Under thermal equilibrium, the product of the number of holes and electrons is constant and independent of the amount of donor and acceptor impurity doping.
\[\boxed{n \cdot p = n_i^2}\] \[\begin{aligned} n & = \text{number of free electrons per unit volume}\\ p & = \text{number of holes per unit volume}\\ n_i & = \text{intrinsic concentration}\\ \end{aligned}\]
While considering the conductivity of the doped semiconductors, only the dominant majority charge carriers have to be considered
N-type semiconductor: \[\begin{aligned} n_N & = N_D + p_N \approx N_D \\ p_N & = \dfrac{n_i^2}{n_N}~(\text{from mass action law})\\ & \approx \dfrac{n_i^2}{N_D} \end{aligned}\]
P-type semiconductor:
\[\begin{aligned} p_P & = N_A + n_P \approx N_A \\ n_p & = \dfrac{n_i^2}{p_P}~(\text{from mass action law})\\ & \approx \dfrac{n_i^2}{N_A} \end{aligned}\]
\[\begin{aligned} N_D & = \text{conc. of donor atoms} \\ n_N & = \text{electron conc. in N-type} \\ p_N & = \text{hole conc. in N-type} \\ N_A & = \text{conc. of acceptor atoms} \\ p_p & = \text{hole conc. in P-type} \\ n_P & = \text{electron conc. in P-type} \end{aligned}\]
Conductivity of N-type semiconductor: \[\sigma_N = qn_N\mu_n \approx qN_D\mu_n \quad \text{since}~n_N \approx N_D\]
Conductivity of P-type semiconductor: \[\sigma_P= qn_P\mu_p \approx qN_A\mu_p \quad \text{since}~p_P \approx N_A\]
If conc. of donor atoms added to a P-type semiconductor exceeds the conc. of acceptor atoms i.e. \(N_D >> N_A\) then P-type is converted to N-type
Similarly, if \(N_A >> N_D\), N-type converted to P-type
Find the conductivity of silicon
in intrinsic condition at a room temp. of \(300^{\circ}\)K
with donor impurity of 1 in \(10^8\)
with acceptor impurity of 1 in \(5\times 10^{7}\)
with both impurities present simultaneously
Given that \(n_i\) for silicon at \(300^{\circ}\)K is \(1.5 \times 10^{10}~\mathrm{cm}^{-3}\), \(\mu_n = 1300~\mathrm{cm^2/V-s}\), \(\mu_p=500~\mathrm{cm^2/V-s}\), number of Si atoms per cm3 = \(5\times 10^{22}\)
In intrinsic condition, \(n=p=n_i\) \[\begin{aligned} \sigma_i &=q n_i\left(\mu_n+\mu_p\right) =\left(1.6 \times 10^{-19}\right)\left(1.5 \times 10^{10}\right)(1300+500) \\ &=4.32 \times 10^{-6} \mathrm{~S} / \mathrm{cm} \end{aligned}\]
Number of silicon atoms \(/ \mathrm{cm}^3=5 \times 10^{22}\) Hence, \(N_D=\frac{5 \times 10^{22}}{10^8}=5 \times 10^{14} \mathrm{~cm}^{-3}\) Further, \(n \approx N_D\) Therefore, \[\begin{aligned} p &=\frac{n_i^2}{n} \approx \frac{n_i^2}{N_D} =\frac{\left(1.5 \times 10^{10}\right)^2}{5 \times 10^{14}}=0.46 \times 10^6 \mathrm{~cm}^{-3} \end{aligned}\] Thus \(p \ll n\). Hence \(p\) may be neglected while calculating the conductivity. \[\begin{aligned} \sigma &=n q \mu_n=N_D q \mu_n \\ &=\left(5 \times 10^{14}\right)\left(1.6 \times 10^{-19}\right)(1300) \\ &=0.104 \mathrm{~S} / \mathrm{cm} . \end{aligned}\]
(c) \(N_A=\frac{5 \times 10^{22}}{5 \times 10^7}=10^{15} \mathrm{~cm}^{-3}\) Further, \(p \approx N_A\) Hence, \[\begin{aligned} n &=\frac{n_i^2}{p} \approx \frac{n_i^2}{N_A} =\frac{\left(1.5 \times 10^{10}\right)^2}{10^{15}}=2.25 \times 10^5 \mathrm{~cm}^{-3} \end{aligned}\] Thus, \(p \gg n\). Hence \(n\) may be neglected while calculating the conductivity. Hence, \[\begin{aligned} \sigma &=p q \mu_P=N_A q \mu_P =\left(10^{15} \times 1.6 \times 10^{-19} \times 500\right) \\ &=0.08 \mathrm{~S} / \mathrm{cm} . \end{aligned}\] (d) With both types of impurities present simultaneously, the net acceptor impurity density is, \[\begin{aligned} N_A{ }^{\prime} &=N_A-N_D=10^{15}-5 \times 10^{14}=5 \times 10^{14} \mathrm{~cm}^{-3} \\ \sigma &=N_A{ }^{\prime} q \mu_p =\left(5 \times 10^{14}\right)\left(1.6 \times 10^{-19}\right)(500) \\ &=0.04 \mathrm{~S} / \mathrm{cm} . \end{aligned}\]