For you guys who studied semiconductor physics must be familiar with the equation:

$$np=n_i^2$$

I can understand why this is true for the intrinsic case (the broken bonds would always provide electron and hole in pairs )

But why is this still true for doped semiconductors? Take Si for example, $n=p=n_i=10^{10}$ in intrinsic case (that's we all know). However, if you dope $10^{15} N_d$ into the material, then $n\sim10^{15}$ and $p \sim 10^5$. The highlighted part is my confusion! Why does $p$ become smaller? Where do the holes go?

## Best Answer

Electrons and holes occupy their states according to the Fermi-Dirac distribution, which has a single parameter $E_f$, the Fermi level (assume a fixed temperature). Provided $E_f$ is in within the band gap and far from the band edges, the (energy integral of) Fermi-Dirac takes an exponential form $\propto e^{E_f}$ for electrons and $\propto e^{-E_f}$ for holes. Note that electrons and holes share the same Fermi level as we are in chemical equilibrium. You can already see that the product of electrons and holes is independent from $E_f$ in above conditions. Therefore $pn$ remains equal to $p_0n_0 = n_i^2$. In this picture, doping the semiconductor merely moves the Fermi level up (in case of donors) and down (in case of acceptors). Enforcing charge neutrality yields: $n=N_d^+ + p$. But $p$ is really small here, therefore $n\approx N_d^+$ and $p\approx n_i^2/N_d^+$.

To answer your question "where do holes go": holes are absence of electrons in the valence band, or in general the number of empty states in the valence band. When you put in donor dopants you inject electrons, which in turn fill some of these empty states and annihilate holes.