Q.In an intrinsic semiconductor the energy gap $E_g$ is $1.2\ \text{eV}$. Its hole mobility is much smaller than electron mobility and independent of temperature. What is the ratio between conductivity at 600 K and that at 300 K? Assume that the temperature dependence of intrinsic carrier concentration $n_i$ is given by

[!FORMULA] $n_i = n_0 \exp\left(-\dfrac{E_g}{2 k_B T}\right)$

where $n_0$ is a constant.

Concept and Intuition

In an intrinsic (pure) semiconductor like Si or Ge, conduction electrons are created only when thermal energy breaks a covalent bond, promoting an electron from the valence band to the conduction band and leaving a hole behind. The number of such carriers depends strongly on temperature. At $T = 0\,\mathrm{K}$, there is no thermal energy available, so the valence band is completely full and the conduction band is completely empty — exactly the band picture of an insulator.

Step-by-Step Solution

1. Carrier generation in an intrinsic semiconductor is thermally activated: $n_i \propto e^{-E_g/2k_BT}$.
2. As $T \to 0$, $n_i \to 0$ since the exponential term vanishes.
3. Zero free carriers means zero conductivity, which is the defining behaviour of an insulator. …

Concept and Intuition

In thermal equilibrium (whether the material is intrinsic or doped), the product $n_en_h$ stays constant at a given temperature and equals $n_i^2$ — this is the semiconductor analogue of the law of mass action, valid for both intrinsic and extrinsic semiconductors.

Step-by-Step Solution

1. The mass-action law states $n_e n_h = n_i^2$ at thermal equilibrium.
2. Solving for $n_i$: $n_i = \sqrt{n_e n_h}$

Common Mistakes …

Concept and Intuition

In any semiconductor at thermal equilibrium, the product of electron and hole concentrations equals the square of the intrinsic carrier concentration: $np = n_i^2$. This law holds whether the material is intrinsic or doped, so doping that raises hole concentration must correspondingly lower electron concentration.

Step-by-Step Solution

1. Intrinsic concentration: $n_i = 1.5\times10^{16}\ \mathrm{m^{-3}}$, so $n_i^2 = 2.25\times10^{32}\ \mathrm{m^{-6}}$.
2. New hole concentration: $p = 3\times10^{22}\ \mathrm{m^{-3}}$. …

Concept and Intuition

In an intrinsic (pure) semiconductor, conduction occurs because thermal energy breaks covalent bonds between atoms, freeing valence electrons to move into the conduction band. Each time an electron is freed this way, it leaves behind a vacancy in the bond — a hole — which itself behaves like a mobile positive charge carrier. Because electron-hole pairs are created together, increasing temperature always increases the concentrations of both types of carrier simultaneously (unlike in an extrinsic/doped semiconductor at intermediate temperatures, but even there, at higher temperatures intrinsic generation eventually dominates).

Step-by-Step Solution

1. Semiconductors conduct via two types of charge carriers: free electrons (in the conduction band) and holes (vacancies in the valence band).
2. Raising temperature increases the thermal energy available to break covalent bonds.
3. Each broken bond generates one free electron and simultaneously leaves one hole — this is pair generation. …

Concept and Intuition

In any semiconductor at thermal equilibrium, the product of electron and hole concentrations equals the square of the intrinsic carrier concentration, $n_e n_h = n_i^2$, regardless of doping (this is analogous to the equilibrium constant in a chemical reaction — doping shifts the individual concentrations but their product stays fixed by the material and temperature). Indium is a trivalent (acceptor) dopant, so doping with it creates a $p$-type semiconductor: it greatly increases the hole concentration $n_h$ far above $n_i$, and correspondingly decreases the minority electron concentration $n_e$ far below $n_i$, keeping the product constant.

Step-by-Step Solution

1. Write the mass-action law: $n_e\,n_h = n_i^2$. …

Concept and Intuition

In a semiconductor, the valence band and conduction band are separated by a small energy gap ($E_g \sim 1$ eV for Si/Ge). At $T = 0$ K there is no thermal energy to excite electrons across this gap, so the conduction band is empty and the material behaves as a perfect insulator. As $T$ increases, a fraction of valence electrons (following a Boltzmann-like/Fermi-Dirac distribution) gain enough thermal energy to jump the gap, populating the conduction band and leaving holes behind — hence conductivity (and free-electron count) rises with temperature. This is fundamentally different from a metal, where the conduction band is either partially filled or overlaps the valence band, giving a very large free-electron density essentially independent of this gap-crossing mechanism, and always far larger than a semiconductor's.

Step-by-Step Solution

1. Evaluate (A): No free electrons at 0 K — TRUE, because there's no thermal energy to promote electrons across the gap.
2. Evaluate (B): No free electrons at any temperature — FALSE, because at $T>0$ some electrons do cross the gap.
3. Evaluate (C): Number of free electrons increases with temperature — TRUE, directly from the thermal excitation mechanism. …

Concept and Intuition

Doping a semiconductor with pentavalent (donor) atoms creates extra free electrons roughly equal in number to the donor atom concentration (since each donor atom contributes one free electron and donor concentration vastly exceeds the intrinsic carrier concentration). The mass-action law, $n_en_h = n_i^2$, always holds regardless of doping, letting us find the (now minority) hole concentration.

Step-by-Step Solution

1. Total atom density $=8\times10^{28}\,\text{m}^{-3}$; doping level $=2$ ppm $=2\times10^{-6}$.
2. Donor concentration $N_D = 2\times10^{-6}\times 8\times10^{28} = 1.6\times10^{23}\,\text{m}^{-3}$.
3. Since $N_D \gg n_i (=10^{16})$, essentially all donors ionize and dominate: $n_e \approx N_D = 1.6\times10^{23}\,\text{m}^{-3}$. …

Concept and Intuition

For an intrinsic (undoped) semiconductor, both electrons and holes contribute to conduction, so the conductivity combines both carrier types: $\sigma=n_i e(\mu_e+\mu_h)$, where $n_i$ is the intrinsic carrier concentration and $\mu_e,\mu_h$ are the electron and hole mobilities. Once $\sigma$ (hence resistivity $\rho=1/\sigma$) is known, an ordinary rod-resistance formula $R=\rho l/A$ gives the resistance from its physical dimensions.

Step-by-Step Solution

1. Given: $n_i=2.5\times10^{19}\ \text{m}^{-3}$, $\mu_e=0.39\ \text{m}^2/\text{V·s}$, $\mu_h=0.19\ \text{m}^2/\text{V·s}$, $l=0.925$ cm, $A=1$ mm².
2. Conductivity: $\sigma=n_i e(\mu_e+\mu_h)=(2.5\times10^{19})(1.6\times10^{-19})(0.58)$.
3. $n_i e=2.5\times1.6=4.0$ (the powers of ten cancel: $10^{19}\times10^{-19}=1$). So $\sigma=4.0\times0.58=2.32$ S/m.
4. Resistivity: $\rho=1/\sigma=1/2.32\approx0.431\ \Omega\cdot\text{m}$ — this matches germanium's well-known intrinsic resistivity (~0.4–0.5 $\Omega\cdot$m at room temperature), confirming the numbers used are physically consistent. …