Determ ine the thermal-equilibrium electron and hole concentrations in silicon at T = 300 K for given doping concentrations. (a) Let Nd = 10^16 cm^-3 and Na = 0. (b) Let Nd = 5 × 10^15 cm^-3 and Na = 2 × 10^15 cm^-3. Recall that ni = 1.5 × 10^10 cm^-3 in silicon at T = 300 K.

Question

Determine the thermal-equilibrium electron and hole concentrations in silicon at $T=300 \mathrm{~K}$ for given doping concentrations. (a) Let $N_{d}=10^{16} \mathrm{~cm}^{-3}$ and $N_{a}=0$ . (b) Let $N_{d}=5 \times 10^{15} \mathrm{~cm}^{-3}$ and $N_{a}=2 \times 10^{15} \mathrm{~cm}^{-3} .$

Recall that $n_{i}=1.5 \times 10^{10} \mathrm{~cm}^{-3}$ in silicon at $T=300 \mathrm{~K}$ .

Recall that [latex]n_{i}=1.5 imes 10^{10} \mathrm{~cm}^{-3}[/latex] in silicon at [latex]T=300 \mathrm{~K}[/latex].

Accepted Answer

(a) From Equation (4.60), the majority carrier electron concentration is

[latex]\begin{array}{c}n_{0}=\frac{\left(N_{d}-N_{a} ight)}{2}+\sqrt{\left(\frac{N_{d}-N_{a}}{2} ight)^{2}+n_{i}^{2}} \ \end{array}[/latex] (4.60)

[latex]\begin{array}{c}n_{0}=\frac{10^{16}}{2}+\sqrt{\left(\frac{10^{16}}{2} ight)^{2}+\left(1.5 imes 10^{10} ight)^{2}} \cong 10^{16} \mathrm{~cm}^{-3}\end{array}[/latex]

The minority carrier hole concentration is found to be

[latex]p_{0}=\frac{n_{i}^{2}}{n_{0}}=\frac{\left(1.5 imes 10^{10} ight)^{2}}{10^{16}}=2.25 imes 10^{4} \mathrm{~cm}^{-3}[/latex]

(b) Again, from Equation (4.60), the majority carrier electron concentration is

[latex]\begin{array}{c}n_{0}=\frac{\left(N_{d}-N_{a} ight)}{2}+\sqrt{\left(\frac{N_{d}-N_{a}}{2} ight)^{2}+n_{i}^{2}} \ \end{array}[/latex] (4.60)

[latex]n_{0}=\frac{5 imes 10^{15}-2 imes 10^{15}}{2}+\sqrt{\left(\frac{5 imes 10^{15}-2 imes 10^{15}}{2} ight)^{2}+\left(1.5 imes 10^{10} ight)^{2}} \cong 3 imes 10^{15} \mathrm{~cm}^{-3}[/latex]

The minority carrier hole concentration is

[latex]p_{0}=\frac{n_{i}^{2}}{n_{0}}=\frac{\left(1.5 imes 10^{10} ight)^{2}}{3 imes 10^{15}}=7.5 imes 10^{4} \mathrm{~cm}^{-3}[/latex]

Comment

In both parts of this example, [latex]\left(N_{d}-N_{a} ight) \gg n_{i}[/latex], so the thermal-equilibrium majority carrier electron concentration is essentially equal to the difference between the donor and acceptor concentrations. Also, in both cases, the majority carrier electron concentration is orders of magnitude larger than the minority carrier hole concentration.

Question 4.9: Determ ine the thermal-equilibrium electron and hole concent...

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