Level 4 — ApplicationBand Theory & Carrier Physics

Band Theory & Carrier Physics

⏱ 60 minutes50 marksprintable — key stays hidden on paper

Level: 4 (Application — novel problems, no hints) Time limit: 60 minutes Total marks: 50

Use the constants: $k = 8.617\times10^{-5}\ \text{eV/K}$ , $k = 1.381\times10^{-23}\ \text{J/K}$ , $q = 1.602\times10^{-19}\ \text{C}$ . For silicon at 300 K take $n_i = 1.5\times10^{10}\ \text{cm}^{-3}$ unless told otherwise. Show all working.

Question 1 — Doping, mass action & Fermi level (12 marks)

A silicon sample at 300 K is doped with $N_D = 2\times10^{16}\ \text{cm}^{-3}$ donors and $N_A = 5\times10^{15}\ \text{cm}^{-3}$ acceptors, all fully ionised.

(a) Determine the equilibrium electron and hole concentrations $n_0$ and $p_0$ . State which are majority/minority carriers. (5)

(b) Calculate the position of the Fermi level relative to the intrinsic level, $E_F - E_i$ , in eV. (3)

(c) The intrinsic carrier concentration obeys $n_i \propto T^{3/2}\exp(-E_g/2kT)$ . Taking $E_g = 1.12\ \text{eV}$ , estimate the factor by which $n_i$ increases when the temperature rises from 300 K to 400 K. (4)

Question 2 — Transport: drift, diffusion, Einstein (12 marks)

In an n-type silicon bar at 300 K the electron concentration varies linearly from $n = 1\times10^{17}\ \text{cm}^{-3}$ at $x=0$ to $n = 6\times10^{16}\ \text{cm}^{-3}$ at $x = 2\ \mu\text{m}$ . Electron mobility is $\mu_n = 1350\ \text{cm}^2/\text{V·s}$ .

(a) Using the Einstein relation, find the electron diffusion coefficient $D_n$ at 300 K. (3)

(b) Compute the electron diffusion current density $J_{n,\text{diff}}$ (magnitude and direction). (5)

(c) An electric field $E = 500\ \text{V/cm}$ is applied along $+x$ . At $x=0$ compute the electron drift current density and compare it in magnitude to the diffusion current at that point. (4)

Question 3 — Direct vs indirect gap & band structure (9 marks)

(a) GaAs ( $E_g = 1.42$ eV, direct) and Si ( $E_g = 1.12$ eV, indirect) are candidate LED materials. Explain, using momentum conservation, why GaAs is far superior for light emission. (4)

(b) Compute the wavelength of the photon emitted by band-to-band recombination in GaAs. State whether it is visible. (3)

(c) A designer claims a photon emitted from Si band-edge recombination would have wavelength ~1100 nm and be equally efficient. Identify the physical error in the efficiency claim. (2)

Question 4 — Non-equilibrium & recombination (10 marks)

A p-type silicon wafer with $N_A = 1\times10^{17}\ \text{cm}^{-3}$ is uniformly illuminated, generating excess carriers $\Delta n = \Delta p = 1\times10^{14}\ \text{cm}^{-3}$ . Minority-carrier lifetime is $\tau_n = 2\ \mu\text{s}$ .

(a) State the equilibrium majority and minority carrier concentrations before illumination. (3)

(b) By what percentage does the majority-carrier concentration change under illumination, and by what factor does the minority-carrier concentration change? Comment on which effect matters for device operation. (4)

(c) When the light is switched off, the excess electrons decay as $\Delta n(t) = \Delta n_0 e^{-t/\tau_n}$ . Find the time for the excess minority concentration to fall to $1\times10^{13}\ \text{cm}^{-3}$ . (3)

Question 5 — Conductivity synthesis (7 marks)

A semiconductor sample is compensated so that $n_0 = p_0$ . Given $n_i = 1.5\times10^{10}\ \text{cm}^{-3}$ , $\mu_n = 1350$ , $\mu_p = 480\ \text{cm}^2/\text{V·s}$ :

(a) Determine $n_0$ and $p_0$ , and hence the resistivity $\rho$ of the sample. (5)

(b) Explain qualitatively why this intrinsic-like resistivity falls sharply as temperature rises, in contrast to a metal. (2)

Answer keyMark scheme & solutions

Question 1

(a) Net donor doping $N_D - N_A = 2\times10^{16} - 5\times10^{15} = 1.5\times10^{16}\ \text{cm}^{-3}$ . (1) Since $N_D-N_A \gg n_i$ , n-type: $n_0 \approx 1.5\times10^{16}\ \text{cm}^{-3}$ . (1) Mass action: $p_0 = n_i^2/n_0 = (1.5\times10^{10})^2/(1.5\times10^{16}) = 1.5\times10^{4}\ \text{cm}^{-3}$ . (2) Electrons = majority, holes = minority. (1)

(b) $E_F - E_i = kT\ln(n_0/n_i) = 0.02585\ln(1.5\times10^{16}/1.5\times10^{10})$ $= 0.02585\times\ln(10^6) = 0.02585\times13.816 = 0.357\ \text{eV}$ . (3) (above $E_i$ )

(c) Ratio $=\left(\frac{400}{300}\right)^{3/2}\exp\!\left[-\frac{E_g}{2k}\left(\frac1{400}-\frac1{300}\right)\right]$ . $(4/3)^{1.5}=1.5396$ . (1) Exponent $=-\frac{1.12}{2\times8.617\times10^{-5}}(0.0025-0.003333) = -6499.4\times(-8.333\times10^{-4}) = 5.416$ . (2) $e^{5.416}=225.0$ ; total $\approx 1.5396\times225.0 \approx 346$ . So $n_i$ rises ~350×. (1)

Question 2

(a) Einstein: $D_n = \frac{kT}{q}\mu_n = 0.02585\times1350 = 34.9\ \text{cm}^2/\text{s}$ . (3)

(b) Gradient $\frac{dn}{dx} = \frac{(6\times10^{16}-1\times10^{17})}{2\times10^{-4}\ \text{cm}} = \frac{-4\times10^{16}}{2\times10^{-4}} = -2\times10^{20}\ \text{cm}^{-4}$ . (2) $J_{n,\text{diff}} = qD_n\frac{dn}{dx} = 1.602\times10^{-19}\times34.9\times(-2\times10^{20})$ $= -1118\ \text{A/cm}^2$ . Magnitude $\approx 1.12\times10^{3}\ \text{A/cm}^2$ . (2) Sign: current flows in $-x$ direction (electrons diffuse $+x$ from high to low, conventional current opposite). (1)

(c) At $x=0$ , $n=10^{17}$ : $J_{n,\text{drift}} = qn\mu_n E = 1.602\times10^{-19}\times10^{17}\times1350\times500$ $= 1.081\times10^{4}\ \text{A/cm}^2$ . (3) Drift ( $1.08\times10^4$ ) exceeds diffusion ( $1.12\times10^3$ ) by ~10×. (1)

Question 3

(a) In an indirect-gap material (Si) the conduction-band minimum and valence-band maximum are at different crystal momenta $k$ . Radiative recombination must conserve both energy and momentum; a photon carries negligible momentum, so a phonon must also participate — a low-probability 3-body process. In GaAs (direct) both extrema are at the same $k$ , so electron and hole recombine emitting only a photon: high probability, efficient light emission. (4: momentum conservation 2, phonon requirement 1, conclusion 1)

(b) $\lambda = \frac{hc}{E_g} = \frac{1240\ \text{eV·nm}}{1.42} = 873\ \text{nm}$ . (2) This is near-infrared, not visible (>700 nm). (1)

(c) Wavelength value is fine, but Si is indirect-gap so band-edge radiative recombination is intrinsically inefficient (phonon-assisted); the efficiency claim ignores momentum-conservation suppression. (2)

Question 4

(a) Majority holes: $p_0 \approx N_A = 1\times10^{17}\ \text{cm}^{-3}$ . (1) Minority electrons: $n_0 = n_i^2/p_0 = (1.5\times10^{10})^2/10^{17} = 2.25\times10^{3}\ \text{cm}^{-3}$ . (2)

(b) Majority change: $\Delta p/p_0 = 10^{14}/10^{17} = 10^{-3} = 0.1\%$ — negligible. (2) Minority: $n = n_0+\Delta n = 2.25\times10^3 + 10^{14} \approx 10^{14}$ ; factor $=10^{14}/2.25\times10^3 \approx 4.4\times10^{10}$ — enormous. (1) Device operation (currents, injection) is governed by the huge fractional change in minority carriers. (1)

(c) $10^{13} = 10^{14}e^{-t/\tau}$ ⇒ $t = \tau\ln(10) = 2\times10^{-6}\times2.3026 = 4.61\ \mu\text{s}$ . (3)

Question 5

(a) $n_0=p_0=n_i = 1.5\times10^{10}\ \text{cm}^{-3}$ . (1) $\sigma = q n_i(\mu_n+\mu_p) = 1.602\times10^{-19}\times1.5\times10^{10}\times(1350+480)$ $= 1.602\times10^{-19}\times1.5\times10^{10}\times1830 = 4.397\times10^{-6}\ \text{S/cm}$ . (3) $\rho = 1/\sigma = 2.27\times10^{5}\ \Omega\cdot\text{cm}$ . (1)

(b) $n_i$ rises exponentially with $T$ (more electrons thermally excited across the gap), so $\sigma$ rises steeply and $\rho$ falls. In a metal carrier density is fixed and increased phonon scattering reduces mobility, so metal resistivity rises with T. (2)

[
 {"claim":"Q1a hole concentration p0=1.5e4","code":"ni=1.5e10; n0=1.5e16; p0=ni**2/n0; result = abs(p0-1.5e4)<1"},
 {"claim":"Q1b EF-Ei approx 0.357 eV","code":"import math; v=0.02585*math.log(1.5e16/1.5e10); result = abs(v-0.357)<0.005"},
 {"claim":"Q1c ni ratio ~346","code":"import math; r=(4/3)**1.5*math.exp(-1.12/(2*8.617e-5)*(1/400-1/300)); result = abs(r-346)<10"},
 {"claim":"Q2b diffusion current magnitude ~1118 A/cm2","code":"q=1.602e-19; Dn=0.02585*1350; grad=(6e16-1e17)/2e-4; J=q*Dn*grad; result = abs(abs(J)-1118)<20"},
 {"claim":"Q3b wavelength 873 nm","code":"lam=1240/1.42; result = abs(lam-873)<3"},
 {"claim":"Q4c decay time 4.61 us","code":"import math; t=2e-6*math.log(10); result = abs(t-4.605e-6)<5e-8"},
 {"claim":"Q5 resistivity ~2.27e5 ohm-cm","code":"q=1.602e-19; sig=q*1.5e10*(1350+480); rho=1/sig; result = abs(rho-2.27e5)<3e3"}
]