# Majority Vs Minority Carriers

Why are electrons considered to be the majority carriers in N type semi-conductors?
An explanation from the website www.t-pub.com[/url] ([url]http://www.tpub.com/content/neets/14179/css/14179_26.htm[/URL]) is:
Since N-type semiconductor has a surplus of electrons, the electrons are considered MAJORITY carriers, while the holes, being few in number, are the MINORITY carriers. But aren't the number of holes and electrons equal? Or have i misunderstood the explanation (most likely)?

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Mapes
Homework Helper
Gold Member
The number of holes and electrons aren't equal in doped semiconductors. The overall charge is zero, but that's because the dopant atoms have more or less protons than the host atoms.

Mapes is right on...see here for one explanation:
http://en.wikipedia.org/wiki/Acceptor_(semiconductors [Broken])

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The number of holes and electrons aren't equal in doped semiconductors. The overall charge is zero, but that's because the dopant atoms have more or less protons than the host atoms.

Okay, tell me if i have got the right idea now:
When we dope silicon with Boron, a Si=Si bond is broken to give a Si^-Si^ (^ refers to a hole). However, the electron corresponding to the hole is used in a valence bonding with Boron rather than in the conduction band. So, we have two holes, one new valence bond but NO additional conducting electron

BUT:

When we dope silicon with Phosphorous, the extra phosphorous electron gets delocalised into the conduction band giving a free electron but no hole.
Here is my question: Why didn't the loss of an electron from phosphorous result in a formation of hole carrier on phosphorous atom which makes the number of carrier holes and conducting electrons equal?
Is it because, the hole on phosphorous is not mobile thus not making it a carrier?

Okay, tell me if i have got the right idea now:
When we dope silicon with Boron, a Si=Si bond is broken to give a Si^-Si^ (^ refers to a hole). However, the electron corresponding to the hole is used in a valence bonding with Boron rather than in the conduction band. So, we have two holes, one new valence bond but NO additional conducting electron

BUT:

When we dope silicon with Phosphorous, the extra phosphorous electron gets delocalised into the conduction band giving a free electron but no hole.
Here is my question: Why didn't the loss of an electron from phosphorous result in a formation of hole carrier on phosphorous atom which makes the number of carrier holes and conducting electrons equal?
Is it because, the hole on phosphorous is not mobile thus not making it a carrier?

when a semiconducter is doped with phosphorus atom containing 5 electron in its outermost shell.Then its 4 electrons are bonded with si atom & one electron left free .This electron is not bonded to any atom so on supply of energy =1.1ev it gets out from the in fluence of pho.atom ,in case of p type semi conducter boron is doped containing 3 electron in o/s .This 3 electron forms bond with 3 atom of si &now si need 1 electron more two complete its octate so this need of electron create a vacancy there resulting in hole.........,,,,,,