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vamshi
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« on: May 30, 2006, 01:24:27 PM »

rearding semiconductors where is fermi level located in an intrinsic,p type,n type

semiconductors and  why?
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Ante
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« Reply #1 on: June 06, 2006, 05:11:50 AM »

Hi,

Here is an explanation for you: http://hyperphysics.phy-astr.gsu.edu/hbase/solids/fermi.html
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prateeksikka
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« Reply #2 on: September 15, 2006, 04:55:10 AM »

fermi level is the level of energy  where there is a 50% possibility to locate an electron  or simply
its the energy which most of` the electrons have .
it depends on doping concentration ....
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prateek

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« Reply #3 on: January 31, 2011, 08:07:07 PM »

Regarding semiconductors where is fermi level(http://en.wikipedia.org/wiki/Fermi_level)located in an intrinsic,p type,n type  semiconductors and  why?
 
In an intrinsic semiconductor, n = p. If we use the band-symmetry approximation, which assumes that there are equal number of states in
equal-sized energy bands at the edges of the conduction and valence bands,
n = p implies that there is an equal chance of finding an electron at the conduction band edge as there is of finding a hole at the valence band edge:
                     f(EC) = 1 − f(EV )
  
    For an n-type semiconductor, there are more electrons in the conduction band than there are holes in the valence band. This also implies that the probability of finding an electron near the conduction band edge is larger than the probability of finding a hole at the valence band edge. Therefore, the Fermi level is closer to the conduction band in an n-type semiconductor.

n − type : f(EC) > (1 − f(EV )) ) |EC − EF | < |EF − EV | ) EF > Ei (13)

       For a p-type semiconductor, there are more holes in the valence band than there are electrons in the conduction band. This also implies that the probability of finding an electron near the conduction band edge is smaller than the probability of finding a hole at the valence band edge. Therefore, the Fermi level is closer to the valence band in an n-type semiconductor.

p − type : f(EC) < (1 − f(EV )) ) |EC − EF | > |EF − EV | ) EF < Ei

Please visit spam removed for more information like this in our articles and support section.
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gogo2520
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« Reply #4 on: January 31, 2011, 11:36:01 PM »

AmericanCircuitTech That was a loaded question, a waste of are time, your a jerk.
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indulis
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« Reply #5 on: February 01, 2011, 07:56:29 AM »

Quote
Please visit spam removed for more information like this in our articles and support section.

There are no articles "like this" there, and some of what is there is highly questionable.
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Hero999
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« Reply #6 on: February 01, 2011, 12:14:30 PM »

Note how the link to Wikipedia goes nowhere?

That's because it's spam.
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opank
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« Reply #7 on: February 03, 2011, 09:08:59 AM »

OK i will try to help .!!!!
as we know the Fermi level is a limit to the transfer of electrons from the valence band to the conduction band. so when the electron is below the Fermi level position then he will remain Valence electron band but when it is above the Fermi level position then he will be in the conduction band but if electron is located at the Fermi level position then he has the same chance to be in the conduction band / valence band.

as we also know that the position of the Fermi level is very dependent of the doping elements were given at a time element that is type (electron majority carriers), the Fermi levels will be more toward the conduction band, the more doping is given, the more she approached conduction band. contrary to p-type
to more clearly you can see the image that I attach

source: microelectronics an integrated approach (international edition)

I hope this answer can help you
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KevinIV
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« Reply #8 on: February 03, 2011, 02:34:28 PM »

Fermi level and energy bands have to play small roles. Current value is regardless of valence bands or conduction bands or the energy required. The reason is because energy levels applied to conductors and semiconductors are very high. Electrons can move to any energy level with a change in temperature.
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