Energy Bands, Doping and the p-n Junction: Quick Revision

A brief note on Energy Bands, Doping and the p-n Junction. Read it before a test. Try each question at the end before you open its answer.

Energy bands

• Bands: the valence band is the highest full band, the conduction band is above it, and the band gap $E_g$ lies between them.
• Conductor: partly filled or overlapping bands. Insulator: $E_g$ above about $3$ eV, for example diamond at $5.5$ eV. Semiconductor: $E_g$ below about $3$ eV, silicon at $1.12$ eV and germanium at $0.67$ eV.
• Hole: a missing electron in the valence band. It acts as a positive charge $+e$.
• Pure crystal: $n_e = n_h = n_i$. The carrier number grows as $e^{-E_g/2kT}$, so the resistance falls when it is heated.
• Light: a photon makes a pair if $\dfrac{1240\ \text{eV nm}}{\lambda} \ge E_g$.

Doping

• n-type: donor atoms from group 15 (P, As, Sb). $n_e \approx N_D$, electrons are the majority carriers.
• p-type: acceptor atoms from group 13 (B, Al, Ga, In). $n_h \approx N_A$, holes are the majority carriers.
• Law of mass action: $n_e\, n_h = n_i^2$. Conductivity: $\sigma = e\,(n_e \mu_e + n_h \mu_h)$.
• A doped crystal is neutral. The ions are fixed and only the carriers move.

The p-n junction

• Depletion layer: a thin layer of fixed ions with no free carriers. It makes a built-in field from n to p and a barrier of about $0.7$ V (Si) or $0.3$ V (Ge).
• Forward bias (p to the positive end): the barrier falls, the layer narrows, and the current grows exponentially.
• Reverse bias: the barrier rises, the layer widens, and only a tiny minority-carrier current $I_s$ flows.
• Diode equation: $I = I_s\left(e^{V/\eta V_T} - 1\right)$, with $V_T = 25.85$ mV at $300$ K.
• Dynamic resistance: $r_d = \dfrac{\eta V_T}{I}$. In a circuit, a conducting silicon diode drops about $0.7$ V.

A worked example

A silicon crystal has $N_A = 4.0 \times 10^{21}\ \text{m}^{-3}$ and $n_i = 1.0 \times 10^{16}\ \text{m}^{-3}$. Find $n_h$ and $n_e$.

1. The holes. $n_h \approx N_A = 4.0 \times 10^{21}\ \text{m}^{-3}$.
2. The electrons. $n_e = \dfrac{n_i^2}{n_h} = \dfrac{(1.0 \times 10^{16}\ \text{m}^{-3})^2}{4.0 \times 10^{21}\ \text{m}^{-3}} = 2.5 \times 10^{10}\ \text{m}^{-3}$.

Practise with new numbers

Check yourself

1. Why does the resistance of a semiconductor fall when it is heated?

2. A p-n junction is forward biased by $0.4$ V. The barrier was $0.7$ V. What is the barrier now, and what happens to the layer?

3. A reverse-biased diode still carries a small current. Where does it come from?

Sources

• NCERT Physics Class 12, Semiconductor Electronics: Materials, Devices and Simple Circuits.
• OpenStax University Physics Volume 3, the chapter Condensed Matter Physics.