Why the product of hole and free electron density is constant in semiconductor?

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 ni = B T3/2 e-Eg/2kT

Since, both p and n are equal, we can write, np = ni2

Now, when some doping materials are added with doping concentration ND >> ni, it is assumed that now free electron density n = ND

In the book, it says that though impure atoms are now added in the intrinsic semiconductor but we still can write, np = ni2

My question is, since we have added impurity which provides more free electrons in the semiconductor, shouldn't the value of np > ni2?

• 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. – Jon Custer Nov 6 '18 at 15:00
• can you elaborate it please? – Sabbir Ahmed Nov 7 '18 at 3:12
• 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. – Mateusz Glinkowski Nov 7 '18 at 15:23