Why Do Minority Carriers Persist in P-Type Material?

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SUMMARY

The discussion centers on the persistence of minority carriers, specifically electrons, in p-type semiconductor materials. In equilibrium, while minority carriers do recombine with excess holes, they are continuously thermally generated, maintaining a non-zero concentration. For instance, in silicon (Si) at 27°C, the intrinsic carrier concentration is approximately 1010 cm-3. Doping silicon with donor ions increases electron concentration while reducing hole concentration to about 104 cm-3, illustrating the balance between thermal generation and recombination.

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why do minority carriers exist? in a p-type material, why don't the minority carriers recombine (and get annihlated) with excess holes?
 
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in P type if there are any minority carriers (electrons) in Valence band where majority charge carriers (holes) are present then they will recombine but minority carriers in p type which are electrons will be in conduction band while holes will be in valence band
 
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Aranion said:
why don't the minority carriers recombine with excess holes?

They do. But you have to understand that this is an equilibrium situation where electron-hole pairs are also continuously being thermally generated. So yes, the concentration of minority carries DOES indeed get seriously reduced in a doped semiconductor, but not reduced to zero.

For example the "intrinsic" carrier concentration in Si is about 10^{10} cm^{-3} at about 27C. This means that electrons and holes are being thermally generated (continuously) and also recombining (continuously) such that the steady state concentration of each is approx 10^{10} cm^{-3}. The product of electron times hole concentration will be approx n p = 10^{20}.

Now if we dope the Si with say 10^{16} cm^{-3} donor ions then the electron concentration will increase to approx 10^{16}. As a result the amount of recombination will increase whereas the amount of thermal generation will remain largely unchanged and the product will still be n p = 10^{20}. So this means we'll now have p = 10^4 cm^{-3}, which is one million times lower concentration of holes (minority carriers) than in the undoped silicon.
 
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thank you guys... i get it now..
 

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