a thinner region, higher barrier goes with a wider region.

Does the depletion region have any free charge carriers?

No. The depletion region is 'depleted' of free electrons and holes. It contains only fixed (immobile) positive donor ions on the n-side and fixed negative acceptor ions on the p-side. Because there are no mobile carriers, this region acts like an insulator and sets up the internal electric field and potential barrier.

How does temperature affect the depletion region?

Raising temperature generates more electron-hole pairs (more intrinsic carriers, ni increases). More carriers are available near the junction, which slightly reduces the built-in potential barrier V0 and tends to make the depletion region a little narrower. Higher temperature also increases the reverse saturation current, because minority carriers become more plentiful.

Does depletion width depend on doping level?

Yes. A more heavily doped side has more ions packed close together, so fewer atomic layers are needed to build up the same charge, making the depletion region penetrate less into the heavily doped side. The depletion region extends more into the lightly doped side. So width depends on both the applied bias AND the doping concentration.

⚠️ The NEET trap
✗ Forward bias widens the depletion region because current is flowing.
✓ Forward bias NARROWS the depletion region; reverse bias widens it. Current flowing has nothing to do with widening the wall.
🧠 Students confuse 'more current' with 'bigger region'. It is the opposite: the region shrinks so current can flow. Widening only happens in reverse bias, where current is almost zero. Read the direction of bias, not the current.

Real NEET questions

NEET 2020

The increase in the width of the depletion region in a p-n junction diode is due to:

A · Both forward bias and reverse bias
B · Increase in forward current
C · Forward bias only
D · Reverse bias only ✓
Solution: Step 1: Recall the built-in field points from the n-side (positive ions) to the p-side (negative ions). Step 2: In reverse bias the external field is applied in the same direction as this built-in field, so the total field increases. Step 3: The stronger field sweeps more majority carriers away from the junction, exposing more immobile ions on both sides. Step 4: More exposed ions means a WIDER depletion region. Forward bias does the opposite (it narrows the region). So the width increases only in reverse bias. Answer: D.
ReNEET 2026

Three identical p-n junction diodes D1, D2 and D3 are connected across a battery (each in series with a 1 kilo-ohm resistor). If the widths of the depletion regions of D1, D2, D3 are W1, W2, W3 respectively, then the correct option is:

A · W1 > W2 > W3
B · W3 = W1 > W2
C · W3 > W2 > W1 ✓
D · W2 > W1 = W3
Solution: Step 1: Identify the bias of each diode from the circuit. D1 is forward biased, D2 is unbiased (no net voltage), D3 is reverse biased. Step 2: Apply the rule. Forward bias narrows the region, so W1 is smallest. Reverse bias widens it, so W3 is largest. Step 3: The unbiased diode keeps its equilibrium width, which is in between, so W2 is intermediate. Step 4: Order from largest to smallest: W3 > W2 > W1. Answer: C.

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Frequently asked

In one line, what changes the depletion width?

The applied bias: forward bias makes it thinner, reverse bias makes it thicker.

What is the depletion width roughly?

About one-tenth of a micrometre (0.1 micrometre) at equilibrium, per NCERT. It changes with applied bias.

Why does reverse current stay almost zero even though the barrier is high?

Because the tiny reverse (saturation) current is carried only by minority carriers, and their number is very small and set by temperature, not by the applied voltage.

Does the barrier voltage V0 depend on the material?

Yes. It is about 0.3 V for germanium and about 0.7 V for silicon at room temperature.

Is the depletion region an insulator or a conductor?

It behaves like an insulator because it has no free charge carriers, only fixed ions.

Related concepts:
Depletion Region and Potential Barrier in a p-n JunctionForward Bias of a p-n Junction Diode ExplainedReverse Bias of a p-n Junction Diode ExplainedDifference Between Forward Bias and Reverse BiasV-I Characteristics of a p-n Junction Diode (Graph)
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