Exercises — Compare band gaps - conductor - semiconductor - insulator

Use the classification table: $E_g\approx 0\to$ conductor; $\sim 0.1$–$3\text{eV}\to$ semiconductor; $\gtrsim 4\text{eV}\to$ insulator.

• $0\text{eV}$ → conductor (bands overlap; always free carriers).
• $0.67\text{eV}$ → semiconductor (this is germanium).
• $1.12\text{eV}$ → semiconductor (this is silicon).
• $5.5\text{eV}$ → insulator (this is diamond).

False. A full band has, for every electron drifting right, one drifting left → the currents cancel to zero. You need empty states to accelerate into, and at $T=0$ nothing is promoted across the $0.5\text{eV}$ gap, so there are none. No current.

What: evaluate the exponent, then convert. $\frac{E_g}{2kT}=\frac{1.12}{2\times 0.02585}=\frac{1.12}{0.0517}=21.66$ Why: the carrier fraction is $e^{-21.66}$. Convert: $e^{-21.66}=10^{-21.66/2.303}=10^{-9.41}\approx 3.9\times 10^{-10}$. So only about 4 in $10^{10}$ states-worth of the budget makes it across. Tiny — yet enough for silicon electronics.

Exponent for Ge: $\frac{0.67}{0.0517}=12.96$, so the factor is $e^{-12.96}$. Ratio Ge to Si: $\frac{n_i^{Ge}}{n_i^{Si}}=e^{-(0.67-1.12)/(2\times 0.02585)}=e^{+(0.45)/0.0517}=e^{8.70}$ Convert: $e^{8.70}=10^{8.70/2.303}=10^{3.78}\approx 6\times 10^3$. Meaning: germanium has roughly 6000× more intrinsic carriers than silicon at room temperature — the smaller gap wins massively because it sits in an exponent.

What: take the ratio of $n_i$ at two temperatures. $\frac{n_i(400)}{n_i(300)}=\exp \left[\frac{E_g}{2k}\left(\frac{1}{300}-\frac{1}{400}\right)\right]$ Why: the prefactor $\sqrt{N_c{N}_v}$ has only mild $T^{3/2}$ dependence; the exponential dominates, so we ratio it. First $\frac{E_g}{2k}=\frac{1.12}{2\times 8.617\times 10^{-5}}=6499\text{K}$. Then $\frac{1}{300}-\frac{1}{400}=0.003333-0.002500=0.0008333{\text{K}}^{-1}$. Product: $6499\times 0.0008333=5.416$. $\frac{n_i(400)}{n_i(300)}=e^{5.416}=10^{5.416/2.303}=10^{2.35}\approx 225$ Meaning: a mere $100\text{K}$ rise multiplies free carriers by ~225×. This is why semiconductor devices are so temperature-sensitive.

What: invert the ratio formula to solve for $E_g$. $1000=\exp \left[\frac{E_g}{2k}\left(\frac{1}{300}-\frac{1}{600}\right)\right]$ Take $\ln$: $\ln 1000=6.908$. The temperature bracket: $\frac{1}{300}-\frac{1}{600}=0.003333-0.001667=0.001667{\text{K}}^{-1}$. So $\frac{E_g}{2k}\times 0.001667=6.908\Rightarrow \frac{E_g}{2k}=4144\text{K}$. Then $E_g=2\times 8.617\times 10^{-5}\times 4144=0.714\text{eV}$. Meaning: about $0.71\text{eV}$ — close to germanium. A "×1000 on doubling $T$" fingerprint pins the gap.

What: in a metal $E_g\approx 0$, so $e^{-E_g/2kT}=e^0=1$ at every $T$ → carrier count $n$ is constant. Then $\sigma =nq\mu$ moves only with $\mu$. A $9\%$ mobility drop means: $\frac{{\sigma }_{330}}{{\sigma }_{300}}=\frac{{\mu }_{330}}{{\mu }_{300}}=0.91$ Meaning: conductivity falls to $91\%$ — resistance rises. This is the metal behaviour: heating helps nothing on carrier count and only adds scattering. See Temperature Dependence of Resistance in Metals.

Step A — carrier factor. $\frac{E_g}{2k}=6499\text{K}$; $\frac{1}{300}-\frac{1}{330}=0.003333-0.003030=0.0003030{\text{K}}^{-1}$. Product $=6499\times 0.0003030=1.969$, so $n$ grows by $e^{1.969}=7.16$. Step B — mobility factor. $\mu$ falls to $0.80$. Step C — net conductivity. $\frac{{\sigma }_{330}}{{\sigma }_{300}}=\underset{\text{carriers}}{\underbrace{7.16}}\times \underset{\text{mobility}}{\underbrace{0.80}}=5.73$ Meaning: conductivity rises ~5.7×. The exponential carrier gain crushes the modest mobility loss — the opposite of the metal. Same $\sigma =nq\mu$ equation, opposite winner.

What: in equilibrium the product of electron and hole concentrations is fixed at $n_i^2$, regardless of doping — that is the mass-action law. $p=\frac{n_i^2}{n}=\frac{(1.0\times 10^{10}{)}^2}{1.0\times 10^{16}}=\frac{1.0\times 10^{20}}{1.0\times 10^{16}}=1.0\times 10^4{\text{cm}}^{-3}$ Why: doping pushed $n$ up by $10^6$, so $p$ must fall by $10^6$ to keep the product constant. Meaning: electrons ($10^{16}$) outnumber holes ($10^4$) by $10^{12}$ → this is a strongly n-type material; electrons are the majority carriers and dominate conduction.

What: we need $n_i$ to grow by $10^{16}/10^{10}=10^6$. Set the ratio of exponentials to $10^6$: $10^6=\exp \left[6499\left(\frac{1}{300}-\frac{1}{T}\right)\right]$ Take $\ln$: $\ln (10^6)=6\times 2.303=13.82$. So $6499\left(\frac{1}{300}-\frac{1}{T}\right)=13.82\Rightarrow \frac{1}{300}-\frac{1}{T}=0.002126$. Then $\frac{1}{T}=0.003333-0.002126=0.001207\Rightarrow T\approx 828\text{K}$. Meaning: you would need to heat silicon to about 828 K ($\sim 555{}^{\circ }\text{C}$) for intrinsic carriers to match a light doping. This is exactly why we dope: it delivers a $10^{16}$ carrier level at room temperature that thermal excitation alone could only reach by nearly melting the device. Doping engineers carriers; temperature is a blunt, unstable tool.