I understood the formation of depletion region in a PN junction with no external bias applied.Immobile ions caused due to diffusion of holes and electrons causes electric field and hence the depletion region. Electric field pulls the majority carries back which is the drift current and the PN junction reaches equilibrium.

My question is when external bias is applied,PN junction is forward biased,why doesn't more immobile ions are formed as holes from P side will move towards the N side leaving an immobile negative ion and electron will move towards the P side leaving behind the immobile positive ion,which can cause the depletion width to increase until it balances the external electric field applied due to bias.More immobile ions will be created in forward bias PN junction and hence more the depletion region?


1 Answer 1


First, note that during reverse-bias, the depletion region becomes wider, while under forward bias the depletion region narrows down to nearly zero width.

why doesn't more immobile ions are formed

Answer: immobile ions are permanent. They are the dopant atoms which were added during manufacture of the diode. In doped silicon, during electric currents, no immobile ions are formed. This is because no phosphorus or boron atoms are being injected into the diode!!!

In p-type silicon, the number of immobile negative ions is constant. Electric currents do not move the immobile ions, nor do they alter the dopant density. And in n-type silicon, the number of immobile positive ions is constant as well. "Immobile" means unmoving. Unmoving under bias, unmoving under high current, unmoving under zero current.

Perhaps you have a "wrong picture" of carrier-motion in doped semiconductor?

The silicon is full of dopant atoms having permanent charge: the immobile ions. The silicon also contains an equal number of movable charge-carriers, mobile silicon ions with charge polarity opposite to the dopant. So, in phosphorus-doped silicon, the material contains positive phosphorus atoms as well as a "gas" made up of movable electrons. The electrons wander around freely, but they can only move in closed loops (complete circuits.) So, whenever an electron drifts farther away from a positive phosphorus atom, another electron moves closer to take its place, and the material remains neutral.

The mobile electrons don't leave any ions behind as they jump between lattice locations. The mobile electrons were always strongly negative, so when they occupy a lattice location, they act as a negative silicon ion. Then, when they move away from a lattice location, that location again becomes a neutral silicon atom: uncharged. (It does not become positive charged.)

Again, the material as a whole is only neutral because the number of positive phosphorus atoms is equal to the number of movable electrons. Also, the dopant density is usually very low, perhaps one mobile electron per 10^6 silicon atoms (plus also one immobile phosphorus atom per 10^6 silicon atoms, of course.) Doped semiconductor behaves like a sealed container with a very few charged particles wandering around inside. This is very different than metals, where every lattice location donates one or more movable charges.

To visualize n-type silicon, imagine a sparse, low-density gas made of electrons, mixed equally with a sparse, low-density "stain" made of positive phosphorus ions.

[edit] Note that all of the above applies to currents in conductive, carrier-filled silicon. Depletion-zones are different; they're insulating. Their population of mobile carries has been swept away, and only the immobile ions are left behind. Perhaps your question is about the boundary at the edge of the depletion region?

This boundary is not fixed, and the boundary location can move depending on applied e-fields (bias voltage.) But the motions of this boundary are a separate phenomenon from the usual electric currents occurring in conductive silicon. At the edge of the depletion zone, an e-field can sweep the "electron gas" out of the silicon, causing the depletion zone to expand. In that case there is no closed circuit, and when one electron moves away from a positive phosphorus atom, no other electron comes in to take its place. As an "electric current," this somewhat resembles the displacement current in capacitor dielectric. It's an increasing polarization, but rather than the dielectric atoms becoming more polarized (as with conventional capacitors,) the conductive region retreats and the insulating region expands, and the boundary between conductor and insulator is moving.

  • \$\begingroup\$ I agree the immobile ions are permanent.But these immobile ions only cause the electric field and the depletion region in unbiased PN junction right?If the mobile electron dont leave any ion then how the depletion region is formed? \$\endgroup\$
    – Shubham
    Commented Mar 5, 2017 at 6:53
  • \$\begingroup\$ In n-type silicon, if only one mobile electron moves, then another mobile electron must simultaneously move opposite. (Or, a closed loop of electrons can all move.) The rule for complete circuits is obeyed, and the material remains neutral. That's electric current in semiconductor. But, if all mobile electrons are swept out of the n-type silicon, then it becomes a positively-charged insulator. That's not DC, instead it's a brief event, "capacitor charging," only occurs during widening of the depletion zone. \$\endgroup\$
    – wbeaty
    Commented Mar 5, 2017 at 7:05

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