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I would like to compare my result in an order of magnitude. So, How can I estimate the strength of the electric field in a typical Si PN-junction?

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Hi Stanpol. Welcome to Physics.SE. I think your question is like challenging other users for doing your work. It phrases, "Can you estimate..." <-- sometimes, it can be mistaken easily ;-) –  Waffle's Crazy Peanut Mar 22 '13 at 13:16
    
Hi, thanks for your comment. How do you think I should ask this type of questions? "Hey, look, check my estimation?" :) –  Stanpol Mar 22 '13 at 13:20
    
I think it's much better now. I've changed the order into a request. Hope I didn't phrase it differently..? ;-) –  Waffle's Crazy Peanut Mar 22 '13 at 13:23
    
Everything is fine, thanks :) –  Stanpol Mar 22 '13 at 13:24

2 Answers 2

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The maximum electric field develops near the depletion layer around the p-n interface. The maximum value, which occurs at $x=o$ (the interface) is given by the equation

$E_{max}=\frac{2(V_{bi}-V-kT/e)}{W}$

where:

$W$: is the width of the depletion layer,

$V_{bi}$: is the difference between the highest and lowest values of the bottom of C-B in the n-type, or the between the highest and lowest values of the top of the V_B in the p-type.

$T$: is the temperature of the semiconductor material

$V$: is the applied voltage

$k$: is Boltzmann’s constant

$e$: the electric charge on the electron

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You can estimate this quantity by dividing the order of magnitude of the potential difference (e.g. a volt) by the order of magnitude of the width of the depletion region (e.g. a micron).

You may be surprised regarding how large the correct value is.

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I am surprised, that's why I am asking... –  Stanpol Mar 22 '13 at 13:23
    
The electric field is large (on the order of a V / micron as noted above). –  Joshua Barr Mar 22 '13 at 13:26

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