Question

In a semiconductor device, the Fermi-energy level is 0.35 eV above the valence band energy. The effective density of states in the valence band at T = 300 K is $1 \times 10^{19} \text{ cm}^{-3}$. The thermal equilibrium hole concentration in silicon at 400 K is ___________ $\times 10^{13} \text{ cm}^{-3}$ (rounded off to two decimal places).
Given kT at 300 K is 0.026 eV.

Hint:

The carrier concentration formulas ($n = N_c \exp(-\frac{E_c - E_F}{kT})$ and $p = N_v \exp(-\frac{E_F - E_v}{kT})$) are fundamental. Be careful with the energy differences in the exponent. Also, remember the temperature dependence: $N_c, N_v \propto T^{3/2}$ and $n_i^2 \propto T^3 \exp(-E_g/kT)$.

Answer 1

Correct Answer: 60

1. Compute \(k\) (in eV/K) from the given \(kT\) at 300 K: \[ k \;=\; \frac{kT(300\ \mathrm{K})}{300} \;=\; \frac{0.026\ \text{eV}}{300\ \text{K}} \;=\; 8.666\overline{6}\times10^{-5}\ \text{eV/K}. \] 2. Compute \(kT\) at \(T=400\ \mathrm{K}\): \[ kT(400) = k\times 400 = 8.666\overline{6}\times10^{-5}\times 400 = 0.034666\overline{6}\ \text{eV}. \] 3. Scale the valence-band effective density of states to 400 K: \[ N_v(400) = N_v(300)\left(\frac{400}{300}\right)^{3/2} = 1\times10^{19}\left(\frac{4}{3}\right)^{3/2} \approx 1.5396007178\times10^{19}\ \mathrm{cm^{-3}}. \] 4. Compute the hole concentration at 400 K using \[ p(400)=N_v(400)\exp\!\Big(-\frac{E_F-E_v}{kT(400)}\Big). \] Substitute values: \[ p(400) \approx 1.5396007178\times10^{19}\times \exp\!\Big(-\frac{0.35}{0.034666\overline{6}}\Big). \] Evaluate the exponent and product: \[ \frac{0.35}{0.034666\overline{6}}\approx 10.0950,\qquad \exp(-10.0950)\approx 4.129\times10^{-5}, \] so \[ p(400)\approx 1.5396007178\times10^{19}\times 4.129\times10^{-5} \approx 6.348983539\times10^{14}\ \mathrm{cm^{-3}}. \] 5. Express the result in the requested form \((\text{value})\times 10^{13}\ \mathrm{cm^{-3}}\): \[ p(400)=6.348983539\times10^{14}\ \mathrm{cm^{-3}} =63.48983539\times10^{13}\ \mathrm{cm^{-3}}. \]