By the end of this chapter you should be able to:
- Describe electron-hole pair generation and recombination in intrinsic silicon, and relate a broken covalent bond to a band-to-band transition.
- Explain why hole conduction is a bookkeeping description of valence-band electron motion, and why the hole nevertheless has a well-defined charge, mobility and effective mass.
- State the temperature dependence of \(n_i\) and evaluate it for silicon at temperatures away from 300 K.
- Distinguish donor and acceptor doping, place \(E_D\) and \(E_A\) in the gap, and justify the assumption of complete ionisation at room temperature.
- Apply charge neutrality to compensated material containing both donors and acceptors.
- Derive and apply the mass-action law \(np = n_i^2\), and calculate minority-carrier concentrations for realistic doping levels.
- Identify the freeze-out, extrinsic and intrinsic regions on a plot of carrier concentration against temperature, and give the temperature limits of the extrinsic region.
Chapter 1 ended with an odd claim: a semiconductor is useful not because it conducts well but because it conducts badly in a controllable way. This chapter makes that claim quantitative. Pure silicon at room temperature contains roughly \(1.0 \times 10^{10}\) mobile electrons per cubic centimetre against \(5.0 \times 10^{22}\) atoms, which is one broken bond in every five million million. That is a very small number, and it is the reason doping works: an impurity present at one atom in a million contributes carriers that outnumber the intrinsic ones by a factor of a million, so the impurity, not the silicon, decides what the material does.
We start with the pure crystal, where thermal energy breaks bonds in pairs and the two halves of each pair — the freed electron and the vacancy it leaves — both carry current. The vacancy has to be handled carefully, because treating it as a positive particle is a bookkeeping device rather than a statement about what is physically moving, and a student who does not grasp that will be permanently confused by hole mobility. We then add pentavalent and trivalent impurities, place the levels they create inside the forbidden gap, and arrive at the two results that the whole of the rest of the course uses without further comment: charge neutrality, and the mass-action law \(np = n_i^2\).
1 Intrinsic Silicon and Thermal Generation
A semiconductor is called intrinsic when its electrical behaviour is determined by the host crystal alone, with any residual impurity present in concentrations too small to matter. Modern float-zone silicon reaches impurity levels below \(10^{12}\ \text{cm}^{-3}\), which is about one foreign atom in \(5\times10^{10}\) — a purity attained by no other manufactured material, and one that had to be achieved before the transistor could be more than a laboratory curiosity.
At absolute zero the intrinsic crystal is a perfect insulator. Every one of the four valence electrons per atom is locked in a covalent bond, the valence band is exactly full, the conduction band is exactly empty, and Chapter 1's argument applies: no empty state adjacent in energy to an occupied one means no current, whatever the field. Raise the temperature and the lattice atoms vibrate about their equilibrium sites with a mean thermal energy of order \(kT\), which is only 25.9 meV at 300 K. That average is far below the 1.12 eV needed to break a bond, but energies are distributed, not uniform, and the high-energy tail of the distribution is populated. Every so often a bond receives enough energy from the vibrating lattice to break.
When it does, two things are created at once and neither can be created alone. An electron leaves the bond and moves off through the crystal as a free carrier — in band language, it has been promoted from the valence band to the conduction band. The bond it left is now incomplete: there is a vacant state in the valence band. This paired creation is electron-hole pair generation, and it is why in intrinsic material the electron and hole concentrations are necessarily equal:
\(n\) is the electron concentration in the conduction band and \(p\) the hole concentration in the valence band, both per cubic centimetre. \(n_i\) is the intrinsic carrier concentration, equal to \(1.0\times10^{10}\ \text{cm}^{-3}\) for silicon at 300 K. Older texts quote \(1.5\times10^{10}\); the difference is a matter of measurement rather than of principle, and it shifts a built-in potential by 21 mV.
The reverse process runs at the same time. A conduction electron wandering through the crystal eventually encounters a broken bond and drops into it, releasing its energy as heat or, in a direct-gap material, as a photon. This is recombination, and it destroys one electron and one hole together. At a fixed temperature the generation rate \(G\), which depends on temperature but not on how many carriers happen to be present, is balanced by the recombination rate \(R\), which is proportional to the product of the two populations because a recombination event needs one of each:
In equilibrium \(np = G/r\), a quantity fixed by temperature alone. In intrinsic material \(n = p = n_i\), so \(G/r = n_i^2\), and therefore \(np = n_i^2\) in any sample at that temperature, doped or not. That is the mass-action law, derived here in three lines and used for the rest of the course; the last section returns to it with worked numbers.
Equilibrium is dynamic, not static. At 300 K silicon generates and recombines about \(10^{16}\) pairs per cubic centimetre per second, and the population of \(10^{10}\) is a steady state maintained by two large opposing flows, not a fixed set of carriers. Any individual electron stays free only for its lifetime \(\tau\), which Chapter 4 measures in microseconds. It matters that the balance is dynamic: shine light on the sample and \(G\) rises, the populations climb until \(rnp\) matches the new \(G\), and the photoconductor of Chapter 14 works.
2 The Hole: Bookkeeping That Behaves Like a Particle
The freed electron is straightforward: it is a real particle in an almost-empty band, it responds to a field, and it drifts opposite to the field because its charge is negative. The vacancy is where students lose their footing, so it is worth being exact about what is claimed.
Nothing physical moves as a hole. What moves is a valence electron. Consider a bond broken at site A, and apply a field pointing from left to right. An electron in the intact bond at site B, immediately to the right of A, is pushed leftwards by the field (its charge is negative) and can hop into the vacancy at A. Site A is now a complete bond and site B is incomplete. Repeat, and the vacancy travels steadily to the right — in the direction of the field — while every actual displacement is an electron moving left.
We could describe this exactly, by tracking each of the \(4\times10^{22}\) valence electrons per cubic centimetre and computing the net current from a nearly full band. Nobody does. It is arithmetically far easier to ignore the \(10^{22}\) electrons that are present and track the \(10^{10}\) states that are absent, and once we do so the absence behaves in every measurable respect like a particle. Assign it:
- a charge of \(+q = +1.602\times10^{-19}\) C, because the neutral crystal minus one electron leaves a net positive region;
- a drift direction along the field rather than against it, which the charge sign now delivers automatically;
- an effective mass \(m_p^* \approx 0.56\,m_0\) in silicon, set by the curvature of the valence band, so that \(F = m_p^* a\) works;
- a mobility \(\mu_p = 480\ \text{cm}^2/\text{V}\!\cdot\!\text{s}\) at 300 K, about a third of the electron's 1350, because the hop requires cooperation from a nearly full band while a conduction electron has empty states all around it.
That last point is the physical content behind "holes are slower than electrons". It is not that the hole is a heavy particle labouring through the lattice; it is that valence-band conduction is a congested process. The consequence runs right through the course: it is why an NPN transistor outperforms a PNP, why n-channel MOSFETs are made smaller than p-channel ones for the same drive current, and why CMOS logic in Chapter 25 sizes its p-device roughly two to three times wider than its n-device.
The current carried by a valence band containing one empty state at \(\mathbf{k}\) is identical, to every measurable digit, to the current that would be carried by a single positive particle of charge \(+q\) and effective mass \(m_p^*\) in an otherwise empty band. Because the two descriptions predict the same currents, the same Hall voltage and the same diffusion, either may be used — and the second is far shorter. Every equation from here on treats electrons and holes symmetrically.
One experiment settles that the sign is real and not a convenience. In the Hall effect, a magnetic field perpendicular to a current deflects the carriers to one side of the bar, and the polarity of the transverse voltage that builds up reveals the sign of the charge that is moving. A bar of p-type silicon gives the opposite Hall polarity to an n-type bar carrying current in the same direction. Whatever a hole is, it deflects like a positive charge, and Chapter 3 will find its statistics behave that way too.
3 How Many Carriers, and How Fast Does the Number Grow?
Chapter 3 derives the result properly; the form is quoted here because the temperature behaviour cannot wait. The intrinsic concentration is
\(N_C\) and \(N_V\) are the effective densities of states in the conduction and valence bands, \(2.8\times10^{19}\) and \(1.04\times10^{19}\ \text{cm}^{-3}\) for silicon at 300 K, and both vary as \(T^{3/2}\). The exponent carries \(E_g/2\) rather than \(E_g\) because each broken bond yields two carriers and \(n_i\) counts one of them.
The prefactor \(\sqrt{N_C N_V} = 1.71\times10^{19}\ \text{cm}^{-3}\) rises gently, as \(T^{3/2}\); the exponential rises violently. Everything about a semiconductor's temperature behaviour comes from that mismatch. It is worth noticing that substituting \(E_g = 1.12\) eV and \(kT = 0.0259\) eV gives \(n_i = 6.7\times10^{9}\ \text{cm}^{-3}\), not the measured \(1.0\times10^{10}\). The discrepancy is a factor of 1.5, and it corresponds to an error of only 21 meV in the effective gap. That is the whole lesson about exponentials: a 2 per cent uncertainty in \(E_g\) becomes a 50 per cent uncertainty in \(n_i\), so \(n_i\) should be looked up, not computed, whenever a measured value exists.
| \(T\) (K) | \(E_g(T)\) (eV) | \(n_i\) (cm\(^{-3}\)) | Relative to 300 K |
|---|---|---|---|
| 200 | 1.147 | \(5.3\times10^{4}\) | \(5.3\times10^{-6}\) |
| 250 | 1.137 | \(7.4\times10^{7}\) | \(7.4\times10^{-3}\) |
| 300 | 1.124 | \(1.0\times10^{10}\) | 1 |
| 350 | 1.111 | \(3.5\times10^{11}\) | 35 |
| 400 | 1.097 | \(5.3\times10^{12}\) | 528 |
| 500 | 1.066 | \(2.5\times10^{14}\) | \(2.5\times10^{4}\) |
From 300 K to 400 K — a rise of one hundred degrees, well within what a power device experiences — \(n_i\) grows by a factor of 528. Differentiating the expression near room temperature gives a useful rule of thumb: \(n_i\) doubles for every 8.6 K rise, so \(n_i^2\), and with it the reverse saturation current of a junction, roughly doubles every 4.3 K, or quadruples every 8.6 K. The familiar statement that reverse saturation current doubles every 10 °C applies to the generation current of a real diode, which scales with \(n_i\) rather than \(n_i^2\), and Chapter 6 sorts out which mechanism dominates where.
Silicon contains \(4.99\times10^{22}\) atoms per cm\(^3\) and, when intrinsic at 300 K, \(1.0\times10^{10}\) conduction electrons per cm\(^3\). What fraction of bonds is broken, and what impurity level would be needed to swamp it?
There are four bonds per atom shared between two atoms, so \(2\times4.99\times10^{22} = 9.98\times10^{22}\) bonds per cm\(^3\). The fraction broken is \(1.0\times10^{10}/9.98\times10^{22} = 1.0\times10^{-13}\): one bond in ten million million.
Now add phosphorus at \(10^{16}\ \text{cm}^{-3}\). That is \(10^{16}/4.99\times10^{22} = 2.0\times10^{-7}\) of the atoms, one in five million, or 0.2 parts per million — a level at which most materials would still be called pure. Yet it supplies a million times as many electrons as the silicon does. The doped conductivity works out at 2.0 S/cm against \(2.9\times10^{-6}\) S/cm intrinsic, a factor of \(6.8\times10^{5}\). This single comparison is the reason semiconductor electronics exists, and it is also the reason a fabrication plant is fanatical about contamination: an impurity you did not intend at \(10^{16}\ \text{cm}^{-3}\) does the same thing as one you did.
4 Pentavalent Doping and the Donor Level
Doping is the deliberate substitution of a small fraction of silicon atoms by atoms of a different valency. The impurity must sit substitutionally, on a lattice site, so that it takes part in the bonding; an atom lodged between sites is an interstitial and does something quite different, usually unhelpful.
Take a group V element — phosphorus, arsenic or antimony — with five valence electrons. Four of them form the covalent bonds the site requires, satisfying all four neighbours. The fifth has no bond to join. It is held only by the weak electrostatic attraction of the one extra proton in the impurity nucleus, screened by the silicon around it, so it is bound very loosely indeed.
How loosely is calculable. Treat the fifth electron as orbiting a single positive charge exactly as in hydrogen, but with two corrections: the Coulomb attraction is screened by the relative permittivity of silicon, \(\varepsilon_r = 11.7\), and the electron responds with its effective mass \(m^* \approx 0.26\,m_0\) rather than the free mass. Scaling the 13.6 eV hydrogen result:
About 26 meV, against measured values of 45 meV for phosphorus, 54 meV for arsenic and 39 meV for antimony in silicon. The hydrogenic model is crude — it ignores the detail of the impurity core — but it gets the order of magnitude right and, more usefully, it explains why the number is small: the screening enters as \(\varepsilon_r^2 = 137\).
The same scaling gives the orbit radius as \(0.529\ \text{\AA} \times \varepsilon_r/(m^*/m_0) = 23.8\) Å, some four lattice constants. The fifth electron is smeared over hundreds of silicon atoms even while bound, which is another way of seeing that it barely belongs to its parent atom.
In band terms the impurity introduces an allowed level, the donor level \(E_D\), lying \(E_C - E_D = 45\) meV below the conduction-band edge, as drawn in the middle panel of Figure 2.2. Since \(kT = 25.9\) meV at 300 K, promoting the electron from \(E_D\) into the conduction band costs less than two \(kT\), and essentially every donor is ionised. Solving the neutrality condition properly gives 99.75 per cent ionised at 300 K, 96 per cent at 150 K, 76 per cent at 100 K and only 6.5 per cent at 50 K. Between about 150 K and 500 K the approximation of complete ionisation, \(n \approx N_D\), is accurate to better than 1 per cent, and every calculation in this course assumes it.
Note what the donor has and has not done. It has supplied one mobile electron. It has not supplied any net charge: the phosphorus atom arrived neutral and, having lost an electron to the conduction band, is left as a fixed \(\text{P}^+\) ion embedded in the lattice. The crystal as a whole stays neutral. The distinction between the mobile electron and the immobile ion is not pedantry — it is the entire mechanism of the depletion region in Chapter 5, where the electrons are swept away and the ions, which cannot follow, are left behind as space charge.
5 Trivalent Doping and the Acceptor Level
A group III element — boron, aluminium, gallium or indium — brings three valence electrons to a site that requires four. Three bonds form; the fourth is incomplete, one electron short. The impurity is an acceptor, so called because that incomplete bond will readily accept an electron from a neighbouring complete bond.
The energy to move a valence electron into the vacancy is small for the same screening reason as before, and boron in silicon sits at \(E_A - E_V = 45\) meV, with aluminium at 67 meV and gallium at 72 meV. Boron is the universal p-type dopant in silicon manufacture, partly for that low level and partly because its atomic radius is close enough to silicon's to keep lattice strain manageable at high concentrations.
When a nearby valence electron is thermally promoted into the acceptor level, three things happen simultaneously and must be kept straight. The acceptor becomes a fixed negative ion, \(\text{B}^-\), because it now holds one more electron than its nucleus balances. A hole appears in the valence band, at the site the promoted electron came from, and that hole is mobile. And no electron has been added to the conduction band — this is not pair generation, and \(n\) is unaffected by it.
Material doped with acceptors is p-type: holes are plentiful, electrons scarce, and conduction is predominantly by holes. With complete ionisation, \(p \approx N_A\). The symmetry with the n-type case is exact if you are careful to keep track of which charges move:
| n-type | p-type | |
|---|---|---|
| Dopant group | V — P, As, Sb | III — B, Al, Ga |
| Level in the gap | \(E_D\), 45 meV below \(E_C\) | \(E_A\), 45 meV above \(E_V\) |
| Mobile carrier added | electron, into \(E_C\) | hole, into \(E_V\) |
| Fixed ion left behind | \(\text{P}^+\), positive | \(\text{B}^-\), negative |
| Majority carrier | electrons, \(n \approx N_D\) | holes, \(p \approx N_A\) |
| Minority carrier | holes, \(p = n_i^2/N_D\) | electrons, \(n = n_i^2/N_A\) |
| Mobility of the majority | 1350 cm\(^2\)/V·s (lightly doped) | 480 cm\(^2\)/V·s (lightly doped) |
Two warnings about vocabulary, both of which cost marks every year. First, n-type material is not negatively charged and p-type material is not positively charged. Both are electrically neutral, because every mobile carrier is matched by a fixed ion of opposite sign. The letters n and p refer to the sign of the majority carrier, nothing more. Second, doping does not change \(n_i\), which depends only on \(E_g\), \(T\) and the effective masses. It changes \(n\) and \(p\) individually, in opposite directions, leaving their product alone — which is the subject of the last section.
6 Majority Carriers, Minority Carriers and Neutrality
In any doped sample the two carrier populations differ by orders of magnitude. The abundant species is the majority carrier and the scarce one the minority carrier, and the labels attach to the material rather than to the particle: electrons are majority carriers in n-type silicon and minority carriers in p-type.
The imbalance is dramatic. Silicon doped with \(10^{16}\ \text{cm}^{-3}\) donors holds \(10^{16}\) electrons and, as the next section shows, \(10^{4}\) holes per cubic centimetre — a ratio of \(10^{12}\), or one hole for every million million electrons. It would be reasonable to conclude that the minority carriers can be ignored. That conclusion is wrong, and expensively so. A bipolar transistor's collector current is a minority-carrier current: electrons injected from the emitter into the p-type base are minority carriers there, and the whole device is a scheme for getting them across the base before they recombine. A diode's reverse saturation current is entirely a minority-carrier effect. Minority carriers are few but they are where the interesting physics happens, which is why Chapter 4 spends a section on their lifetime.
The constraint that ties the populations to the doping is charge neutrality. A macroscopic piece of semiconductor cannot support a net charge density — any imbalance sets up an electric field that redistributes carriers in picoseconds until it is cancelled. Counting all four charged species, mobile and fixed:
Positive charges on the left — holes and ionised donors — balance negative charges on the right — electrons and ionised acceptors. Assuming complete ionisation, \(N_D^+ = N_D\) and \(N_A^- = N_A\).
Combining neutrality with the mass-action law \(np = n_i^2\) gives two equations in two unknowns. Eliminating \(p = n_i^2/n\) yields a quadratic whose positive root is
The exact result, valid for any combination of donors and acceptors. Its p-type twin follows by exchanging \(n \leftrightarrow p\) and \(N_D \leftrightarrow N_A\).
Three limits cover almost every practical case. When \(N_D - N_A \gg n_i\), the square root is dominated by the first term and \(n \approx N_D - N_A\): the material is n-type and the net doping is what counts. When \(N_A - N_D \gg n_i\), the mirror result gives \(p \approx N_A - N_D\). When the two dopings are equal, the net vanishes and \(n = p = n_i\): the sample is compensated and behaves, in carrier concentration, exactly like intrinsic material.
Compensation is not a curiosity. Every bipolar transistor is built by it: an n-type collector region is counter-doped with acceptors to make the p-type base, and that base is counter-doped again with a heavier donor concentration to make the n-type emitter. The dopants are not removed at any stage, they are outnumbered. This matters for mobility, because a compensated sample with \(N_D = 10^{17}\) and \(N_A = 9\times10^{16}\) has the same net carrier count as one doped with \(10^{16}\) donors alone, but nearly twenty times as many ionised scattering centres, so its mobility — and its conductivity — is much lower. Chapter 4 quantifies that.
7 The Mass-Action Law and Its Consequences
The first section derived \(np = n_i^2\) from the balance of generation and recombination. The name is borrowed from chemistry, where the equilibrium constant of a reaction fixes the product of the reactant concentrations; here the reaction is a bond breaking and reforming, and \(n_i^2\) plays the role of the equilibrium constant. Three conditions are attached, and every one of them is violated somewhere in this course, so they are worth stating plainly:
- Thermal equilibrium. No applied bias, no illumination, no injection. A forward-biased junction has \(np > n_i^2\) near the junction and an illuminated photodiode has \(np \gg n_i^2\); in those situations the law fails and is replaced by the quasi-Fermi levels of Chapter 6.
- Non-degenerate doping. Below about \(10^{18}\ \text{cm}^{-3}\) in silicon. Above that the Fermi level enters a band, Boltzmann statistics no longer approximate Fermi-Dirac, and the product falls below \(n_i^2\).
- One stated temperature. \(n_i^2\) changes by \(10^{6}\) between 200 K and 400 K, so an answer without a temperature is not an answer.
Within those conditions the law is exact and remarkably useful, because it converts a difficult quantity into an easy one. Minority-carrier concentrations are far too small to measure directly, but the majority concentration is just the doping, and one division gives the other.
Take silicon at 300 K with \(n_i = 1.0\times10^{10}\ \text{cm}^{-3}\), so \(n_i^2 = 1.0\times10^{20}\ \text{cm}^{-6}\).
(a) \(N_D = 10^{16}\ \text{cm}^{-3}\) phosphorus. Since \(N_D \gg n_i\), \(n \approx N_D = 1.0\times10^{16}\ \text{cm}^{-3}\), and \(p = n_i^2/n = 10^{20}/10^{16} = 1.0\times10^{4}\ \text{cm}^{-3}\). Doping has raised \(n\) by a factor of \(10^{6}\) and depressed \(p\) by the same factor, leaving the product untouched. The ratio \(n/p\) is \(10^{12}\).
(b) \(N_A = 5\times10^{16}\ \text{cm}^{-3}\) boron. \(p \approx 5.0\times10^{16}\ \text{cm}^{-3}\) and \(n = 10^{20}/5\times10^{16} = 2.0\times10^{3}\ \text{cm}^{-3}\). Two thousand electrons per cubic centimetre — and yet this is the concentration that sets the reverse saturation current of a junction made from this material.
(c) Compensated: \(N_D = 10^{17}\), \(N_A = 3\times10^{16}\ \text{cm}^{-3}\). The net donor concentration is \(7\times10^{16}\ \text{cm}^{-3}\), which still greatly exceeds \(n_i\), so \(n \approx 7.0\times10^{16}\ \text{cm}^{-3}\) and \(p = 10^{20}/7\times10^{16} = 1.43\times10^{3}\ \text{cm}^{-3}\). The material is n-type even though it contains \(3\times10^{16}\) acceptors, and its mobility will be that of material doped to \(1.3\times10^{17}\) total impurities, not \(7\times10^{16}\).
Check the exact formula on the least favourable case, (a): \(n = 5\times10^{15} + \sqrt{(5\times10^{15})^2 + 10^{20}} = 1.00000\times10^{16}\). The approximation is good to eleven decimal places, which is why nobody uses the quadratic in practice.
Figure 2.3 shows why the whole scheme has temperature limits at both ends. Below about 150 K the donors begin to retain their electrons — \(kT\) falls below \(E_C - E_D\) — and \(n\) collapses towards zero. This is carrier freeze-out, and it is why ordinary silicon circuits stop working in cryogenic apparatus. Above about 550 K, \(n_i\) has risen to a substantial fraction of \(N_D\); by 648 K it equals \(10^{16}\ \text{cm}^{-3}\) and the doping has been swamped. The material is then intrinsic again, junctions cease to rectify, and the device fails. Between the two lies the extrinsic region, a plateau over which \(n = N_D\) almost exactly and the carrier concentration is set by the manufacturer rather than by the weather. Every device in this course operates there, and every rating on a data sheet is a statement about staying inside it.
8 Summary and Key Results
| Quantity or law | Expression | Value or worked figure |
|---|---|---|
| Intrinsic concentration | \(n_i = \sqrt{N_C N_V}\,e^{-E_g/2kT}\) | \(1.0\times10^{10}\) cm\(^{-3}\) for Si; \(2.4\times10^{13}\) for Ge; \(2.1\times10^{6}\) for GaAs |
| Intrinsic condition | \(n = p = n_i\) | Pairs are created and destroyed together, so the two counts cannot differ |
| Temperature dependence | \(n_i \propto T^{3/2} e^{-E_g/2kT}\) | Doubles every 8.6 K near 300 K; \(\times 528\) from 300 K to 400 K |
| Donor level | \(E_C - E_D \approx 13.6\,(m^*/m_0)/\varepsilon_r^2\) eV | 45 meV for P, 54 for As, 39 for Sb; complete ionisation above about 150 K |
| Acceptor level | \(E_A - E_V\) small for the same screening reason | 45 meV for B, 67 for Al, 72 for Ga in silicon |
| Charge neutrality | \(p + N_D^{+} = n + N_A^{-}\) | The crystal stays neutral; n-type and p-type are not charged materials |
| Exact carrier count | \(n = \tfrac{1}{2}(N_D-N_A) + \sqrt{\tfrac{1}{4}(N_D-N_A)^2 + n_i^2}\) | Reduces to \(n \approx N_D - N_A\) whenever the net doping exceeds \(n_i\) |
| Mass-action law | \(np = n_i^2\) in thermal equilibrium | \(N_D = 10^{16} \Rightarrow p = 10^{4}\) cm\(^{-3}\); \(N_A = 5\times10^{16} \Rightarrow n = 2\times10^{3}\) cm\(^{-3}\) |
| Operating window | Freeze-out – extrinsic – intrinsic | For \(N_D = 10^{16}\): frozen below 150 K, flat to about 550 K, swamped at 648 K |
9 Common Mistakes
Each donor contributes one mobile electron and one fixed positive ion, so the crystal remains exactly neutral; the same is true of acceptors with the signs reversed. The letters refer to the sign of the majority carrier and to nothing else. Charge neutrality is not an approximation but a constraint enforced in picoseconds by the field any imbalance would create, and it is the equation you write first in almost every problem in Part 1.
The law is a statement of thermal equilibrium. Forward-bias a junction and the injected minority carriers push \(np\) above \(n_i^2\) near the junction by the factor \(e^{qV/kT}\), which at 0.6 V is \(1.2\times10^{10}\). Illuminate a sample and the same thing happens. Applying the mass-action law to a working diode gives a minority concentration wrong by ten orders of magnitude. Use it to find the equilibrium starting point, then let the bias move the populations away from it.
In compensated material the dopants cancel in the carrier count and add in the scattering. With \(N_D = 10^{17}\) and \(N_A = 3\times10^{16}\), the electron concentration is \(7\times10^{16}\) cm\(^{-3}\), the difference; but the mobility is set by the total ionised impurity concentration \(1.3\times10^{17}\) cm\(^{-3}\), the sum. Two different questions, two different arithmetic operations, and confusing them makes the conductivity wrong by a factor of two or more.
10 Chapter Review
1. A silicon sample at 300 K is doped with \(2\times10^{17}\) cm\(^{-3}\) arsenic. Find \(n\) and \(p\), state which are majority and minority carriers, and give the ratio between them.
Arsenic is pentavalent, so it is a donor and the material is n-type. Its level lies 54 meV below \(E_C\), well within reach of \(kT = 25.9\) meV, so assume complete ionisation: \(n \approx N_D = 2.0\times10^{17}\) cm\(^{-3}\). Electrons are the majority carriers. From \(np = n_i^2\) with \(n_i = 1.0\times10^{10}\), \(p = 1.0\times10^{20}/2.0\times10^{17} = 5.0\times10^{2}\) cm\(^{-3}\); holes are the minority carriers. The ratio \(n/p = 4.0\times10^{14}\). Note that \(2\times10^{17}\) is approaching the degenerate limit of about \(1.4\times10^{18}\) cm\(^{-3}\), so the mass-action law is still valid here but would not be an order of magnitude higher.
2. Explain, without using the word "hole", how a p-type sample carries current. Then explain why the hole description is preferred.
Every bond in the crystal is complete except at the acceptor sites, where one electron is missing. Under an applied field, a valence electron adjacent to an incomplete bond is pushed against the field direction and hops into the vacancy, completing that bond and leaving its own site incomplete. The next electron along repeats the move. Net charge transport therefore occurs by successive short displacements of valence electrons, all of them moving against the field. The hole description is preferred because tracking \(4\times10^{22}\) valence electrons per cubic centimetre is intractable, whereas tracking the \(10^{16}\) vacancies is easy, and the vacancy's motion is exactly reproduced by a fictitious particle of charge \(+q\), effective mass \(0.56\,m_0\) and mobility 480 cm\(^2\)/V·s. The two descriptions give identical currents; the second has \(10^{6}\) times fewer objects to count.
3. A sample contains \(N_D = 4\times10^{16}\) and \(N_A = 6\times10^{16}\) cm\(^{-3}\). Determine its type and both carrier concentrations at 300 K.
Acceptors exceed donors, so the net doping is \(N_A - N_D = 2\times10^{16}\) cm\(^{-3}\) of acceptors and the material is p-type. Since \(2\times10^{16} \gg n_i = 10^{10}\), the square root in the exact expression is dominated by the doping term and \(p \approx 2.0\times10^{16}\) cm\(^{-3}\). Then \(n = n_i^2/p = 1.0\times10^{20}/2.0\times10^{16} = 5.0\times10^{3}\) cm\(^{-3}\). The point to notice is that the sample contains \(10^{17}\) impurity atoms in total but behaves, in carrier count, like one containing \(2\times10^{16}\); its mobility, however, is degraded as though it were doped to \(10^{17}\), because every one of those ions scatters.
4. Why does the exponent in the expression for \(n_i\) contain \(E_g/2\) rather than \(E_g\)?
The probability that a bond of energy \(E_g\) is broken carries the Boltzmann factor \(\exp(-E_g/kT)\), and each breakage produces one electron and one hole, so the product \(np\) is proportional to \(\exp(-E_g/kT)\). In intrinsic material \(n = p = n_i\), so \(n_i^2 \propto \exp(-E_g/kT)\) and \(n_i \propto \exp(-E_g/2kT)\). The half arises from taking the square root of a product, not from any electron needing only half the gap energy: no electron ever occupies an energy inside the gap, because there are no states there.
5. A silicon device doped at \(10^{16}\) cm\(^{-3}\) is specified for −55 °C to +125 °C. Relate both limits to Figure 2.3, and say what would change if the device were germanium.
The cold limit, −55 °C = 218 K, sits comfortably above the freeze-out knee near 150 K, where donors begin to retain their electrons; below about 100 K only 76 per cent of donors are ionised and the resistivity climbs steeply. The hot limit, 125 °C = 398 K, is where \(n_i = 5\times10^{12}\) cm\(^{-3}\), still only \(5\times10^{-4}\) of the doping, so the material is safely extrinsic; the practical constraint at that end is not loss of doping but leakage current, which scales with \(n_i^2\) and has risen by a factor of \(2.8\times10^{5}\) since 300 K. For germanium the whole picture shifts down in temperature: with the same doping, \(n_i\) reaches \(10^{16}\) cm\(^{-3}\) at only 464 K (191 °C) against silicon's 648 K, and leakage becomes intolerable well before that, which is why germanium parts are rated to about 75–100 °C.