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.

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.

The diode equation

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

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:

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.