Why Do Minority Carriers Persist in P-Type Material?

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Minority carriers, such as electrons in p-type materials, persist due to a balance between thermal generation and recombination processes. Even though these carriers recombine with excess holes, they are continuously generated thermally, preventing their concentration from reaching zero. In doped semiconductors, the concentration of minority carriers is significantly reduced but remains measurable. For instance, doping silicon with donor ions increases electron concentration while decreasing hole concentration dramatically. This equilibrium explains the presence of minority carriers in p-type materials despite their tendency to recombine.
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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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