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I'm having trouble grasping the concept of extrinsic/doped semiconductors. For example, suppose that the semiconductor i n-typed with a certain density of donors.

At room temperature, about 300 K, I understand that you can assume the donors to be fully ionized, which I interpret as all valence electrons are in the conductor band leaving the donor band filled with holes. If this is correct, then there shouldn't be any holes in the valence band, right?

If I haven't understood it correctly, how do you find the electron- and hole density at 300 Kelvin given a certain density of donors?

Thanks in advance!

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  • $\begingroup$ Only the dopants are ionized, moving electrons (in n-type material) from localized states on the dopant atoms to the conduction band (extended states). $\endgroup$ – Jon Custer Oct 5 '17 at 18:46
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When you add some other atoms i.e, dopants into a pure crystal (intrinsic semiconductor), we get extrinsic semiconductor. At 300K, intrinsic semiconductors conduct very little electric currents so we use extrinsic semiconductors which conduct high currents even at 300K. And obviously, for extrinsic semiconductors there must donors and accepters present in the conduction and valence bands respectively.

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  • $\begingroup$ Thanks for the clarification! But how do you find the density of holes and electrons? $\endgroup$ – CoderOgden Oct 5 '17 at 18:33
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It is like the mass action law in chemistry, for example the ion products of OH$^-$ and H3O$^+$ in water. The product of the electron and hole concentrations is independent of doping, and thus equal to $n_i^2 =p_i^2$, their intrinsic concentrations squared. This is relatively easily derived: the product of the Boltzmann factors does not depend on the position of the chemical potential (the "Fermi level" at finite temperature).

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There will be thermally generared electron hole pairs even in extrinsic semiconductors just like intrinsic semiconductors.And as it is an extrinsic n-type s/c (let's say) there will be excess electrons in the lattice which on getting sufficient ionisation energy (this energy is less than the energy required to jump the forbidden band) will get excited to the conduction band.

Let's suppose a pure silicon crystal has 5×10 ^28 atoms m^-3 .it is doped by 1 PPM concentration of pentavalent As.Let'so calculate the number of electrons and holes given that n i (intrinsic electron conecentration)=1.5×10^16 /m^3.

Note that thermally generated electrons (n i=1.5 ×10^16 /m^3) are negligible small as compared to those produced by doping.since n e×n h =( n i)^2 ,the number of holes n h is given by n h=(1.5×10^16)^2 /5×10^22 = ~4.5×10^9 / m^3

Guess this would help!

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