Formation of Ions in diode

Question

I have some basic diode questions. When we forward bias a diode the electrons in the N region cross over the depletion region and enter the P region. Here they are supposed to recombine with the holes which are abundant in the P region. However my issue is this. When electron and holes recombine dont they form an ion? And ions are not mobile. So what mistake am I making here?

Answer 1

This answer will describe what happens on the P side of PN junction. The same concepts are readily applied to the N side with appropriate changes in signs of mobile and static charge.

PN junction schematics:

Forward bias:

Under forward bias conditions (P side biased positively relative to N side), electrons are injected from the N side into the depletion region. They are swept by the electric field across the depletion region and are injected into the P side quasi-neutral region:

Due to this injection, the concentration of electrons is increased dramatically near the edge of the depletion region on the P side. There is no net electric field in the quasi-neutral region which can affect these electrons anymore, however they keep diffusing away from the depletion region due to concentration gradients.

While diffusing through the quasi neutral P side, excess electrons have a chance to "meet" a hole and perform a recombination. The hole is just a vacant electron state some atom or ion - yes, the recombination can either ionize the acceptor atom, or neutralize an ionized silicon atom. We say "electron recombines with hole", but in reality it is just that the electron is trapped in the vacant place in the lattice.

Anyway, it seems like this process leads to a loss of mobile carriers and your question is: "if the mobile carriers are lost, how the current is sustained?". Am I right?

The answer:

You forget about charge conservation: if the electrons which are being injected into P region were recombining with the holes that initially present there, there would be a net increase in the negative charge on the P side. This mechanism can't sustain a steady state current which is flowing through the diode.

What happens is that for each hole which recombines with excess electron on the P side, there is one additional hole which is being supplied from the P side metal contact. This means that there is a diffusion current of electrons which is supplied from the N side, but there is also a current of holes which compensates for the recombination, which is supplied from the P side contact (I say "P side contact", but it is a power supply who is responsible for supplying this current).

The current densities are:

On the above graphs you can see that the sum of electrons current and holes current is constant throughout the diode. This means that holes which are lost in recombination are compensated by the current from the P side metal contact - the net charge density remains the same in quasi-neutral P region (neutral), and the amount of free charge carriers remains the same.

In summary:

In steady state forward bias:

• The current density is constant throughout the device
• The charge neutrality is preserved in quasi-neutral regions

Answer 2

From Wikipedia:

When N-doped and P-doped pieces of semiconductor are placed together to form a junction, electrons [move towards] the P-side and holes [move towards] the N-side. Departure of an electron from the N-side to the P-side leaves a positive donor ion behind on the N-side, and likewise the hole leaves a negative acceptor ion on the P-side.

[During this transference when] the ... electrons come [in] contact with [the] holes, [they] are eliminated by recombination. [Similarly in the case] of [the] N side. The net result is the diffused electron and holes are gone, leaving the charged ions adjacent to the interface in a region ... called the "Depletion Layer".

Rest of the ions are responsible for the electric field