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I am confused about a couple of points:

When the electrons diffuse from N side to P side, and holes diffuse from P side to N side, negative charged region is formed at the junction in P type and positive charged region is formed near the junction in the N type.

The question is:

  1. Why is the positive charge not distributed in the entire N type region (as there are free electrons for conduction in the N type region)?
  2. Why are the charges fixed at junction despite the presence of holes and free electrons?
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  • \$\begingroup\$ A few pointers for you: You only need to use one question mark at the end of a question. Observing English capitalization and grammar rules will go a long way to getting your questions the attention you would like. Almost all of your previous questions have been edited by other volunteers to help you out. \$\endgroup\$ – JYelton Feb 7 '14 at 16:35
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Why is the positive charge not distributed in the entire N type region (as there are free electrons for conduction in the N type region)?

It actually is. The diagrams are shown that way because we're primarily interested in the electric field produced by that charge, which is strongest near the junction, since that's where it's nearest the corresponding negative charge.

Why are the charges fixed at junction despite the presence of holes and free electrons?

Once the diffusion has stabilized, there are no free carriers in the central region, which becomes insulating, or "intrinsic" — in other words, it has the characteristics of pure (undoped) silicon. This insulating barrier becomes the dielectric between the two conductive regions, effectively forming a capacitor.

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  • \$\begingroup\$ That means that if we connect the outer end of the P type semiconductor with the outer end of the N type semiconductor , curent should flow ? \$\endgroup\$ – user28804 Feb 7 '14 at 17:01
  • \$\begingroup\$ You'd like to think so, but no. Additional junctions are formed between the semiconductor and whatever wire you might use to create such a connection, and it will always turn out that the fields at these junctions cancel the original P-N field. \$\endgroup\$ – Dave Tweed Feb 7 '14 at 17:04

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