JEE Physics Semiconductor Electronics Notes Set 01

Energy Bands in Solids

When atoms come together to form a crystal, their discrete energy levels spread into wide bands. The band made from the outermost (valence) electrons is the valence band, and the band above it that holds free electrons is the conduction band. The empty region between them is the forbidden gap \( E_g \).

In a conductor the two bands overlap (or the conduction band is partly filled), so electrons move freely. In an insulator the gap is large (above about 3 eV), so almost no electron can reach the conduction band. In a semiconductor the gap is small (about 0.7 eV for germanium and 1.1 eV for silicon), so thermal energy can lift some electrons across it.

Intrinsic Semiconductors

A pure crystal such as Si or Ge is intrinsic. Each electron raised to the conduction band leaves a hole in the valence band, so electron and hole densities are equal: \( n_e = n_h = n_i \). The density \( n_i \) rises rapidly with temperature, roughly as \( e^{-E_g/2kT} \), so the resistance of a semiconductor falls when it is heated. This is a negative temperature coefficient of resistance, the opposite of a metal.

The total current is the sum of the electron and hole contributions. Using mobilities \( \mu_e \) and \( \mu_h \):

\[ \sigma = e\,(n_e \mu_e + n_h \mu_h), \qquad I = e A (n_e v_e + n_h v_h) \]

Extrinsic Semiconductors and Doping

Adding a small amount of a suitable impurity (doping) raises conductivity enormously.

Both types are electrically neutral overall. At a given temperature the mass action law holds: \( n_e n_h = n_i^2 \). If one carrier density is raised by doping, the other falls so that the product stays constant.

The p-n Junction

When p-type and n-type crystals are joined, electrons diffuse from n to p and holes from p to n. They recombine near the junction and leave behind fixed ions: positive donor ions on the n side and negative acceptor ions on the p side. This region without free carriers is the depletion layer, and it sets up a barrier potential (about 0.7 V for Si and 0.3 V for Ge) with a field directed from n to p.

At equilibrium, the diffusion current (majority carriers) is exactly balanced by the drift current (minority carriers), so the net current is zero.

Biasing a Junction Diode

The diode therefore conducts in only one direction. In forward bias a silicon diode starts conducting at the knee (cut-in) voltage of about 0.7 V, and a germanium diode at about 0.3 V. If the reverse voltage becomes large enough, the junction breaks down and the current rises suddenly.

Rectifiers

A rectifier converts alternating current into direct current by using the one-way property of a diode. A half wave rectifier uses one diode and conducts only in alternate half cycles, so the output ripple has the same frequency as the input. A full wave rectifier (centre-tap or bridge) uses two or four diodes and conducts in both half cycles, so the ripple frequency is twice the input frequency.

A capacitor connected across the load acts as a filter and smooths the output. The ripple is reduced by a larger capacitor, a larger load resistance and a higher input frequency.

Special Purpose Diodes

A Zener diode is heavily doped, has a thin depletion layer and is operated in reverse breakdown. The voltage across it stays almost constant over a wide range of current, so it is used as a voltage regulator. It is connected in reverse bias with a series resistor \( R_S \) in parallel with the load, and the current splits as \( I_S = I_Z + I_L \).

A photodiode is operated in reverse bias; light of energy greater than the band gap creates extra carriers and increases the reverse current in proportion to the intensity. A solar cell works without bias and converts light into electrical energy; pairs formed in the depletion region are separated by the junction field. A light emitting diode (LED) is forward biased; recombination releases photons of energy close to \( E_g \), and the wavelength is found from \( \lambda\,(\text{Å}) = 12400/E_g\,(\text{eV}) \).

Worked Examples

Example 1. Si has \( n_i = 1.5 \times 10^{16} \) m\( ^{-3} \). Doping with indium makes \( n_h = 4.5 \times 10^{22} \) m\( ^{-3} \). Find \( n_e \).
By the mass action law, \( n_e = n_i^2/n_h = (1.5 \times 10^{16})^2/(4.5 \times 10^{22}) = 5 \times 10^{9} \) m\( ^{-3} \).

Example 2. A 9 V source feeds a 1 k\( \Omega \) load through a 100 \( \Omega \) resistor and a 6 V Zener diode across the load. Find the Zener current.
Source current = (9 − 6)/100 = 30 mA. Load current = 6/1000 = 6 mA. Zener current = 30 − 6 = 24 mA.

Example 3. Find the largest wavelength that a germanium photodiode (\( E_g = 0.75 \) eV) can detect.
\( \lambda = 12400/0.75 \approx 16500 \) Å.