Minority charge carriers - what are these?

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SUMMARY

The discussion clarifies the concept of minority charge carriers in N-type semiconductors, specifically addressing the confusion surrounding the relationship between electrons and holes. In N-type materials, donor atoms provide extra electrons that become delocalized in the conduction band, while the holes created are localized to the donor atoms, making them minority carriers. This distinction is crucial as it highlights that minority carriers refer to free charges, not the charges associated with lattice ions. The role of thermal energy in generating electron-hole pairs is also emphasized, reinforcing the understanding of minority carrier dynamics.

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Kenny_L
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Hi all, this question comes from an old related thread .. https://www.physicsforums.com/threads/help-in-semiconductors-minority-and-majority-carriers.613598/

I'm also interested in the definition of these 'minority charge carriers' in semiconductors.

I don't understand the explanations in the old thread.

The way I've been seeing it is... take an N-type semiconductor for example, which has impurity atoms introduced (to say a Silicon lattice) so that a donor impurity atom will have 1 extra electron than a silicon atom. However, that extra 1 electron is accounted for (neutralised) because the donor impurity atom is neutral to begin with. So even if that electron becomes 'free', then there will be a free hole to go with it. So that's a one-to-one correspondence. So the free hole won't be a 'minority' in that case (since the hole concentration still equals electron concentration in terms of 'free' carriers).

Then, somebody indicates that temperature (thermal energy) is able to break covalent bonds, resulting in an electron-hole pair. And they say that this generates minority carriers (ie. holes being minority carriers). However, my view is that when covalent bonds are broken due to temperature, there's still going to be one free electron to every free hole, right? So this means that the holes are never going to be in the 'minority', right?

Can someone help to set things straight - to explain why holes are 'minority' carriers in N-type semiconductors? The way I see it right now is... holes don't seem to be in the minority if there's always going to be one-to-one correspondence between electrons and holes. There also doesn't appear to be any diagrams in books etc that actually show clearly how (for example) the holes actually become a 'minority'. Thanks for your help in advance!
 
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Kenny_L said:
Hi all, this question comes from an old related thread .. https://www.physicsforums.com/threads/help-in-semiconductors-minority-and-majority-carriers.613598/

I'm also interested in the definition of these 'minority charge carriers' in semiconductors.

I don't understand the explanations in the old thread.

The way I've been seeing it is... take an N-type semiconductor for example, which has impurity atoms introduced (to say a Silicon lattice) so that a donor impurity atom will have 1 extra electron than a silicon atom. However, that extra 1 electron is accounted for (neutralised) because the donor impurity atom is neutral to begin with. So even if that electron becomes 'free', then there will be a free hole to go with it. So that's a one-to-one correspondence. So the free hole won't be a 'minority' in that case (since the hole concentration still equals electron concentration in terms of 'free' carriers).

Then, somebody indicates that temperature (thermal energy) is able to break covalent bonds, resulting in an electron-hole pair. And they say that this generates minority carriers (ie. holes being minority carriers). However, my view is that when covalent bonds are broken due to temperature, there's still going to be one free electron to every free hole, right? So this means that the holes are never going to be in the 'minority', right?

Can someone help to set things straight - to explain why holes are 'minority' carriers in N-type semiconductors? The way I see it right now is... holes don't seem to be in the minority if there's always going to be one-to-one correspondence between electrons and holes. There also doesn't appear to be any diagrams in books etc that actually show clearly how (for example) the holes actually become a 'minority'. Thanks for your help in advance!
The minority/majority carriers refer to the free charges, not the charges on any ions in the lattice. I.e. in an n-type doped material, the donor donates an electron to the conduction band, where the electron is then delocalised, in contrast the hole left behind is localised to the donor atom. This differs to when a valence band electron is excited into the conduction band, as in this case the hole is also free due to being part of the band rather than associated with a particular atom.
 
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Vagn said:
The minority/majority carriers refer to the free charges, not the charges on any ions in the lattice. I.e. in an n-type doped material, the donor donates an electron to the conduction band, where the electron is then delocalised, in contrast the hole left behind is localised to the donor atom. This differs to when a valence band electron is excited into the conduction band, as in this case the hole is also free due to being part of the band rather than associated with a particular atom.

Hi Vagn! Thanks so much - actually, incredibly much - for your comments about this. I had always been trying to understand this kind of thing. Thanks for helping me to understand the concept of minority charge carriers. Greatly appreciated!
 

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