Interleaved — Phase 2

Hardware interleaved practice

printable — key stays hidden on paper

Instructions: Solve all problems. Each targets a different subtopic — read carefully and choose the correct method before computing. Use $k_BT \approx 0.0259$ eV at $T = 300$ K, $n_i = 1.0\times10^{10}\ \text{cm}^{-3}$ for silicon unless stated otherwise. Show reasoning. Total: 50 marks.

1. A silicon sample is doped with phosphorus at $N_D = 5\times10^{16}\ \text{cm}^{-3}$ . Assuming full ionization at 300 K, find the electron and hole concentrations. State which are majority/minority carriers. (5 marks)

2. Explain qualitatively why a material with a band gap of $9\ \text{eV}$ (diamond) is an insulator while one with $\approx 1.1\ \text{eV}$ (silicon) is a semiconductor at room temperature. Reference the population of the conduction band. (4 marks)

3. A PN junction has $N_A = 1\times10^{17}\ \text{cm}^{-3}$ (p-side) and $N_D = 1\times10^{16}\ \text{cm}^{-3}$ (n-side). Compute the built-in potential $V_{bi}$ at 300 K. (5 marks)

4. For an electron mobility $\mu_n = 1350\ \text{cm}^2/\text{V·s}$ at 300 K, use the Einstein relation to find the diffusion coefficient $D_n$ . (4 marks)

5. GaAs and Si both have band gaps near $1\ \text{eV}$ , yet GaAs is used in LEDs and Si is not. Explain the physical distinction responsible, and name which is direct and which is indirect. (5 marks)

6. A silicon diode has saturation current $I_S = 1\times10^{-14}\ \text{A}$ . Using the Shockley equation, find the current at a forward bias of $V = 0.6\ \text{V}$ , $T = 300$ K, ideality factor $n=1$ . (6 marks)

7. In an n-type sample, an electric field of $E = 500\ \text{V/cm}$ is applied. Electron concentration is $n = 5\times10^{16}\ \text{cm}^{-3}$ , $\mu_n = 1350\ \text{cm}^2/\text{V·s}$ . Compute the drift current density $J_n$ . (5 marks)

8. Sketch and describe qualitatively what happens to the depletion region width and majority-carrier flow when a PN junction is switched from forward to reverse bias. (5 marks)

9. An electron concentration profile varies linearly from $n_1 = 1\times10^{17}\ \text{cm}^{-3}$ at $x=0$ to $n_2 = 2\times10^{16}\ \text{cm}^{-3}$ at $x = 2\ \mu\text{m}$ . With $D_n = 35\ \text{cm}^2/\text{s}$ , find the diffusion current density. (6 marks)

10. A semiconductor at 300 K has $n_i = 1.0\times10^{10}\ \text{cm}^{-3}$ . It is doped p-type with $N_A = 1\times10^{15}\ \text{cm}^{-3}$ . Using the mass action law, find the minority electron concentration. (5 marks)

Answer keyMark scheme & solutions

1. (Subtopic 2.1.6/2.1.12 — carrier concentration & majority/minority) Full ionization ⇒ majority electrons $n \approx N_D = 5\times10^{16}\ \text{cm}^{-3}$ . Minority holes via mass action: $p = n_i^2/n = (10^{10})^2/(5\times10^{16}) = 2\times10^{3}\ \text{cm}^{-3}$ . Electrons = majority, holes = minority. Why this method: donor doping fixes majority carrier directly; minority requires mass action, not direct doping.

2. (Subtopic 2.1.2/2.1.3 — band gap & conductivity) Thermal excitation across the gap follows $\propto e^{-E_g/2k_BT}$ . At 300 K, $k_BT\approx0.0259$ eV. For $E_g = 9$ eV the exponent is astronomically negative → essentially no carriers in the conduction band → insulator. For $E_g=1.1$ eV, a small but non-negligible population is excited → measurable conductivity → semiconductor. Why: conceptual comparison — recognize the exponential gap dependence, not a plug-in.

3. (Subtopic 2.2.6 — built-in potential) $V_{bi} = k_BT/q \cdot \ln\!\left(\frac{N_A N_D}{n_i^2}\right) = 0.0259\ln\!\left(\frac{10^{17}\cdot10^{16}}{10^{20}}\right)$ $= 0.0259 \ln(10^{13}) = 0.0259 \times 29.93 = 0.775\ \text{V}.$ Why: junction problem with both dopings → the $V_{bi}$ log formula (distinguish from single-side carrier calc).

4. (Subtopic 2.1.10 — Einstein relation) $D_n = \frac{k_BT}{q}\mu_n = 0.0259 \times 1350 = 34.97\ \text{cm}^2/\text{s} \approx 35\ \text{cm}^2/\text{s}.$ Why: mobility→diffusion link is Einstein; note $k_BT/q$ in volts times mobility.

5. (Subtopic 2.1.5 — direct vs indirect) GaAs is direct (conduction-band minimum aligned in $k$ -space with valence-band maximum): electron–hole recombination emits a photon efficiently. Si is indirect: recombination requires a phonon to conserve momentum, so radiative emission is inefficient. Hence GaAs → LEDs, Si → not. Why: the LED clue signals band-gap type, not gap magnitude.

6. (Subtopic 2.2.10 — Shockley equation) $I = I_S\left(e^{V/(nk_BT/q)} - 1\right) = 10^{-14}\left(e^{0.6/0.0259}-1\right).$ $0.6/0.0259 = 23.17$ , $e^{23.17}\approx 1.16\times10^{10}$ . $I \approx 10^{-14}\times1.16\times10^{10} = 1.16\times10^{-4}\ \text{A} \approx 0.116\ \text{mA}.$ Why: forward bias current → Shockley exponential; the "−1" is negligible here.

7. (Subtopic 2.1.8 — drift current) $J_n = q n \mu_n E = (1.6\times10^{-19})(5\times10^{16})(1350)(500).$ $= 1.6\times10^{-19}\times5\times10^{16} = 8\times10^{-3}$ ; $\times1350 = 10.8$ ; $\times500 = 5400\ \text{A/cm}^2$ . $J_n = 5.4\times10^{3}\ \text{A/cm}^2.$ Why: applied field → drift ( $qn\mu E$ ), NOT diffusion (no gradient given).

8. (Subtopic 2.2.7/2.2.8 — bias behavior)

Forward bias: external voltage opposes $V_{bi}$ , lowers barrier, depletion region narrows, majority carriers flow readily → large current.
Reverse bias: voltage adds to $V_{bi}$ , raises barrier, depletion region widens, majority flow blocked, only tiny reverse saturation current from minority carriers. Why: qualitative comparison — recognize barrier/width behavior differs by sign of bias.

9. (Subtopic 2.1.9 — diffusion current) Gradient: $\dfrac{dn}{dx} = \dfrac{n_2-n_1}{\Delta x} = \dfrac{(2\times10^{16}-1\times10^{17})}{2\times10^{-4}\ \text{cm}} = \dfrac{-8\times10^{16}}{2\times10^{-4}} = -4\times10^{20}\ \text{cm}^{-4}.$ $J_n = qD_n\frac{dn}{dx} = (1.6\times10^{-19})(35)(-4\times10^{20}) = -2240\ \text{A/cm}^2.$ Magnitude $\approx 2.24\times10^{3}\ \text{A/cm}^2$ (electrons diffuse toward low-concentration region; current sign per convention). Why: a spatial gradient (not a field) ⇒ diffusion formula. Convert μm→cm carefully.

10. (Subtopic 2.1.7 — mass action law) $p \approx N_A = 10^{15}$ . Then $n = n_i^2/p = (10^{10})^2/10^{15} = 10^{5}\ \text{cm}^{-3}$ . Why: acceptor doping fixes holes; electrons via $np=n_i^2$ .

[
  {
    "claim": "Problem 3: V_bi = 0.775 V",
    "code": "import math\nkT=0.0259\nNA=1e17; ND=1e16; ni=1e10\nVbi=kT*math.log(NA*ND/ni**2)\nresult = abs(Vbi-0.775)<0.005"
  },
  {
    "claim": "Problem 6: Shockley current approx 1.16e-4 A",
    "code": "import math\nIS=1e-14; V=0.6; kT=0.0259\nI=IS*(math.exp(V/kT)-1)\nresult = abs(I-1.16e-4)/1.16e-4 < 0.05"
  },
  {
    "claim": "Problem 1: minority holes = 2e3 cm^-3",
    "code": "ni=1e10; n=5e16\np=ni**2/n\nresult = abs(p-2e3)<1"
  }
]