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I have a question about how the PN junction works in a diode. According to http://hyperphysics.phy-astr.gsu.edu/hbase/solids/pnjun.html:

"Filling a hole makes a negative ion and leaves behind a positive ion on the n-side. A space charge builds up, creating a depletion region which inhibits any further electron transfer unless it is helped by putting a forward bias on the junction."

My question is: why does the filling a hole make a negative ion? Shouldn't the hole cancel out with the electron giving a neutrally charge ion? And the consequently, there would be no e-field created? I'm not sure what I'm missing here... Why are there positive and negative ions in the depletion region? I thought there'd be neutrally charge ions...

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A hole is not the result of a charge-neutral atom losing an electron. A hole is created when an "acceptor" atom is located in a silicon crystal but that atom does not have as many electrons available for bonding as do the silicon atoms. Silicon atoms bond by sharing a pair of electrons, each atom contributes one electron to the bond. The acceptor atom leaves one bond unfilled, and it's that unfilled bond that constitutes the hole. Note that although this hole exists the atoms are completely charge-neutral. It's easy for a wandering electron to get stuck in the hole, and when that happens the acceptor atom actually has one more electron than it normally would...thus it has become a negative ion. The captured electron came from somewhere...some atom that was also previously charge-neutral...so that atom has become a positive ion. Since we have an immobile negative ion and an immobile positive ion, an electric field exists between them.

As holes are filled with wayward electrons the e-field increases in strength until it prevents any more movement of electrons. At this point the depletion region has been created. This region is depleted of free (mobile) charge carriers but the impurity (non-silicon) atoms are ionized.

I've mentioned silicon but the same thing can be done with some other materials, such as germanium and gallium-arsenide.

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    \$\begingroup\$ Si has valency 4, i.e there are 4 bonds. A Donor (Phosphorus) atom has valency 5 and an Acceptor (like boron) has valency 3. All covalent bonds occur because of electron sharing, just not Si. You are conflating concepts. \$\endgroup\$ – placeholder Jul 27 '13 at 16:23
  • \$\begingroup\$ Correction, you aren't conflating, I miswrote, but the way it is written can lead people to think that it's the presence/absence in ONE bond that causes the acceptor/donor action. \$\endgroup\$ – placeholder Jul 27 '13 at 16:40
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I'm not sure what I'm missing here... Why are there positive and negative ions in the depletion region?

Joe gives a good overview of how a hole is created when an acceptor atom captures an electron that was previously moving freely in the valence band. Similarly a conduction-band electron is created when a donor atom releases an electron into the conduction band.

I am guessing that what you're missing in this is that the acceptor is an impurity atom in the crystal, and it is fixed in location (unless we start talking about the extremely high temperatures that are used during the diffusion process to actually get the impurity atoms into the crystal in the first place). Whereas the hole is free to move around. So the hole is free to be pushed around by the built-in field of the junction, but the acceptor (now a negatively charged ion) is stuck where it is.

I thought there'd be neutrally charge ions.

The word "ion" means an atom with a net positive or negative charge. If it weren't charged one way or the other, we'd just call it an "atom" and not an ion.

So why do the acceptors become ions instead of just staying neutral?

One way to look at it is that the Fermi-Dirac statistics tell us what is the probability of any particular state being filled with an electron at any particular time. Since acceptor atoms, for whatever reason, provide states that are near the valence band edge (and so below the Fermi level) those states are going to end up being filled more than 50% of the time.

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  • \$\begingroup\$ What did you say about acceptor atoms? In a non-charged p-type semiconductor the Fermi level lies below the upper edge of the valence band. That’s why there are scores of holes. \$\endgroup\$ – Incnis Mrsi Aug 26 '16 at 12:32
  • \$\begingroup\$ @IncnisMrsi, what you're describing (Fermi level below the valence band edge) only happens in cases of very high dopant concentration and is called degenerate doping. It is not true for all p-type semiconductors. In fact we often identify a degenerate doped p-type material as "p+" rather than just "p" type. \$\endgroup\$ – The Photon Aug 26 '16 at 16:34
  • \$\begingroup\$ Accepted. Ī̲ possibly confused the Fermi level with Fermi energy at 0K. \$\endgroup\$ – Incnis Mrsi Aug 26 '16 at 16:50

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