Effect on the width of the depletion region on doping level of an unbiased PN Jn.

  • Thread starter kihr
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  • #1
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Homework Statement


The problem is what would be the effect on the width of the depletion region of an unbiased PN junction on the doping levels of the P and N sides


Homework Equations



This is a conceptual question. There is no equation involved.

The Attempt at a Solution



Increase in doping level implies more free carriers (electrons in the n side and holes in the p side) available for diffusion. Thus the greater number of such available free carriers (as compared to the number of free carriers available for a lesser doping level) would neutralise the charges on a greater number of ions (positive and negative) in the depletion region. Hence the net effect would be to reduce the width of the depletion region.
I would request for a ratification of the correctness of this understanding. Thanks.
 

Answers and Replies

  • #2
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But what happens to the concentration of ions in the depletion region when you increase the doping level?
 
  • #3
102
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The concentration of ions goes up with increase in the doping level. Does this, therefore, imply that the width of the depletion region will increase for higher doping levels? It looks as if my initial thinking was not right.
 
  • #4
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One more thing. With a higher doping level the drift current should go up due to the higher electric field at the junction, and the diffusion current should go down due to the same reason. The net effect should be that the equilibrium condition would be attained faster in this case than if the junction were lightly doped.
 
  • #5
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I really don't know if you can solve this without using any formulas, since there seems to be a lot of factors. These are the formulas I would use

[itex]W = \sqrt{\frac{2\epsilon_r\epsilon_0 V_j}{q}\left(\frac{1}{N_a} + \frac{1}{N_d} \right)}[/itex], where
[itex]V_j = V_0 = \frac{kT}{q}\ln \left( \frac{N_aN_d}{n_i^2} \right)[/itex] for an unbiased junction

Hope this helps :)
 

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