Extrinsic Semiconductor (N-type)

Question

I'm having trouble grasping the concept of extrinsic/doped semiconductors. For example, suppose that the semiconductor is 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?

Answer 1

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).

Answer 2

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.

Answer 3

There will be thermally generated 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 $5 \cdot 10 ^{28}$ atoms $\text{m}^{-3}$. It is doped by 1 PPM concentration of pentavalent $\text{As}$. Let's calculate the number of electrons and holes given that $n_i$ (intrinsic electron conecentration) $=1.5 \cdot 10^{16}/\text{m}^3$.

Note that thermally generated electrons ($n_i=1.5\cdot 10^{16}/\text{m}^3$) are negligible small as compared to those produced by doping. Since $n_e \cdot n_h = n_i^2$, the number of holes $n_h$ is given by $$n_h=\frac{(1.5\cdot 10^{16})^2}{5\cdot 10^{22}} = ~ 4.5 \cdot 10^9/\text{m}^3$$

Guess this would help!