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I have learned from a text book of electronics that the free electron density, n (which is equal to that of hole, p) of intrinsic semiconductor due to thermal excitation is a function of temperature and the relation is given by $$n_i = B T^{3/2} e^{-\frac{E_g}{2kT}}$$

Since, both $p$ and $n$ are equal, we can write, $np = n_i^2$

Now, when some doping materials are added with doping concentration $N_D\gg n_i$, it is assumed that now free electron density $n=N_D$

In the book, it says that though impure atoms are now added in the intrinsic semiconductor but we still can write, $np = n_i^2$.

My question is, since we have added impurity which provides more free electrons in the semiconductor, shouldn't the value of $np > n_i^2$?

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  • $\begingroup$ Because of detailed balance. Put more free electrons in through doping, and some will recombine with holes - this reaction will proceed until the rate of generation matches the rate of recombination. $\endgroup$
    – Jon Custer
    Nov 6, 2018 at 15:00
  • $\begingroup$ can you elaborate it please? $\endgroup$ Nov 7, 2018 at 3:12
  • $\begingroup$ Well the relation np=ni^2 is still satisfy because when you increse electrons you will decrese holes. However when you inject electrons from high doping material n+ to p doping material (by applying appropriate voltage) you will have np>ni^2 in the nonequilibrium state. Important question here is how mamy free states we have in the conduction band. Because in high doped materials we have degenerated semicondutor. $\endgroup$ Nov 7, 2018 at 15:23

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When we dope the silicon crystal by n-type atoms take P, then we will be having more number of free electrons, which will be moving all around the material. Now when this electron finds a hole then it will recombine but will leave behind another hole. As the concentration of the electrons is more, it will suppress the concentration of the holes by following the law of mass action.

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