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I've read that if there is very low doping in a pn-junction diode, the depletion region will be large because a large volume of depleted semiconductor is needed to generate enough electric field to balance the diffusion current. I have trouble understanding this.

How is it that a large volume of depleted semiconductor is needed to generate enough electric field to balance the diffusion current in case of low doping?

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  • $\begingroup$ As it stands, I can't quite understand what you are trying to ask. This may indicate that this question is too unclear/does not contain enough information to be useful to this site. Please try to improve the question by adding more information or rephrasing some things, pinpointing exactly what you're trying to ask. If this is not done, the question is likely to be closed. In your case, it would be useful to tell us your line of reasoning, so we can see where you have misunderstood. $\endgroup$ – Danu Sep 26 '14 at 15:50
  • $\begingroup$ I'm not sure anyone can 'make' you understand! But, to get you thinking - what is a depletion layer?, what is it depleted of?, and where does the stuff that gets depleted come from? $\endgroup$ – Jon Custer Sep 26 '14 at 23:06
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Let our semiconductor be a crystal of a Group 14 element, for clarity.

When you replace a Group 14 atom with a Group 15 atom, it releases one extra electron to the conduction band, and remains with only 4 valence electrons (instead of the regular 5), and hence acquires a positive charge. This is how a N-type semiconductor is created.

When you replace a Group 14 atom with a Group 13 atom, it brings only 3 electrons instead of 4 allotted for valence band, and one state becomes vacant, i.e. a hole is formed. Likewise, if this “primordial hole”☺ drifts away, the atom remains with 4 (instead of its own 3) valence electrons, and hence acquires a negative charge. This is how a P-type semiconductor is created.

Update: another text about formation of N- and P-type semiconductors is at When recombination in PN junction occurs, which atom becomes an ion?

More concentrated is your N-type (P-type) semiconductor, higher charge density will have respective side of the depletion zone, and thinner will it be for the same voltage drop. You can also refer to Reverse bias P-N junction and Charge density in depletion layer for pn-junction

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