Q.The number of silicon atoms per m³ is $5 \times 10^{28}$. This is doped simultaneously with $5 \times 10^{22}$ atoms per m³ of Arsenic and $5 \times 10^{20}$ per m³ atoms of Indium. Calculate the number of electrons and holes. Given that $n_i = 1.5 \times 10^{16}\ \text{m}^{-3}$. Is the material n-type or p-type?

Since the donor (Arsenic) concentration is far larger than the acceptor (Indium) concentration, the net donor density dominates and sets the electron concentration almost exactly, while the hole concentration follows from the law of mass action $np = n_i^2$.

Step 1 — Identify the doping concentrations

• Donor (Arsenic, pentavalent) concentration: $N_D = 5\times10^{22}\ \text{m}^{-3}$
• Acceptor (Indium, trivalent) concentration: $N_A = 5\times10^{20}\ \text{m}^{-3}$
• Intrinsic carrier concentration: $n_i = 1.5\times10^{16}\ \text{m}^{-3}$

Since $N_D \gg N_A$, the semiconductor is compensated but net donor-dominated.

Step 2 — Net donor concentration

Step 3 — Electron concentration

Because $N_D - N_A \gg n_i$, essentially every net donor electron stays free, so the electron (majority carrier) concentration is:

Step 4 — Hole concentration via the mass-action law

Step 5 — Identify the type

Since $n \gg p$ (electrons vastly outnumber holes), the material is n-type.