Band Theory & Carrier Physics

Question 1

(a) (6 marks) Electron density: $n = \int_{E_c}^{\infty} g_c(E) f(E)\,dE$. For $(E-E_F)\gg k_BT$ (i.e. $E_F$ several $k_BT$ below $E_c$), the "+1" is negligible: $f(E)\approx e^{-(E-E_F)/k_BT}. \quad(2)$ Integrating the parabolic band DOS with this Boltzmann tail gives the standard effective-DOS result: $n = N_c\,e^{-(E_c-E_F)/k_BT}. \quad(2)$ For holes, $1-f(E) = \dfrac{1}{1+e^{(E_F-E)/k_BT}} \approx e^{-(E_F-E)/k_BT}$ for $(E_F-E)\gg k_BT$, giving $p = N_v\,e^{-(E_F-E_v)/k_BT}. \quad(2)$ Condition: $E_c-E_F \gg k_BT$ and $E_F-E_v \gg k_BT$ (Fermi level well inside gap; non-degenerate). (condition stated within marks)

(b) (5 marks) Multiply: $np = N_cN_v\,e^{-(E_c-E_F)/k_BT}e^{-(E_F-E_v)/k_BT}=N_cN_v\,e^{-(E_c-E_v)/k_BT}=N_cN_v\,e^{-E_g/k_BT}. \quad(2)$ $E_F$ cancels → RHS depends only on material + $T$, not doping. (1) For intrinsic material $n=p=n_i$, so $n_i^2 = N_cN_v e^{-E_g/k_BT}$ → $n_i=\sqrt{N_cN_v}\,e^{-E_g/2k_BT}$. (1) Since $np$ has no $E_F$-dependence it equals its intrinsic value $n_i^2$ for any doping (as long as non-degenerate) → $np=n_i^2$. Physically, more electrons enhance recombination and suppress holes in exact proportion. (1)

(c) (5 marks) Intrinsic: $n=p$: $N_c e^{-(E_c-E_i)/k_BT}=N_v e^{-(E_i-E_v)/k_BT}$. (2) Take ln: $\ln N_c -\frac{E_c-E_i}{k_BT}=\ln N_v-\frac{E_i-E_v}{k_BT}$. Solve for $E_i$: $E_i=\frac{E_c+E_v}{2}+\frac{k_BT}{2}\ln\frac{N_v}{N_c}. \quad(2)$ Offset $=\frac{k_BT}{2}\ln(N_v/N_c)=\frac{0.02585}{2}\ln(1.04\times10^{19}/2.8\times10^{19})$ eV $=0.012925\times\ln(0.3714)=0.012925\times(-0.9905)=-0.0128$ eV $\approx -12.8\ \text{meV}$. Negative ⇒ $E_i$ lies below mid-gap (toward valence band, since $N_v<N_c$). (1)

(d) (6 marks) $N_D=10^{17}\gg n_i$ ⇒ $n\approx N_D = 1\times10^{17}\ \text{cm}^{-3}$. (2) $p=n_i^2/n=(10^{10})^2/10^{17}=10^{20}/10^{17}=1\times10^{3}\ \text{cm}^{-3}$. (2) $E_F-E_i=k_BT\ln(n/n_i)=0.02585\ln(10^{17}/10^{10})=0.02585\times\ln(10^7)=0.02585\times16.118=0.4167\ \text{eV}\approx0.417\ \text{eV}$. (1) Majority = electrons; minority = holes. (1)

Question 2

(a) (4 marks) Drift: $J_{n,\text{drift}}=qn\mu_n\mathcal{E}$ (1). Diffusion: $J_{n,\text{diff}}=qD_n\dfrac{dn}{dx}$ (1) (positive because electrons carry $-q$ and flow opposite to their particle flux). Total: $J_n = qn\mu_n\mathcal{E} + qD_n\frac{dn}{dx}. \quad(2)$

(b) (8 marks) Set $J_n=0$: $qn\mu_n\mathcal{E} + qD_n\dfrac{dn}{dx}=0$. (1) With $n=n_i e^{(E_F-E_i)/k_BT}$, differentiate: $\frac{dn}{dx}=n\cdot\frac{1}{k_BT}\left(\frac{dE_F}{dx}-\frac{dE_i}{dx}\right). \quad(2)$ Equilibrium ⇒ $E_F$ constant ⇒ $dE_F/dx=0$. (1) $E_i$ tracks the band edges/electrostatic potential: $E_i=\text{const}-qV(x)$ so $\dfrac{dE_i}{dx}=-q\dfrac{dV}{dx}=q\mathcal{E}$ (with $\mathcal{E}=-dV/dx$). (1) Thus $\dfrac{dn}{dx}=n\cdot\dfrac{-q\mathcal{E}}{k_BT}$. (1) Substitute into $J_n=0$: $qn\mu_n\mathcal{E}+qD_n\left(\frac{-qn\mathcal{E}}{k_BT}\right)=0 \Rightarrow \mu_n=\frac{qD_n}{k_BT}\Rightarrow \boxed{\frac{D_n}{\mu_n}=\frac{k_BT}{q}}. \quad(2)$

(c) (5 marks) $D_n=\mu_n\cdot\frac{k_BT}{q}=1350\times0.02585 = 34.9\ \text{cm}^2/\text{s}$. (2) Gradient: $\dfrac{dn}{dx}=\dfrac{10^{15}-10^{16}}{1\times10^{-4}\ \text{cm}}=\dfrac{-9\times10^{15}}{10^{-4}}=-9\times10^{19}\ \text{cm}^{-4}$. (1) $J_{n,\text{diff}}=qD_n\dfrac{dn}{dx}=1.602\times10^{-19}\times34.9\times(-9\times10^{19})$ $=1.602\times10^{-19}\times(-3.141\times10^{21})=-503\ \text{A/cm}^2$. (1) Magnitude $\approx5.0\times10^{2}\ \text{A/cm}^2$; sign negative ⇒ conventional current flows in $-x$ direction (electrons diffuse from high to low $n$, i.e. $+x$, so conventional current is $-x$). (1)

(d) (3 marks) Band diagram: non-uniform doping bends the bands so that $E_F$ stays flat (horizontal) across the sample. (1) The built-in field pushes electrons one way (drift); the concentration gradient drives diffusion the other way. (1) In equilibrium these are exactly equal and opposite (the Einstein relation guarantees cancellation for each carrier), giving $J=0$; a flat $E_F$ is the signature of equilibrium/zero net current. (1)

Question 3

(a) (5 marks) $n_i=\sqrt{N_cN_v}\,e^{-E_g/2k_BT}$ with $N_c,N_v\propto T^{3/2}$ ⇒ $\sqrt{N_cN_v}\propto T^{3/2}$. (2) So $n_i = C\,T^{3/2}e^{-E_g/2k_BT}$ ⇒ $\dfrac{n_i}{T^{3/2}}=C\,e^{-E_g/2k_BT}$. (1) Take ln: $\ln(n_i/T^{3/2})=\ln C - \dfrac{E_g}{2k_B}\cdot\dfrac1T$. (1) A plot vs $1/T$ is linear with slope $-E_g/2k_B$; multiply slope by $-2k_B$ to extract $E_g$. (1)

(b) (6 marks) Curve shape (3 regions) correctly ordered on $\log n$ vs $1/T$ (increasing $1/T$ to the right): (2)

• Freeze-out (high $1/T$, low $T$): donors not fully ionised; $n$ rises steeply with $T$ as electrons are thermally released from donor levels ($\propto e^{-E_D/2k_BT}$-type). (1.5)
• Extrinsic/saturation (mid range): all donors ionised, $n_i\ll N_D$, so $n\approx N_D$ ≈ constant (flat plateau). (1.5)
• Intrinsic (low $1/T$, high $T$): $n_i(T)$ grows exponentially; once $n_i\gtrsim N_D$, $n\to n_i$ and rises steeply again (slope $-E_g/2k_B$). (1)

(c) (7 marks)

import numpy as np
def n_majority(T, ND, Nc300, Nv300, Eg):
    kB = 8.617e-5  # eV/K
    Nc = Nc300 * (T/300.0)**1.5
    Nv = Nv300 * (T/300.0)**1.5
    ni = np.sqrt(Nc*Nv) * np.exp(-Eg/(2*kB*T))
    return ND/2.0 + np.sqrt((ND/2.0)**2 + ni**2)

Marks: correct $N_c,N_v$ scaling (2), correct $n_i$ (2), correct charge-neutrality formula returned (2). Limits (1): when $n_i\ll N_D$, $n\to N_D$ (extrinsic); when $n_i\gg N_D$, $n\to n_i$ (intrinsic).

[
  {"claim":"Q1c: E_i offset from midgap ≈ -12.8 meV","code":"kT=8.617e-5*300; off=kT/2*log(1.04e19/2.8e19)*1000; result = abs(float(off)-(-12.8))<0.3"},
  {"claim":"Q1d: p = ni^2/ND = 1e3 cm^-3","code":"p=(1.0e10**2)/1.0e17; result = abs(p-1000.0)<1e-6"},
  {"claim":"Q1d: EF-Ei = kT ln(n/ni) ≈ 0.417 eV","code":"kT=8.617e-5*300; d