PN Junction Diode: Forward and Reverse Bias

The junction and the depletion region

A PN junction diode is formed where P-type semiconductor (extra positive charge carriers, or "holes") meets N-type semiconductor (extra electrons). Right at the junction, electrons and holes diffuse across and cancel each other out, leaving a thin depletion region with no free charge carriers, and a small built-in electric field that opposes any further diffusion, holding the junction in equilibrium.

The depletion region isn't empty: it holds fixed ions (negative on the P side, positive on the N side) left behind by the carriers that recombined. Those ions are what create the built-in field and the barrier potential, typically quoted as about 0.7 V for silicon and 0.2 to 0.3 V for germanium.

Forward bias vs. reverse bias

Forward bias connects the positive terminal of a source to the P-side. This pushes against the built-in field, narrows the depletion region, and once the applied voltage exceeds a threshold (around 0.7V for silicon), current flows easily and grows rapidly with voltage.

Reverse bias connects the positive terminal to the N-side instead. This widens the depletion region and reinforces the built-in field, blocking current almost entirely: only a tiny leakage current flows, largely independent of voltage until breakdown.

Forward and reverse bias compared

	Forward bias	Reverse bias
P side connected to	Positive terminal	Negative terminal
Depletion region	Narrows	Widens
Current	Large, rises steeply past about 0.7 V (Si)	Tiny (μA or less) until breakdown
Carried mainly by	Majority carriers	Minority carriers

The diode equation

The current through a diode as a function of the voltage across it is:

I = I_0\left(e^{V/\eta V_T} - 1\right)

where $I_0$ is the small reverse saturation current, $\eta$ is an ideality factor (1 for an ideal diode), and $V_T = kT/q$ is the thermal voltage, about 26 mV at room temperature. Under forward bias, the exponential term dominates and the $-1$ becomes negligible.

Worked example: the "60mV per decade" rule

For an ideal diode ( $\eta=1$ ) at room temperature, by what factor does the current increase when the forward voltage rises from 0.60V to 0.66V?

Solution: Since forward current is approximately $I \approx I_0 e^{V/V_T}$ , the ratio between two currents depends only on the voltage difference:

\dfrac{I_2}{I_1} = e^{(V_2-V_1)/V_T} = e^{0.06/0.026} = e^{2.31}

which works out to about 10, a 60 mV increase in forward voltage multiplies the current roughly tenfold. This is exactly why a diode's forward voltage barely moves even as the current through it changes by orders of magnitude: the relationship is logarithmic, not linear.

Worked example: a diode in a simple circuit

A 5 V supply, a 1 kΩ resistor and a silicon diode are connected in series. Find the current when the diode is forward biased, and when it is reversed.

Solution: In circuit problems a forward-biased silicon diode is usually modelled as a fixed 0.7 V drop. KVL around the loop gives:

I = \dfrac{5 - 0.7}{1000} = 4.3\text{ mA}

Reversed, the diode blocks: the current is essentially zero, so there is no drop across the resistor and the full 5 V appears across the diode.

The diode as a rectifier

Because a diode conducts in one direction only, it turns AC into pulsating DC. A half-wave rectifier (one diode) passes only the positive half-cycles. A full-wave rectifier (two diodes with a centre-tapped transformer, or a four-diode bridge) flips the negative half-cycles too.

That changes the ripple frequency of the output. From a 50 Hz mains supply, a half-wave rectifier's output repeats at 50 Hz, while a full-wave rectifier's output repeats at 100 Hz, which is easier to smooth with a capacitor.

Common mistakes

Using 0.7 V for every diode. That's silicon. Germanium is about 0.2 to 0.3 V, and an ideal diode in a question means 0 V drop.
Saying the reverse current is exactly zero. A small, nearly constant leakage current flows until breakdown. It's zero only in the ideal model.
Mixing up which way the depletion region changes. Forward bias narrows it and reverse bias widens it.
Giving the full-wave ripple frequency as 50 Hz. Full-wave output pulses twice per input cycle, so it's twice the supply frequency.

Where this comes up in exams

The JEE Main 2026 and NEET (UG) 2026 syllabi list the semiconductor diode's I-V characteristics in forward and reverse bias, the diode as a rectifier, and the LED, photodiode, solar cell and Zener diode. GATE 2027 EC lists the P-N junction under Electronic Devices. The JEE Advanced 2026 syllabus has no semiconductor topics.

Checked against JEE (Main) 2026 syllabus (NTA), NEET (UG) 2026 syllabus (NMC, published by NTA), GATE 2027 EC syllabus and JEE (Advanced) 2026 syllabus. Syllabi can change from year to year, so confirm with the latest official notification for your exam.

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