Electron concentration with equal P and B doping

Phosphorus is a donor and boron an acceptor. When both are present, the electrons given by the donors fill the acceptor states — this is compensation.

Given: $N_d = 10^{16}\ \text{cm}^{-3}$, $N_a = 10^{16}\ \text{cm}^{-3}$, silicon at 300 K, $n_i = 1.45\times10^{10}\ \text{cm}^{-3}$.

Charge neutrality and mass action

$$n + N_a = p + N_d,\qquad np = n_i^2$$

Since $N_a = N_d$, the neutrality equation gives $n = p$. Then

$$n^2 = n_i^2 \;\Rightarrow\; n = n_i$$ $$n = p = 1.45\times10^{10}\ \text{cm}^{-3}$$

Answer: electron concentration $n = n_i \approx 1.45\times10^{10}\ \text{cm}^{-3}$ (equal to the hole concentration).

Remarks

• The sample is fully compensated: it behaves like intrinsic material in carrier numbers, and its Fermi level returns to $E_{Fi}$.
• It is not identical to pure Si: it contains $2\times10^{16}$ ionised impurities per cm³, which scatter carriers, so the mobilities (and hence conductivity) are lower than in truly intrinsic silicon.