Understanding Semiconductors: The Power of the Mass-Action Law

Demonstrative Video

Mass Action Law

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

Charge densities in doped semiconductors

\[\begin{aligned} n_N & = N_D + p_N \approx N_D \\ p_N & = \dfrac{n_i^2}{n_N}~(\text{from mass action law})\\ & \approx \dfrac{n_i^2}{N_D} \end{aligned}\]
N-type semiconductor:

P-type semiconductor:

\[\begin{aligned} p_P & = N_A + n_P \approx N_A \\ n_p & = \dfrac{n_i^2}{p_P}~(\text{from mass action law})\\ & \approx \dfrac{n_i^2}{N_A} \end{aligned}\]

\[\begin{aligned} N_D & = \text{conc. of donor atoms} \\ n_N & = \text{electron conc. in N-type} \\ p_N & = \text{hole conc. in N-type} \\ N_A & = \text{conc. of acceptor atoms} \\ p_p & = \text{hole conc. in P-type} \\ n_P & = \text{electron conc. in P-type} \end{aligned}\]

Extrinsic Conductivity

\[\sigma_N = qn_N\mu_n \approx qN_D\mu_n \quad \text{since}~n_N \approx N_D\]
Conductivity of N-type semiconductor:
\[\sigma_P= qn_P\mu_p \approx qN_A\mu_p \quad \text{since}~p_P \approx N_A\]
Conductivity of P-type semiconductor:
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

Problem

Find the conductivity of silicon
1. in intrinsic condition at a room temp. of \(300^{\circ}\)K
2. with donor impurity of 1 in \(10^8\)
3. with acceptor impurity of 1 in \(5\times 10^{7}\)
4. 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 = \(5\times 10^{22}\)

Solution

\[\begin{aligned} \sigma_i &=q n_i\left(\mu_n+\mu_p\right) =\left(1.6 \times 10^{-19}\right)\left(1.5 \times 10^{10}\right)(1300+500) \\ &=4.32 \times 10^{-6} \mathrm{~S} / \mathrm{cm} \end{aligned}\]
In intrinsic condition,
\[\begin{aligned} p &=\frac{n_i^2}{n} \approx \frac{n_i^2}{N_D} =\frac{\left(1.5 \times 10^{10}\right)^2}{5 \times 10^{14}}=0.46 \times 10^6 \mathrm{~cm}^{-3} \end{aligned}\]
\[\begin{aligned} \sigma &=n q \mu_n=N_D q \mu_n \\ &=\left(5 \times 10^{14}\right)\left(1.6 \times 10^{-19}\right)(1300) \\ &=0.104 \mathrm{~S} / \mathrm{cm} . \end{aligned}\]
\(p\)\(p \ll n\) Therefore, Further, Hence, Number of silicon atoms

\[\begin{aligned} n &=\frac{n_i^2}{p} \approx \frac{n_i^2}{N_A} =\frac{\left(1.5 \times 10^{10}\right)^2}{10^{15}}=2.25 \times 10^5 \mathrm{~cm}^{-3} \end{aligned}\]

\[\begin{aligned} N_A{ }^{\prime} &=N_A-N_D=10^{15}-5 \times 10^{14}=5 \times 10^{14} \mathrm{~cm}^{-3} \\ \sigma &=N_A{ }^{\prime} q \mu_p =\left(5 \times 10^{14}\right)\left(1.6 \times 10^{-19}\right)(500) \\ &=0.04 \mathrm{~S} / \mathrm{cm} . \end{aligned}\]

\[\begin{aligned} \sigma &=p q \mu_P=N_A q \mu_P =\left(10^{15} \times 1.6 \times 10^{-19} \times 500\right) \\ &=0.08 \mathrm{~S} / \mathrm{cm} . \end{aligned}\]

\(n\)\(p \gg n\) Hence, Further, (c)