Simulation · Physics · Class 12
p-n junction under bias
From the lesson Diode under forward and reverse bias in Semiconductor Electronics: Materials, Devices and Simple Circuits. Change the values and watch what happens.
p-n junction under biasPhysics · Class 12
The idea behind it
NCERT §14.6
- 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.