Band Theory & Carrier Physics

Q1 (10)

(a) Multiply: $np = N_C N_V\, e^{-(E_C-E_F)/k_BT}\,e^{-(E_F-E_V)/k_BT} = N_C N_V\,e^{-(E_C-E_V)/k_BT}$ The $E_F$ terms cancel, leaving $np = N_C N_V e^{-E_g/k_BT}$ with $E_g = E_C - E_V$. (2) Since this is independent of $E_F$ and equals $n_i^2$ (definition, intrinsic where $n=p=n_i$): $n_i = \sqrt{N_C N_V}\,e^{-E_g/2k_BT}$ (2)

(b) $E_F$ appears with $+$ in $n$ and $-$ in $p$; raising $E_F$ increases $n$ but decreases $p$ by the same exponential factor, so the product cancels the $E_F$ dependence (1). $np$ therefore depends only on band structure ($N_C,N_V,E_g$) and $T$ (1). $n_i$ is pinned by the band gap and temperature — it is a material/temperature constant, not a doping quantity (1). (3)

(c) Set $n=p$: $N_C e^{-(E_C-E_i)/k_BT} = N_V e^{-(E_i-E_V)/k_BT}$. Take logs and solve: $E_i = \frac{E_C+E_V}{2} + \frac{k_BT}{2}\ln\!\frac{N_V}{N_C}$ (2) $E_i$ is at midgap exactly when $N_C = N_V$ (log term vanishes). (1)

Q2 (8)

(a) Conductor: conduction and valence bands overlap (or partially filled band), $E_g \approx 0$, e.g. copper (1). Semiconductor: small gap, $E_g \approx 0.5$–$2\ \text{eV}$, e.g. Si $1.12\ \text{eV}$ (1). Insulator: large gap $E_g > 5\ \text{eV}$, e.g. diamond $\sim5.5\ \text{eV}$ or SiO₂ $\sim9\ \text{eV}$ (1). Conduction requires electrons in conduction band / holes in valence band; large gap ⇒ negligible thermal excitation (1). (4)

(b) The factor $e^{-E_g/2k_BT}$ is exponentially small when $E_g \gg 2k_BT$. At 300 K, $2k_BT\approx0.052\ \text{eV}$; for insulators $E_g/2k_BT$ is huge, giving essentially zero free carriers; for semiconductors the exponent is modest so appreciable $n_i$ exists. (2)

(c) Direct gap: conduction band minimum and valence band maximum at the same $k$ — electron–hole recombination emits a photon directly (momentum conserved). Indirect gap: minima at different $k$ — recombination needs a phonon to conserve momentum, so it is inefficient for light emission. This is why LEDs/lasers use direct-gap materials (GaAs) not Si. (2)

Q3 (12)

(a) $n \approx N_D = 1\times10^{16}\ \text{cm}^{-3}$ (donor >> $n_i$) (2) $p = n_i^2/n = (1.5\times10^{10})^2/10^{16} = 2.25\times10^{20}/10^{16} = 2.25\times10^{4}\ \text{cm}^{-3}$ (2)

(b) Ratio $= 10^{16}/2.25\times10^{4} \approx 4.4\times10^{11}$, about 11–12 orders of magnitude (1). Electrons (majority) dominate conduction overwhelmingly (2). (3)

(c) Net doping $N_D - N_A = 10^{16} - 6\times10^{15} = 4\times10^{15}\ \text{cm}^{-3}$ (n-type) (2). $n \approx 4\times10^{15}\ \text{cm}^{-3}$ (1). $p = n_i^2/n = 2.25\times10^{20}/4\times10^{15} = 5.625\times10^{4}\ \text{cm}^{-3}$ (2). (5)

Q4 (12)

(a) $f(E) = \dfrac{1}{1+e^{(E-E_F)/k_BT}}$ (1) For $E-E_F \gg k_BT$, the exponential dominates: $f(E)\approx e^{-(E-E_F)/k_BT}$ (Boltzmann tail) (2). At $E = E_F$: exponent $=0$, so $f = 1/(1+1) = 1/2$ (1). (4)

(b) Equilibrium electron current $J_n = q\mu_n n E + qD_n \dfrac{dn}{dx} = 0$ (1). With built-in field $E = -\dfrac{1}{q}\dfrac{dE_F... }{}$ — use $n = n_i e^{(E_F-E_i)/k_BT}$; in equilibrium $E_F$ flat, non-uniform $E_i$. Field from $E = -\dfrac{d\phi}{dx}$, and $\dfrac{dn}{dx} = \dfrac{n}{k_BT}\dfrac{d(E_F-E_i)}{dx}$. With $qE = \dfrac{dE_i}{dx}$ (band bending) (2): $q\mu_n n E = qD_n\frac{n}{k_BT}\,qE \;\Rightarrow\; \mu_n = \frac{qD_n}{k_BT}$ Hence $\dfrac{D_n}{\mu_n} = \dfrac{k_BT}{q}$ (3). (6)

(c) $D_n = \mu_n \dfrac{k_BT}{q} = 1350 \times 0.02585 = 34.9\ \text{cm}^2/\text{s}$ (since $k_BT/q = 0.02585\ \text{V}$). (2)

Q5 (10)

(a) $J_n = q\mu_n n E + qD_n\dfrac{dn}{dx}$. $q$ electron charge magnitude, $\mu_n$ mobility, $n$ electron density, $E$ field (drift term), $D_n$ diffusion coefficient, $dn/dx$ concentration gradient (diffusion term). (3)

(b) Sample answer (award for correct exponential form, sqrt, and units handling):

import numpy as np
def n_intrinsic(Nc, Nv, Eg, T):
    kB = 8.617e-5        # eV/K
    return np.sqrt(Nc*Nv)*np.exp(-Eg/(2*kB*T))

(4) — 2 for sqrt(Nc*Nv), 2 for exp(-Eg/(2*kB*T)).

(c) Generation: thermal/optical energy creates electron–hole pairs (excites electrons across gap) (1). Recombination: an electron drops from CB to VB annihilating a hole, releasing energy (photon/phonon) (1). In equilibrium generation rate = recombination rate, keeping $np = n_i^2$ constant (1). (3)

Q6 (8)

(a) Freeze-out (low T): dopants not fully ionised, $n$ rises with $T$ (1). Extrinsic/saturation (mid T): all dopants ionised, $n \approx N_D$ roughly flat (1). Intrinsic (high T): $n_i$ exceeds $N_D$, $n$ rises steeply/exponentially as intrinsic generation dominates (1). Correct ordering & shape (1). (4)

(b) $\dfrac{n_i(400)}{n_i(300)} = \exp\!\left[-\dfrac{E_g}{2k_B}\left(\dfrac1{400}-\dfrac1{300}\right)\right]$. $-\dfrac{E_g}{2k_B} = -\dfrac{1.12}{2\times8.617\times10^{-5}} = -6499\ \text{K}$ (1). $\left(\dfrac1{400}-\dfrac1{300}\right) = -8.333\times10^{-4}\ \text{K}^{-1}$ (1). Exponent $= -6499 \times (-8.333\times10^{-4}) = 5.416$ (1). Ratio $= e^{5.416} \approx 225$ (1). (4)

[
{"claim":"Q3a minority hole p = 2.25e4 cm^-3","code":"ni=1.5e10; ND=1e16; p=ni**2/ND; result = abs(p-2.25e4)<1"},
{"claim":"Q3c compensated: n=4e15, p=5.625e4","code":"ni=1.5e10; net=1e16-6e15; n=net; p=ni**2/n; result = abs(n-4e15)<1e6 and abs(p-5.625e4)<1"},
{"claim":"Q4c D_n = 34.9 cm^2/s","code":"mu=1350; Vt=0.02585; Dn=mu*Vt; result = abs(Dn-34.9)<0.2"},
{"claim":"Q6b ni ratio ~225","code":"import sympy as sp; Eg=sp.Rational(112,100); kB=8.617e-5; exp_arg=-Eg/(2*kB)*(sp.Rational(1,400)-sp.Rational(1,300)); ratio=sp.exp(exp_arg); result = abs(float(ratio)-225)<8"}
]