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?

  • \$\begingroup\$ Defintely they make an ion with no mobile carriers (that's we called Depletion Layer) \$\endgroup\$
    – Atom
    Sep 11, 2013 at 12:15
  • \$\begingroup\$ I am not talking about the ions in the depletion region. I am talking about the ones formed when the diode is forward biased and the electrons and holes recombine \$\endgroup\$ Sep 11, 2013 at 12:27
  • \$\begingroup\$ The entire diode is neutral as a whole, taking into consideration the depletion region also. So why do you say donor atom is ionized with a +1 charge. Also just to be clear I am talking about the region outside the depletion region and not the depletion region itself \$\endgroup\$ Sep 11, 2013 at 14:14
  • \$\begingroup\$ I deleted my earlier erroneous comment. A hole exists when a acceptor atom is in the silicon lattice, and if an electron is trapped in the hole it will form a negative ion. Even though the ion is not mobile, the trapped electron is weakly held and the applied voltage, along with thermal energy, will soon free it to move a little further toward the positive (anode) voltage. \$\endgroup\$
    – Joe Hass
    Sep 11, 2013 at 17:22

2 Answers 2


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:

enter image description here

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:

enter image description here

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:

enter image description here

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
  • \$\begingroup\$ Really cool. I guess I didnt think of it. So the hole combines with the negative ion on the P side and all becomes neutral again correct? \$\endgroup\$ Sep 16, 2013 at 11:41
  • \$\begingroup\$ @user2714712 What you're saying seems incorrect. I expanded my answer to provide a bit more intuition. For more than intuition you are welcome to read the link in the answer. \$\endgroup\$
    – Vasiliy
    Sep 16, 2013 at 12:32

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

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    \$\begingroup\$ I really don't understand the "phantom downvoters" on this SE. If an answer offends your elite sensibilities, at least a few words of wisdom should be bestowed when smiting with the downvote axe. Or post your own perceived-as-gospel answer. How can you expect newcomers to participate if punitive downvotes are poured asunder as deterrents to participation? \$\endgroup\$
    – Ron J.
    Sep 11, 2013 at 12:41
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    \$\begingroup\$ Plagiarized from Wikipedia without attribution of source: Para 1: "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." Para 2: "Following transfer, the diffused electrons come into contact with holes on the P-side and are eliminated by recombination. Likewise for the diffused holes on the N-side.". Plagiarism site flagged this post. SHASWAT is back with his habit of stealing without providing source. \$\endgroup\$ Sep 11, 2013 at 12:45
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    \$\begingroup\$ -1 for the reason explained by Anindo. \$\endgroup\$ Sep 11, 2013 at 12:49
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    \$\begingroup\$ @RonJ: Discussion of the way the site works really belongs on meta. Downvoting costs the downvoter reputation points so is probably not done without thought. I've been downvoted many times and don't resent it. I take it as a cue to review my posting carefully and improve or delete it. \$\endgroup\$ Sep 11, 2013 at 13:46
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    \$\begingroup\$ @SHASWAT, please memorize the citation format that RedGrittyBrick has added. In the future, you must cite your sources in this manner. Otherwise, you will again be "smitten with the downvote axe". \$\endgroup\$ Sep 12, 2013 at 3:11

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