Level 4 — ApplicationMaterials & Atomic Structure

Materials & Atomic Structure

⏱ 60 minutes60 marksprintable — key stays hidden on paper

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

Question 1 — Carrier Concentration & Conductivity (14 marks)

A slab of intrinsic silicon at $T = 300\text{ K}$ has an intrinsic carrier concentration $n_i = 1.5 \times 10^{10}\ \text{cm}^{-3}$ . The electron mobility is $\mu_n = 1350\ \text{cm}^2/\text{V·s}$ and the hole mobility is $\mu_p = 480\ \text{cm}^2/\text{V·s}$ . The electronic charge is $q = 1.6 \times 10^{-19}\ \text{C}$ .

(a) Compute the intrinsic conductivity $\sigma_i$ of the silicon in $(\Omega\cdot\text{cm})^{-1}$ . (4)

(b) The slab is now doped uniformly with phosphorus at $N_D = 5 \times 10^{16}\ \text{cm}^{-3}$ (full ionisation). State the majority and minority carrier types, then compute the majority carrier concentration and the minority carrier concentration using $n \cdot p = n_i^2$ . (5)

(c) Compute the conductivity of the doped sample and state, with a numerical ratio, by what factor it changed relative to the intrinsic case. (5)

Question 2 — Bonding & Lattice Geometry (12 marks)

(a) Silicon has 14 electrons. Write out its shell occupancy using the Bohr model, and hence explain why silicon forms four covalent bonds. (4)

(b) In the silicon diamond-cubic lattice each atom sits at the centre of a regular tetrahedron of nearest neighbours. Given the nearest-neighbour bond length $d = 0.235\ \text{nm}$ , calculate the cubic lattice constant $a$ , using the geometric relation $d = \dfrac{\sqrt{3}}{4}\,a$ . (4)

(c) Explain why this rigid, fully-bonded tetrahedral structure means pure silicon behaves as an insulator at $0\text{ K}$ but as a semiconductor at room temperature. (4)

Question 3 — Thermal Effects (12 marks)

The intrinsic carrier concentration of a semiconductor follows

$n_i(T) = C\,T^{3/2}\exp\!\left(-\frac{E_g}{2kT}\right)$

where $k = 8.62 \times 10^{-5}\ \text{eV/K}$ .

(a) For silicon, $E_g = 1.12\ \text{eV}$ . Taking the ratio $\dfrac{n_i(350)}{n_i(300)}$ , and treating the $T^{3/2}$ pre-factor and $C$ as constant across this range (so only the exponential varies), compute this ratio. (6)

(b) Repeat part (a) for germanium, $E_g = 0.66\ \text{eV}$ , over the same temperature range. (4)

(c) Using your two results, explain in terms of leakage current why silicon is preferred over germanium for devices operating at elevated temperature. (2)

Question 4 — Compound Semiconductor Selection (12 marks)

An engineer must choose a semiconductor for each of the following applications. For each, name one suitable material from {Si, Ge, GaAs, GaN, SiC} and justify the choice in one or two sentences using a physical property.

(a) A blue LED. (3) (b) A high-frequency, high-electron-mobility RF amplifier. (3) (c) A high-power, high-temperature power module for an electric-vehicle inverter. (3) (d) A low-cost mass-produced logic microprocessor. (3)

Question 5 — Extrinsic Doping Analysis (10 marks)

A silicon sample is doped with both boron (acceptors) at $N_A = 2 \times 10^{16}\ \text{cm}^{-3}$ and phosphorus (donors) at $N_D = 3 \times 10^{16}\ \text{cm}^{-3}$ , both fully ionised. Take $n_i = 1.5 \times 10^{10}\ \text{cm}^{-3}$ .

(a) State whether the sample is n-type or p-type, and give the net doping concentration. (3)

(b) Using charge neutrality with net doping $N = N_D - N_A$ and $n p = n_i^2$ , compute the equilibrium electron and hole concentrations. (5)

End of paper

Answer keyMark scheme & solutions

Question 1

(a) Intrinsic: $n = p = n_i$ . $\sigma_i = q\,n_i(\mu_n + \mu_p)$

Formula (1)
$\mu_n+\mu_p = 1830\ \text{cm}^2/\text{V·s}$ (1)
$\sigma_i = (1.6\times10^{-19})(1.5\times10^{10})(1830)$ (1)
$= 4.392\times10^{-6}\ (\Omega\cdot\text{cm})^{-1}$ (1)

(b) Phosphorus = donor → majority = electrons (n-type), minority = holes (1). Full ionisation: $n \approx N_D = 5\times10^{16}\ \text{cm}^{-3}$ (2). $p = n_i^2/n = (1.5\times10^{10})^2 / (5\times10^{16}) = 2.25\times10^{20}/5\times10^{16} = 4.5\times10^{3}\ \text{cm}^{-3}$ (2).

(c) Since $n \gg p$ , $\sigma \approx q n \mu_n$ (1): $\sigma = (1.6\times10^{-19})(5\times10^{16})(1350) = 10.8\ (\Omega\cdot\text{cm})^{-1}$ (2) Ratio $= 10.8 / 4.392\times10^{-6} \approx 2.46\times10^{6}$ (2) — conductivity increased ~2.5 million times, showing the dramatic effect of doping.

Question 2

(a) Shells: $2, 8, 4$ (K=2, L=8, M=4) (2). The outer (valence) shell has 4 electrons; it shares each with a neighbour to reach a stable octet of 8, hence four covalent bonds (2).

(b) $d = \frac{\sqrt3}{4}a \Rightarrow a = \frac{4d}{\sqrt3}$ (1) $= \frac{4 \times 0.235}{1.732}$ (1) $= \frac{0.940}{1.732}$ (1) $= 0.5427\ \text{nm}$ (1). (Accept ≈0.543 nm.)

(c) At $0\text{ K}$ all four valence electrons of every atom are locked in covalent bonds — no free carriers → insulator (2). At room temperature thermal energy breaks a small fraction of bonds, generating electron–hole pairs that carry current → semiconductor (2).

Question 3

(a) With pre-factor cancelled, $\frac{n_i(350)}{n_i(300)} = \exp\!\left[-\frac{E_g}{2k}\left(\frac1{350}-\frac1{300}\right)\right]$ (2) $\frac{E_g}{2k} = \frac{1.12}{2(8.62\times10^{-5})} = 6497.7$ K (1) $\frac1{350}-\frac1{300} = 0.0028571 - 0.0033333 = -4.7619\times10^{-4}$ (1) Exponent $= -6497.7 \times (-4.7619\times10^{-4}) = 3.094$ (1) Ratio $= e^{3.094} \approx 22.1$ (1)

(b) $\frac{E_g}{2k} = \frac{0.66}{2(8.62\times10^{-5})} = 3828.3$ K (1) Exponent $= 3828.3 \times 4.7619\times10^{-4} = 1.8230$ (1) Ratio $= e^{1.823} \approx 6.19$ (2)

(c) Silicon's larger bandgap makes $n_i$ (and thus thermally-generated leakage current) far less temperature-sensitive; germanium's smaller gap causes leakage to rise steeply with heat, so silicon holds up better at elevated temperature (2).

Question 4

(a) GaN — wide bandgap (~3.4 eV) gives photon emission in the blue/UV range (3). (b) GaAs — very high electron mobility (~8500 cm²/V·s), ideal for high-frequency RF (3). (c) SiC — wide bandgap, high thermal conductivity and high breakdown field suit high-power, high-temperature switching (3). (d) Si — cheap, abundant, mature fabrication and stable native oxide for mass logic ICs (3). (Accept equivalent well-justified choices; marks for correct material + valid physical reason.)

Question 5

(a) $N_D > N_A$ so net donors → n-type (1). Net $N = N_D - N_A = 3\times10^{16} - 2\times10^{16} = 1\times10^{16}\ \text{cm}^{-3}$ (2).

(b) Since $N \gg n_i$ : $n \approx N = 1\times10^{16}\ \text{cm}^{-3}$ (3). $p = n_i^2/n = 2.25\times10^{20}/1\times10^{16} = 2.25\times10^{4}\ \text{cm}^{-3}$ (2).

(c) The 2×10¹⁶ acceptors cancel (compensate) 2×10¹⁶ of the donors; only the difference contributes free carriers, so effective doping is reduced to the net value 1×10¹⁶ (2).

[
  {"claim":"Q1a intrinsic conductivity ~4.392e-6",
   "code":"q=1.6e-19; ni=1.5e10; sig=q*ni*(1350+480); result = abs(sig-4.392e-6) < 1e-9"},
  {"claim":"Q1b minority hole conc = 4.5e3",
   "code":"ni=1.5e10; ND=5e16; p=ni**2/ND; result = abs(p-4.5e3) < 1"},
  {"claim":"Q1c doped conductivity = 10.8 and ratio ~2.46e6",
   "code":"q=1.6e-19; sig=q*5e16*1350; sigi=q*1.5e10*1830; result = abs(sig-10.8)<1e-6 and abs(sig/sigi-2.459e6) < 5e3"},
  {"claim":"Q2b lattice constant a ~0.5427 nm",
   "code":"a=4*0.235/sqrt(3); result = abs(float(a)-0.5427) < 0.001"},
  {"claim":"Q3a Si ratio ~22.1",
   "code":"k=8.62e-5; Eg=1.12; r=exp(-Eg/(2*k)*(Rational(1,350)-Rational(1,300))); result = abs(float(r)-22.1) < 0.3"},
  {"claim":"Q3b Ge ratio ~6.19",
   "code":"k=8.62e-5; Eg=0.66; r=exp(-Eg/(2*k)*(Rational(1,350)-Rational(1,300))); result = abs(float(r)-6.19) < 0.15"},
  {"claim":"Q5b p = 2.25e4",
   "code":"ni=1.5e10; n=1e16; p=ni**2/n; result = abs(p-2.25e4) < 1"}
]