Lesson 8 of 12 · 7 min
Diode under forward and reverse bias
NCERT §14.6
Rohan clips one diode to a cell and a bulb. One way round, the bulb glows. Flip the diode, and it stays dark.
The lesson in notes
In short
A semiconductor diode is a p-n junction with metal contacts at its ends, a two-terminal device. In its symbol, the arrow points in the conventional direction of current under forward bias.
Forward bias: p-side to the battery's positive terminal, n-side to the negative. The applied voltage drops almost entirely across the depletion region, whose resistance far exceeds that of the p and n regions.
Forward bias opposes the built-in potential V₀, so the depletion layer narrows and the barrier falls to (V₀ − V).
A small V lowers the barrier a little and only the most energetic carriers cross, so current is small. A larger V lowers it further, many more carriers cross, and the current rises.
Electrons cross into the p-side and holes into the n-side, where each is a minority carrier: this is minority carrier injection. Their concentration peaks at the junction edges and they diffuse away from it, and these diffusion currents make up the forward current, typically in mA.
Reverse bias: n-side positive, p-side negative. The applied voltage adds to the barrier, raising it to (V₀ + V) and widening the depletion region.
This all but stops diffusion (electrons n → p, holes p → n), so the diffusion current falls enormously compared with forward bias.
Minority carriers that wander near the junction are swept across by its field to their majority side. This drift current is only a few µA, because minority carriers are scarce; under forward bias it is still there but negligible beside the mA injected current.
The reverse current hardly depends on voltage: even a small reverse bias sweeps across every minority carrier that arrives, so the current is limited by the supply of minority carriers, not by V.