Extrinsic Semiconductor

Why doping is necessary

• The conductivity of an intrinsic semiconductor depends on temperature, and at room temperature it is very small — only the few thermally generated electron-hole pairs are available to carry current. No useful electronic device can be built from a material whose carrier supply is that small and that temperature-sensitive.
• Adding a small amount of a suitable impurity increases the conductivity manifold, and — more importantly — makes it controllable, since the doping level is fixed by the manufacturer rather than by the weather.
• CBSE asks this as "why is an intrinsic semiconductor deliberately converted into an extrinsic semiconductor" and as "why is doping a necessity for making electronic devices". Both want the same two points: intrinsic conductivity is very small, and doping raises it greatly and controllably.

Doping — the definition to quote

• Doping is the deliberate addition of a small, controlled amount (typically a few parts per million) of a suitable impurity to a pure semiconductor in order to increase its conductivity. The impurity atoms are called dopants, and the resulting material is an extrinsic (or impurity, or doped) semiconductor.
• The dopant must not distort the original lattice: it occupies only a very few of the original lattice sites, and for that its atoms must be nearly the same size as the Si or Ge atoms they replace. This is why the dopants are taken from the groups immediately next to group IV.
• Two types of dopant are used for tetravalent Si or Ge:

n-type semiconductor

• When a pentavalent atom occupies a Si lattice site, four of its five valence electrons form covalent bonds with the four neighbouring Si atoms. The fifth electron has no bond to join.
• That fifth electron sees the four bonding electrons as part of the effective core of its parent atom, so it is only very weakly bound. The energy needed to set it free is very small — about $0.01$ eV in germanium and $0.05$ eV in silicon — so even at room temperature it becomes free to move through the lattice.
• Compare this with the energy needed to jump the forbidden gap in the intrinsic material: about $0.72$ eV in Ge and about $1.1$ eV in Si. The dopant electron is released some fifty to a hundred times more cheaply, which is the whole point of doping.
• Because the pentavalent atom donates one extra electron for conduction, it is called a donor impurity. The number of electrons supplied this way depends strongly on the doping level and is independent of any rise in ambient temperature, whereas the intrinsically generated pairs increase only weakly with temperature.
• In the doped crystal, the total electron count $n_e$ comes from the donors plus intrinsic generation, while $n_h$ comes from intrinsic generation only. Worse for the holes: the larger electron population raises the recombination rate, so the hole count is reduced still further. Hence

$$n_e \gg n_h$$

• Electrons are the majority carriers and holes the minority carriers — this is an n-type semiconductor.
• Diagram in words: the usual square bond grid, with one lattice site occupied by a core marked $+5$ (or, in the simplified version, a fixed core carrying one extra effective positive charge) and a single loose electron drawn beside it, not tied into any bond.

p-type semiconductor

• When a trivalent atom occupies a Si lattice site, its three valence electrons form covalent bonds with three neighbouring Si atoms. The bond with the fourth neighbour is left with a vacancy — a hole.
• An electron in the outer orbit of some neighbouring atom can jump in to fill that vacancy, leaving a hole at its own site instead. The hole is thus free to move through the crystal and is available for conduction.
• When the trivalent atom accepts that fourth electron it becomes effectively negatively charged, so it is called an acceptor impurity, and it can be drawn as a fixed core with one effective negative charge plus its associated hole.
• One acceptor atom gives one hole. These are additional to the intrinsically generated holes, while conduction electrons come from intrinsic generation alone, and recombination reduces their number from $n_i$ down to $n_e$. Hence

$$n_h \gg n_e$$

• Holes are the majority carriers and electrons the minority carriers — this is a p-type semiconductor.
• Diagram in words: the square bond grid with one site occupied by a core marked $+3$ (or a fixed core with one effective negative charge), and one of its four bonds shown incomplete, with a small open circle marking the hole.

Charge neutrality

• The crystal as a whole stays electrically neutral, whichever way it is doped. The charge on each additional mobile carrier is exactly equal and opposite to the charge of the ionised core it came from: a free electron for each positive ionised donor, a hole for each negative ionised acceptor.
• The standard assertion-reason: in an n-type semiconductor the electron density exceeds the hole density, yet the crystal maintains overall charge neutrality, precisely because the charge of the donated electrons is equal and opposite to that of the ionised donors. Both statements are true and the second explains the first.
• Note also that "doping increases conductivity" is true, but "doping always increases the number of electrons" is false — trivalent doping increases holes and reduces electrons.

Minority carriers are actively suppressed

• In an extrinsic semiconductor the majority carriers are so abundant that a thermally generated minority carrier is very likely to meet one and recombine.
• So the dopant, by adding a large number of one type of carrier, indirectly reduces the concentration of the other type below its intrinsic value.

Energy band diagrams of extrinsic semiconductors

• Doping introduces additional allowed energy states inside the forbidden gap: a donor level $E_D$ for pentavalent doping and an acceptor level $E_A$ for trivalent doping.
• n-type at $T > 0$ K: draw the conduction band (mostly empty, with a few filled dots near its bottom) and the valence band (completely full, dots throughout), separated by the gap $E_g$. Draw a dashed donor level $E_D$ just below $E_C$ — about $0.01$ eV below it in Ge, $0.05$ eV in Si. Electrons from $E_D$ move into the conduction band with a very small supply of energy.
• p-type at $T > 0$ K: draw the conduction band (essentially empty) and the valence band (nearly full, with a few empty circles near its top), separated by $E_g$. Draw a dashed acceptor level $E_A$ just above $E_V$. With a very small supply of energy an electron from the valence band jumps up into $E_A$, ionising the acceptor negatively and leaving a hole behind in the valence band. Equivalently: the hole from $E_A$ sinks down into the valence band. Electrons rise up and holes fall down when they gain external energy.
• At room temperature most donor atoms and most acceptor atoms are ionised, while only about $10^{12}$ Si atoms per cubic metre are. So the conduction-band electrons of an n-type sample come overwhelmingly from the donors, and the valence-band holes of a p-type sample come overwhelmingly from the acceptors.
• Worked example CBSE has set: germanium ($E_g \approx 0.72$ eV) doped with antimony, a pentavalent atom. The antimony donor level sits about $0.01$ eV below the conduction band, so at ordinary temperatures nearly every antimony atom is ionised and releases its fifth electron straight into the conduction band. This supplies far more free electrons than intrinsic generation ever could, so the conductivity of the doped crystal is greatly increased, and the crystal is n-type with electrons as majority carriers. The same answer serves for silicon doped with antimony.

The mass-action law

$$n_e n_h = n_i^2$$

• The electron and hole concentrations of a semiconductor in thermal equilibrium always satisfy this, whether the sample is intrinsic or extrinsic. Doping raises one and depresses the other, keeping the product fixed at the intrinsic value squared for that temperature.

Worked pattern (NCERT Example 14.2)

• A pure Si crystal has $5\times10^{28}$ atoms m$^{-3}$ and is doped with $1$ ppm of pentavalent As. Given $n_i = 1.5\times10^{16}$ m$^{-3}$, find $n_e$ and $n_h$.
• Donor density: $N_D = 10^{-6}\times5\times10^{28} = 5\times10^{22}$ m$^{-3}$.
• The thermally generated electrons ($n_i \sim 10^{16}$ m$^{-3}$) are utterly negligible beside this, so $n_e \approx N_D = 5\times10^{22}$ m$^{-3}$.
• By the mass-action law,

$$n_h = \dfrac{n_i^2}{n_e} = \dfrac{(1.5\times10^{16})^2}{5\times10^{22}} = \dfrac{2.25\times10^{32}}{5\times10^{22}} \approx 4.5\times10^{9}\ \text{m}^{-3}$$

• Note the scale: doping has pushed the electron count up by six orders of magnitude and the hole count down by seven.

n-type versus p-type

Conductivity at $0$ K and at room temperature

• At absolute zero, both n-type and p-type semiconductors behave as perfect insulators: the valence band is completely filled, the conduction band completely empty, no dopant level is ionised, and $\sigma = 0$. Doping makes no difference at $0$ K.
• At room temperature, essentially all the donor (or acceptor) levels are ionised and a few intrinsic pairs are generated as well, so both types acquire an appreciable conductivity — much higher than that of an equivalent intrinsic sample, because doping supplies far more carriers than pure thermal generation does.

Effect of raising the temperature of an extrinsic sample

• The dopant contribution is already almost fully ionised and does not grow. What grows is intrinsic pair generation, and that always creates one electron and one hole together.
• So heating an n-type sample increases the number of free electrons and the number of holes by the same additional amount. The electron count nevertheless remains far larger throughout, because it started with the whole donor population behind it.

Standard results

• $n_e n_h = n_i^2$ always, in thermal equilibrium.
• $n_e \gg n_h$ (n-type), $n_h \gg n_e$ (p-type); $n_e \approx N_D$ and $n_h \approx N_A$ for normal doping levels.
• Donor level lies just below the conduction band; acceptor level lies just above the valence band; both about $0.01$ eV deep in Ge and $0.05$ eV in Si.
• Band gaps for comparison: Ge $0.72$ eV, Si $1.1$ eV.
• Doping level about $1$ ppm gives $N_D \sim 10^{22}$ m$^{-3}$ against $n_i \sim 10^{16}$ m$^{-3}$.
• The crystal is neutral overall, before and after doping.
• Both n-type and p-type are insulators at $0$ K.