Question

How can p-type and n-type crystal from pure semiconductor be formed? Find the maximum wavelength of electromagnetic radiation which can create a hole-electron pair in germanium of a band-gap 0.65 eV.
OR
How is the conductivity of a metal and semiconductor changed with raising temperature? Calculate the conductivity of an n-type semiconductor from the following data: Density of conduction electrons = 8×10¹³ /cm³, Density of holes = 5×10¹² /cm³, Mobility of conduction electrons = 2.3×10⁴ cm²/V·s, Mobility of holes = 100 cm²/V·s.

Hint:

Doping with a pentavalent impurity gives n-type (extra electrons) and a trivalent impurity gives p-type (holes). Use \( \lambda_{max} = hc/E_g \) for the germanium part, and \( \sigma = e(n_e\mu_e + n_h\mu_h) \) for the conductivity.

Answer 1

Option 1: Formation of p-type / n-type crystals and maximum wavelength in germanium

Step 1 (n-type crystal): Start from a pure (intrinsic) tetravalent semiconductor such as silicon or germanium, in which every atom shares its 4 valence electrons in covalent bonds. Add a small amount of a pentavalent impurity (P, As, Sb or Bi). Four of the impurity's five valence electrons form bonds; the fifth is loosely bound and becomes free even at room temperature. These donated electrons make electrons the majority carriers, giving an n-type crystal.

Step 2 (p-type crystal): To the same pure semiconductor add a trivalent impurity (B, Al, In or Ga). Its 3 valence electrons complete only three of the four bonds; the missing bond is a hole that can accept an electron. These acceptor holes make holes the majority carriers, giving a p-type crystal.

Step 3 (Condition for pair creation): To lift one electron from the valence band to the conduction band (creating one electron-hole pair), the photon energy must be at least equal to the band-gap energy \( E_g \). The largest wavelength corresponds to the smallest energy, i.e. \( E = E_g \): \[ E_g = h\nu = \frac{hc}{\lambda_{max}} \Rightarrow \lambda_{max} = \frac{hc}{E_g} \]

Step 4 (Convert band-gap to joule): \[ E_g = 0.65\ \text{eV} = 0.65 \times 1.6\times10^{-19}\ \text{J} = 1.04\times10^{-19}\ \text{J} \]
Step 5 (Substitute): With \( h = 6.6\times10^{-34}\,\text{J s} \) and \( c = 3\times10^{8}\,\text{m/s} \), \[ hc = 6.6\times10^{-34} \times 3\times10^{8} = 1.98\times10^{-25}\ \text{J m} \] \[ \lambda_{max} = \frac{1.98\times10^{-25}}{1.04\times10^{-19}} = 1.9\times10^{-6}\ \text{m} \]
Step 6 (Result): \( \lambda_{max} \approx 1.9\times10^{-6}\ \text{m} = 1.9\ \mu\text{m} \) (about 19000 Å, in the infrared region).
\[\boxed{\lambda_{max} \approx 1.9\times10^{-6}\ \text{m}\ (1.9\ \mu\text{m})}\]

Option 2: Temperature dependence and conductivity of an n-type semiconductor

Step 1 (Metal): In a metal the number density of free electrons is fixed. On raising the temperature the lattice ions vibrate more strongly, electrons collide more often, the relaxation time \( \tau \) falls, so resistance rises and conductivity decreases. Metals therefore have a positive temperature coefficient of resistance.

Step 2 (Semiconductor): In a semiconductor, raising the temperature breaks many more covalent bonds and lifts electrons across the band-gap, so the carrier density n increases roughly exponentially. This large rise in carrier number outweighs the fall in relaxation time, so resistance falls and conductivity increases. Semiconductors therefore have a negative temperature coefficient of resistance.

Step 3 (Formula): The conductivity of a doped semiconductor carrying both types of charge is \[ \sigma = e\,(n_e\,\mu_e + n_h\,\mu_h) \] where \( n_e, n_h \) are the electron and hole densities and \( \mu_e, \mu_h \) their mobilities.

Step 4 (Products, in CGS-practical units): \[ n_e\,\mu_e = (8\times10^{13})(2.3\times10^{4}) = 1.84\times10^{18} \] \[ n_h\,\mu_h = (5\times10^{12})(100) = 5\times10^{14} \] \[ n_e\mu_e + n_h\mu_h = 1.84\times10^{18} + 0.0005\times10^{18} = 1.8405\times10^{18} \]
Step 5 (Substitute e = 1.6×10^-19 C): \[ \sigma = 1.6\times10^{-19} \times 1.8405\times10^{18} = 0.294\ \Omega^{-1}\text{cm}^{-1} \]
Step 6 (Result): \( \sigma \approx 0.294\ \Omega^{-1}\text{cm}^{-1} = 29.4\ \Omega^{-1}\text{m}^{-1} \) (i.e. 29.4 S/m). The electron term dominates because electrons are the majority carriers and have the far higher mobility, as expected for an n-type sample.
\[\boxed{\sigma \approx 0.294\ \Omega^{-1}\text{cm}^{-1} \approx 29.4\ \text{S/m}}\]