The Workbook

1. A: the bands overlap, so electrons move freely into empty states. A metal. B: a gap of about \(1\) eV, small enough for thermal energy to lift some electrons across. A semiconductor. C: a gap of \(6\) eV, far too large. An insulator.

2. Only A. In B and C every electron is in the full valence band at \(0\) K and nothing can move.

1. \(\lambda_{\max} = \dfrac{1240}{E_g}\) nm. Si: \(\dfrac{1240}{1.1} \approx 1130\) nm. Ge: \(\dfrac{1240}{0.7} \approx 1770\) nm. Diamond: \(\dfrac{1240}{5.5} \approx 225\) nm.

2. Visible photons carry \(1.8\) to \(3.1\) eV. This is more than the gaps of Si and Ge, so both absorb visible light and look dark. It is far less than \(5.5\) eV, so diamond cannot absorb it; the light passes through.

Metal: the number of free electrons is fixed. Heating makes the ions vibrate more, the electrons are scattered more often, and the resistance rises.

Semiconductor: heating lifts many more electrons across the gap, creating electron-hole pairs. The number of carriers grows very fast with temperature. This outweighs the extra scattering, so the resistance falls.

1. \(1.5 \times 10^{16}\) per m\(^3\). Every electron that leaves the valence band leaves one hole behind, so in a pure crystal \(n_h = n_e = n_i\).

2. \(\dfrac{1.5 \times 10^{16}}{5 \times 10^{28}} = 3 \times 10^{-13}\). About one atom in three million million. Pure silicon is a very poor conductor.

A hole is a missing electron in a covalent bond of the valence band, an empty state where an electron should be.

Under a field a neighbouring bound electron jumps into the hole, leaving a new hole where it came from. The empty state moves the opposite way to the electrons, so it behaves like a free particle of charge \(+e\).

\(\sigma = 1.6 \times 10^{-19} \times 1.5 \times 10^{16} \times (0.135 + 0.048) = 2.4 \times 10^{-3} \times 0.183 \approx 4.4 \times 10^{-4}\) S/m.

Resistivity \(\rho = \dfrac{1}{\sigma} \approx 2.3 \times 10^{3}\ \Omega\) m. Copper has \(1.7 \times 10^{-8}\ \Omega\) m, about \(10^{11}\) times less.

Phosphorus and arsenic have five valence electrons. Four form bonds with silicon; the fifth is loosely held and becomes a free electron. They are donors and give n-type silicon.

Boron and indium have three valence electrons. One bond is left short of an electron, a hole. They are acceptors and give p-type silicon.

1. Each donor gives one electron, far more than the intrinsic number: \(n_e \approx N_D = 10^{22}\) m\(^{-3}\). Then \(n_h = \dfrac{n_i^2}{n_e} = \dfrac{(1.5 \times 10^{16})^2}{10^{22}} = \dfrac{2.25 \times 10^{32}}{10^{22}} = 2.25 \times 10^{10}\) m\(^{-3}\).

2. \(\dfrac{n_e}{n_h} = \dfrac{10^{22}}{2.25 \times 10^{10}} \approx 4.4 \times 10^{11}\). Electrons are the majority carriers; holes are the minority. Doping has removed most of the holes.

\(n_h \approx N_A = 5 \times 10^{21}\) m\(^{-3}\). \(n_e = \dfrac{n_i^2}{n_h} = \dfrac{2.25 \times 10^{32}}{5 \times 10^{21}} = 4.5 \times 10^{10}\) m\(^{-3}\).

Holes are the majority carriers and electrons the minority. This is p-type silicon.

1. n-type. The extra level lies just below the conduction band and holds the fifth electrons of donor atoms. Such an electron needs only \(0.05\) eV to enter the conduction band.

2. The donor electron needs \(0.05\) eV, only about twice \(kT\). Random thermal energy easily supplies this, so nearly every donor has given up its electron at \(300\) K.

3. A valence electron needs the full gap, \(1.1\) eV, more than \(40\) times \(kT\). Very few electrons ever collect this much thermal energy, so \(n_i\) stays small.

\(\sigma = 1.6 \times 10^{-19} \times 10^{22} \times 0.135 = 216\) S/m. The hole term is negligible.

\(\dfrac{216}{4.4 \times 10^{-4}} \approx 5 \times 10^{5}\). One donor per five million atoms has raised the conductivity half a million times.

1. \(N_D = \dfrac{5 \times 10^{28}}{10^{6}} = 5 \times 10^{22}\) m\(^{-3}\), so \(n_e \approx 5 \times 10^{22}\) m\(^{-3}\).

2. \(n_h = \dfrac{2.25 \times 10^{32}}{5 \times 10^{22}} = 4.5 \times 10^{9}\) m\(^{-3}\).

No. Each free electron came from a phosphorus atom that is now a fixed positive ion. The extra electrons are balanced by the extra ionic charges, so the crystal as a whole is neutral. "n-type" describes the carriers, not the net charge.

1. Electrons diffuse from the n side, where they are many, into the p side. Holes diffuse from the p side into the n side.

2. On the n side the donors left behind are fixed positive ions; on the p side the acceptors that gained electrons are fixed negative ions. This thin layer of fixed charge, with no free carriers, is the depletion region.

3. The fixed charges set up an electric field from n to p. This field opposes further diffusion and drives a drift current the other way. Diffusion and drift balance, and the junction settles with a potential barrier across it.

1. X is the depletion region. Electrons that diffused in from the n side have filled the holes of acceptor atoms there, leaving the acceptors as negative ions.

2. From the positive ions to the negative ions: from n to p. A field pushes an electron against its direction, so towards the n side, back where it came from.

3. About \(0.7\) V for silicon (\(0.3\) V for germanium).

1. Forward bias. The battery pushes holes and electrons towards the junction, the barrier is lowered by the applied voltage, and X becomes narrower. Once the applied voltage passes about \(0.7\) V, a large current flows.

2. Reverse bias. Carriers are pulled away from the junction, the barrier is raised, and X becomes wider. Only the few minority carriers cross, giving a current of microamperes that hardly depends on the voltage.

1. The current stays near zero until about \(0.7\) V and then rises steeply. The knee voltage is \(0.7\) V.

2. At \(0.8\) V the curve reads \(30\) mA.

3. At \(-30\) V the reverse current is about \(-1\ \mu\)A, thirty thousand times smaller than the forward current. Breakdown begins near \(-50\) V, where the current suddenly grows.

1. \(r = \dfrac{0.80 - 0.75}{(30 - 20) \times 10^{-3}} = \dfrac{0.05}{0.010} = 5\ \Omega\).

2. Over \(20\) V the reverse current changes by perhaps \(1\ \mu\)A: \(r \approx \dfrac{20}{10^{-6}} = 2 \times 10^{7}\ \Omega\), of the order of \(10^{7}\ \Omega\). The diode is nearly an open switch.

The diode takes \(0.7\) V, leaving \(5 - 0.7 = 4.3\) V across the resistor. \(I = \dfrac{4.3}{430} = 0.010\) A \(= 10\) mA.

1. Circuit (a). The triangle of the symbol points in the direction of easy (conventional) current flow. In (a) it points from the positive side of the battery round to the negative; in (b) it points against the battery.

2. (a): \(I = \dfrac{6 - 0.7}{530} = 10\) mA, and the resistor has \(10 \times 10^{-3} \times 530 = 5.3\) V across it. (b): no current, so \(0\) V across the resistor.

3. (a): \(0.7\) V, the forward drop. (b): the whole \(6\) V of the battery appears across the diode, since the resistor drops nothing.

1. It rises, roughly doubling for every \(10\) K. The reverse current is carried by minority carriers, and their number \(n_i^2/N\) grows rapidly with temperature.

2. It falls, by about \(2\) mV per kelvin. More carriers are present on both sides, so a smaller barrier is enough to balance the diffusion.

A pair is created only if the photon energy is at least \(E_g\).

\(1.5\) eV: above \(1.1\) eV and above \(0.7\) eV, so pairs in both Si and Ge. \(0.9\) eV: below \(1.1\) eV, no pair in Si; above \(0.7\) eV, a pair in Ge. Three of the four cases.

In reverse bias the positive terminal is on the n side and pulls electrons away from the junction. The negative terminal on the p side pulls holes away. More donor and acceptor ions are left uncovered on each side.

The layer of fixed charge with no free carriers therefore grows thicker, and the barrier voltage across it grows with the applied voltage.

The junction has a built-in barrier of about \(0.7\) V that stops the majority carriers crossing. A forward voltage cancels this barrier, so electrons from the n side and holes from the p side pour across and a large current flows.

A reverse voltage adds to the barrier and pulls the carriers away from the junction. Only the tiny number of minority carriers can cross, so the current is almost zero. The diode conducts one way only.