Why do doped semiconductors have majority carriers?

  • Thread starter Pranav Jha
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In summary, 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...,,,,,,
  • #1
Pranav Jha
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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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  • #2
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.
 
  • #3
Mapes is right on...see here for one explanation:
http://en.wikipedia.org/wiki/Acceptor_(semiconductors )
 
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  • #4
Mapes said:
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?
 
  • #5
Pranav Jha said:
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...,,,,,,
 

What is the difference between majority and minority carriers?

Majority carriers are the dominant type of charge carriers in a material, meaning they are more abundant and contribute to the majority of the current flow. Minority carriers, on the other hand, are present in lower concentrations and have a minor role in current conduction.

How do majority and minority carriers affect the conductivity of a material?

The presence of majority carriers increases the conductivity of a material, as they are responsible for the bulk of the current flow. Minority carriers, on the other hand, can decrease the conductivity as they have a lower concentration and contribute less to current conduction.

What are some examples of materials with majority and minority carriers?

Semiconductors, such as silicon and germanium, have both majority and minority carriers. In silicon, the majority carriers are electrons, while the minority carriers are holes. In germanium, the majority carriers are holes, while the minority carriers are electrons.

How do majority and minority carriers affect the behavior of a pn junction?

In a pn junction, majority carriers from one side of the junction will diffuse into the other side, leading to a recombination of carriers and the formation of a depletion region. This depletion region creates a barrier for majority carriers, while minority carriers can still diffuse across it. This results in the junction acting as a diode, allowing current flow in one direction but not the other.

What are the implications of majority and minority carriers in electronic devices?

The control and manipulation of majority and minority carriers is crucial in the functioning of electronic devices. For example, in transistors, the majority and minority carriers can be modulated to control the flow of current and amplify signals. In solar cells, minority carriers are generated to create a potential difference and generate electricity from light.

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