Chapter 03 · Modern Physics · Full book
Semiconductor Electronics
Core intuition: Chapter Roadmap
This chapter explains how a slab of nearly-pure silicon—an indifferent
conductor—can be coaxed into doing arithmetic, turning alternating current
into direct current, and emitting light. The thread runs from the energy-band
picture of solids to the devices that built the information age.
- § The Band Picture Why some solids conduct, some insulate, and a narrow band of materials sit in between. The single number that decides it: the energy gap \(E_g\).
- § Intrinsic Semiconductors Pure silicon and germanium: thermally generated electron–hole pairs, the strange but useful notion of a mobile “hole,” and conductivity that rises with temperature.
- § Extrinsic Semiconductors Doping: how one impurity atom in ten million multiplies the carrier population a million-fold, and the mass-action law \(np=n_i^2\) that governs it.
- § The p–n Junction What happens when n-type meets p-type: the depletion region, the built-in barrier, and the one-way valve that results.
- § The Junction Diode Forward and reverse bias, the \(I\)–\(V\) curve, the knee voltage, dynamic resistance, and the Shockley equation behind them.
- § Rectifiers Turning AC into DC—half-wave and full-wave—with their ripple factors and efficiencies derived from scratch, then smoothed by a capacitor filter.
- § Special Diodes The Zener voltage regulator, the photodiode, the LED, and the solar cell: four devices, one junction, run in four directions.
- § Logic Gates The digital layer: OR, AND, NOT, NAND, NOR, and the remarkable fact that one gate alone can build any computer.
- § Beyond the Syllabus: The Transistor A brief, flagged enrichment look at the three-terminal device that amplifies and switches—kept outside the exam floor, included for the curious and for olympiad readers.
Perplexing Questions
Perplexing questions
Before we begin, let these questions sit with you. Do not reach for
formulas. You will return to each one—with full answers—before this
chapter ends.
- The Impurity Paradox: Pure silicon is a stubborn near-insulator. Add just one foreign atom for every ten million silicon atoms—a contamination so slight no chemist would call the crystal impure—and its conductivity leaps by a factor of a million. How can so little change so much?
- The Particle That Is an Absence: In a semiconductor, current is carried partly by “holes”—the absence of an electron. Yet a hole moves, carries a definite positive charge, and responds to fields like a real particle. How can nothing behave like something?
- The One-Way Street in a Symmetric Crystal: A diode passes current freely in one direction and blocks it in the other. But it is a single crystal of silicon, p-type on one side and n-type on the other—both made of the same atoms. Where does the one-way asymmetry come from, if not from any asymmetry in the material itself?
- Why Colour Needs Chemistry, Not Just Voltage: A red LED and a blue LED both glow when you pass current through them. Yet you cannot turn a red LED blue by raising the voltage—you must change the material. And no one has ever built a silicon LED that glows visibly at all. Why is colour fixed by the substance, and why is silicon mute?
- The Diode That Refuses to Let Voltage Rise: A Zener diode, run backwards, clamps the voltage across it to a fixed value no matter how the supply fluctuates. A single two-terminal device holds an output steady against a wandering input. What is it doing that an ordinary resistor cannot?
- One Gate to Build Them All: From copies of a single kind of logic gate—and nothing else—you can assemble any logical function whatsoever, and therefore an entire computer. Which gate has this power, and why does it suffice when a plain AND or OR does not?
Each question targets an idea that, once seen clearly, makes the rest of the
chapter fall into place.
By the end, every one of them will be transparent.
By the end, every one of them will be transparent.
Why Some Solids Conduct: The Band Picture
◈ Bands & doping
live benchSet the gap and watch who crosses it: a conductor with no gap at all, silicon at 1.1 eV letting a trickle through, an insulator that never does — then dope it and watch the carrier count jump.
In Volume 1 and in the chapter on current electricity you treated a conductor's resistance as a measured fact: copper conducts, glass does not, and that was that. We now ask the deeper question—why—and the answer turns out to control everything that follows.
A single atom has sharply defined energy levels: an electron may occupy the \(1s\), \(2s\), \(2p,\dots\) levels and nothing in between. Bring \(N\) atoms together into a crystal, with \(N\sim 10^{23}\), and the Pauli exclusion principle forbids any two electrons from sharing the same quantum state. Each sharp atomic level therefore splits into \(N\) closely-spaced levels. Because \(N\) is astronomically large, these levels merge into a near-continuous energy band.
Two bands decide a material's electrical character:
- the valence band, the highest band that is filled (or nearly filled) with electrons at absolute zero; and
- the conduction band, the next band up, normally empty.
Between them may lie a forbidden energy gap \(E_g\)—a range of energies no electron in the crystal is allowed to have.
The Single Number That Sorts All Solids
An electron can carry current only if it can accelerate—only if there is an empty nearby state for it to move into. A completely full band carries no net current: for every electron drifting one way there is another drifting the opposite way, and the two cancel. Conduction requires either a partly-filled band or electrons promoted across a gap into an empty band. This single criterion sorts every solid into three classes.
- Conductors (metals) The valence and conduction bands overlap, or the valence band is only half full. There is no gap to cross; electrons find empty states immediately above them. \(E_g \approx 0\). Resistivity is tiny, \(\rho \sim 10^{-8}\,\Omega\,\text{m}\), and—crucially—it rises with temperature, because hotter lattice vibrations scatter the already-abundant carriers more often.
- Insulators A wide gap, \(E_g \gtrsim 3\,\text{eV}\) (diamond: \(5.5\,\text{eV}\)). At room temperature the thermal energy \(k_BT \approx 0.026\,\text{eV}\) is far too small to lift any appreciable number of electrons across it. The conduction band stays empty; \(\rho \sim 10^{11}\text{--}10^{19}\,\Omega\,\text{m}\).
- Semiconductors A narrow gap, \(E_g \sim 1\,\text{eV}\) (silicon: \(1.10\,\text{eV}\); germanium: \(0.66\,\text{eV}\)). At \(0\,\text{K}\) a semiconductor is a perfect insulator—its valence band is full and its conduction band empty. But the gap is small enough that at room temperature a tiny fraction of electrons is thermally kicked across it, leaving an equal number of vacancies behind. This double population is what makes the material useful, and it is why a semiconductor's resistivity falls with temperature—the opposite of a metal.
Core intuition: A Full Band Is as Useless as an Empty One
The deepest idea in this chapter is counter-intuitive: a band crammed full of
electrons conducts nothing. Current is net motion, and in a full band
every leftward electron is matched by a rightward one. Only an unfilled
band carries current. That is why metals (partly-filled band) conduct, why
insulators (full valence band, empty conduction band, wide gap) do not, and
why a semiconductor must be warmed or doped—both of which
work by leaving empty states in the valence band and putting a few electrons
in the conduction band.
The temperature behaviour is the cleanest experimental fingerprint: heat a metal and its resistance rises; heat a semiconductor and its resistance falls. A thermometer built from a bead of semiconductor—a thermistor—exploits exactly this.
◇ Figure 1
Fig 1. The energy-band criterion. A conductor has overlapping (or partly
filled) bands; a semiconductor a narrow gap that thermal energy can bridge in
part; an insulator a gap too wide to cross at room temperature.
✓ Checkpoint · Bands and Classification
- Why does a completely filled energy band carry no electric current?For every electron drifting one way there is one drifting the opposite way, so the net current is zero. Conduction needs an unfilled band with empty states to move into.
- A solid has \(E_g = 0.5\,\)eV. Semiconductor or insulator?Semiconductor — the gap is narrow enough (tens of \(k_BT\)) for a thermal carrier population at room temperature.
- When heated, which has rising resistance: a metal or a semiconductor?The metal: its carrier number is fixed, so more lattice scattering raises resistance. The semiconductor's resistance falls (carriers multiply).
- Diamond (\(E_g = 5.5\,\)eV) is both transparent and insulating. Explain both with one number.The wide gap: visible photons (\(1.8\)–\(3.1\,\)eV) cannot excite an electron across \(5.5\,\)eV, so light passes (transparent); and almost no electrons are thermally excited, so it does not conduct (insulator).
Misconception alert: A Semiconductor Is Not “Half a Conductor”
The name suggests something midway between a metal and an insulator, conducting
“a bit.” The truer picture is an insulator with a small gap: at \(T=0\) a
pure semiconductor conducts nothing at all. What sets it apart is not a modest fixed
conductivity but the sign of its temperature response. Heating a metal
raises its resistance (more lattice vibration, more scattering); heating a
semiconductor lowers its resistance, because the exponential \(e^{-E_g/2k_BT}\)
floods the conduction band with carriers far faster than scattering can hinder them.
A material whose resistance falls as it warms is the unmistakable fingerprint of a
semiconductor.
Worked Examples
Example 1 · Why a \(6\,\mathrm{eV}\) gap means “insulator”
Estimate how many times more thermally excited carriers a semiconductor
(\(E_g=1.1\,\mathrm{eV}\), silicon) has than an insulator (\(E_g=6.0\,\mathrm{eV}\),
diamond) at \(T=300\,\mathrm{K}\), using the Boltzmann factor \(e^{-E_g/2k_BT}\).
Solution.
Given: \(E_g^{\mathrm{Si}}=1.1\,\mathrm{eV}\), \(E_g^{\mathrm{dia}}=6.0\,\mathrm{eV}\), \(k_BT=0.0259\,\mathrm{eV}\) at \(300\,\mathrm{K}\).
Find: the ratio of carrier populations.
Setup: The fraction of electrons excited across a gap scales as \(e^{-E_g/2k_BT}\), so the ratio is \(e^{-(E_g^{\mathrm{Si}}-E_g^{\mathrm{dia}})/2k_BT}\).
Solve: \[ \frac{n_{\mathrm{Si}}}{n_{\mathrm{dia}}} = e^{(6.0-1.1)/(2\times0.0259)} = e^{94.7}\approx 1.4\times10^{41}. \] Answer: \(\boxed{\sim10^{41}}\) — an astronomically larger carrier population.
Check: Silicon's own excited fraction is only \(e^{-1.1/0.0517}\approx5.8\times10^{-10}\) (tiny, but non-zero); diamond's is \(\sim10^{-51}\) (effectively zero). Because \(E_g\) sits in the exponent, a factor-of-five change in the gap becomes forty-one orders of magnitude in carriers. ✓
Find: the ratio of carrier populations.
Setup: The fraction of electrons excited across a gap scales as \(e^{-E_g/2k_BT}\), so the ratio is \(e^{-(E_g^{\mathrm{Si}}-E_g^{\mathrm{dia}})/2k_BT}\).
Solve: \[ \frac{n_{\mathrm{Si}}}{n_{\mathrm{dia}}} = e^{(6.0-1.1)/(2\times0.0259)} = e^{94.7}\approx 1.4\times10^{41}. \] Answer: \(\boxed{\sim10^{41}}\) — an astronomically larger carrier population.
Check: Silicon's own excited fraction is only \(e^{-1.1/0.0517}\approx5.8\times10^{-10}\) (tiny, but non-zero); diamond's is \(\sim10^{-51}\) (effectively zero). Because \(E_g\) sits in the exponent, a factor-of-five change in the gap becomes forty-one orders of magnitude in carriers. ✓
Example 2 · Heating silicon from \(300\) to \(400\,\mathrm{K}\)
By what factor does the intrinsic carrier population of silicon (\(E_g=1.1\,\mathrm{eV}\)) rise when it is heated from \(300\,\mathrm{K}\) to \(400\,\mathrm{K}\)? Use \(e^{-E_g/2k_BT}\).
Solution.
Given: \(E_g=1.1\,\mathrm{eV}\); \(k_BT=0.0259\,\mathrm{eV}\) at \(300\,\mathrm{K}\) and \(0.0345\,\mathrm{eV}\) at \(400\,\mathrm{K}\).
Find: the ratio \(n(400)/n(300)\).
Setup: With \(n\propto e^{-E_g/2k_BT}\), take the ratio of the two Boltzmann factors.
Solve: \[ \frac{n(400)}{n(300)}=\frac{e^{-1.1/(2\times0.0345)}}{e^{-1.1/(2\times0.0259)}} =\frac{e^{-15.96}}{e^{-21.28}}=e^{5.32}\approx 2.0\times10^{2}. \] Answer: \(\boxed{\approx 200\times}\).
Check: A mere \(100\,\mathrm{K}\) rise multiplies the carrier count two-hundredfold — exactly why an uncooled semiconductor's conductivity is so temperature-sensitive and why real devices need thermal management. ✓
Find: the ratio \(n(400)/n(300)\).
Setup: With \(n\propto e^{-E_g/2k_BT}\), take the ratio of the two Boltzmann factors.
Solve: \[ \frac{n(400)}{n(300)}=\frac{e^{-1.1/(2\times0.0345)}}{e^{-1.1/(2\times0.0259)}} =\frac{e^{-15.96}}{e^{-21.28}}=e^{5.32}\approx 2.0\times10^{2}. \] Answer: \(\boxed{\approx 200\times}\).
Check: A mere \(100\,\mathrm{K}\) rise multiplies the carrier count two-hundredfold — exactly why an uncooled semiconductor's conductivity is so temperature-sensitive and why real devices need thermal management. ✓
Intrinsic Semiconductors: Electrons and Holes
◈ Conductivity against temperature
live benchHeat the sample and read σ on a log scale: a semiconductor climbs as pairs are generated, a metal sags as the lattice scatters harder — the sign of the slope is the whole diagnosis.
A pure semiconductor—silicon or germanium with no deliberate impurity—is called intrinsic. Each atom has four valence electrons and bonds covalently to four neighbours in a tetrahedral lattice. At absolute zero every valence electron is locked in a bond; the valence band is full, the conduction band empty, and the crystal is an insulator.
Thermal Generation and the Birth of the Hole
Raise the temperature and lattice vibrations occasionally supply enough energy (\(\geq E_g\)) to break a covalent bond. The liberated electron is promoted to the conduction band, where it roams freely. But it leaves behind an empty bond—a hole.
The hole is the chapter's first genuinely new idea. A neighbouring bound electron can hop into the vacancy, which fills the old hole but creates a new one where the electron came from. As electrons hop one way, the vacancy migrates the other way. Rather than track \(10^{23}\) bound electrons, we track the single vacancy and call it a particle: a hole, with charge \(+e\) and its own effective mass, drifting in the direction a positive charge would.
In an intrinsic semiconductor electrons and holes are created strictly in pairs, so \[ n_e = n_h = n_i, \] where \(n_i\) is the intrinsic carrier concentration. At \(300\,\)K, \(n_i(\text{Si}) \approx 1.5\times10^{16}\,\text{m}^{-3}\) and \(n_i(\text{Ge}) \approx 2.4\times10^{19}\,\text{m}^{-3}\)—germanium's narrower gap gives it some \(1600\) times more carriers. Both numbers are minute beside the \(\sim 10^{29}\,\text{m}^{-3}\) atoms present: only about one bond in \(10^{13}\) is broken at any instant in silicon.
Birth of the formula: Conductivity of a Semiconductor
Both carriers respond to an applied field \(\vec{E}\). Electrons drift with
mobility \(\mu_e\) (drift speed per unit field) and holes with mobility
\(\mu_h\). From the drift-current expression of the current-electricity
chapter, the current density contributed by carriers of concentration \(n\),
charge magnitude \(e\), and mobility \(\mu\) is \(j = ne\mu E\). Electrons and
holes drift in opposite directions but, carrying opposite charge, their
currents add:
\[
j = j_e + j_h = e\,(n_e \mu_e + n_h \mu_h)\,E.
\]
Comparing with \(j = \sigma E\) gives the conductivity
\[
\boxed{\;\sigma = e\,(n_e \mu_e + n_h \mu_h)\;}
\]
For an intrinsic sample, \(n_e = n_h = n_i\), so
\[
\sigma_i = e\,n_i\,(\mu_e + \mu_h),
\qquad \rho_i = \frac{1}{\sigma_i}.
\]
Worked anchor (Si, \(300\,\)K): with \(\mu_e = 0.135\) and
\(\mu_h = 0.048\,\text{m}^2\text{V}^{-1}\text{s}^{-1}\),
\[
\sigma_i = (1.6\times10^{-19})(1.5\times10^{16})(0.135+0.048)
\approx 4.4\times10^{-4}\,\text{S m}^{-1},
\]
so \(\rho_i \approx 2.3\times10^{3}\,\Omega\,\text{m}\)—about eleven orders
of magnitude more resistive than copper, yet eight orders less than glass.
A semiconductor genuinely sits between.
Temperature law. The carrier concentration follows
\(n_i \propto T^{3/2}\,e^{-E_g/2k_BT}\), so to a good approximation
\[
\sigma \;\propto\; e^{-E_g/2k_BT}.
\]
The exponential dominates: a modest rise in \(T\) multiplies \(n_i\) rapidly,
and the conductivity climbs even though the mobilities fall slightly. This is
the algebra-level rule NEET and BITSAT need: semiconductor
conductivity rises (resistivity falls) roughly exponentially with
temperature, set by the gap \(E_g\).
Misconception alert: A Hole Is Not a Real Positive Particle
A hole is a bookkeeping device, not a fundamental particle. There is no
positively-charged object sitting in the bond. What physically moves is a
succession of electrons hopping into vacancies. We could, in principle,
describe the whole process by tracking every bound electron—but that is
hopeless arithmetic. Treating the lone vacancy as a particle of charge \(+e\)
gives identical predictions with trivial effort.
Two consequences students miss:
- A hole's “mass” (its effective mass) is a property of the band, not of any real object, and differs from the free-electron mass.
- Holes generally have lower mobility than electrons (\(\mu_h \lt \mu_e\)): the hopping mechanism is clumsier than free-electron drift. This is why electrons usually dominate conduction even when \(n_e = n_h\).
Worked Examples
Example 3 · In intrinsic silicon, who carries more current?
Intrinsic silicon has equal electron and hole populations, \(n_e=n_h=n_i\), but the electron mobility \(\mu_e=0.135\,\mathrm{m^2/(V\,s)}\) exceeds the hole mobility \(\mu_h=0.048\,\mathrm{m^2/(V\,s)}\). What fraction of the current is carried by electrons?
Solution.
Given: \(n_e=n_h\), \(\mu_e=0.135\), \(\mu_h=0.048\,\mathrm{m^2/(V\,s)}\).
Find: the electron share of the current.
Setup: Drift current is \(J=en\mu E\); with equal populations the shares are in the ratio of mobilities, so the electron fraction is \(\mu_e/(\mu_e+\mu_h)\).
Solve: \[ \frac{\mu_e}{\mu_e+\mu_h}=\frac{0.135}{0.135+0.048}=\frac{0.135}{0.183}=0.738. \] Answer: \(\boxed{\text{electrons }73.8\%,\ \text{holes }26.2\%}\).
Check: Equal numbers do not mean equal currents: the nimbler electrons carry the larger share, and \(73.8\%+26.2\%=100\%\) as it must. The hole, though it is the absence of an electron, drifts and conducts like a genuine positive carrier — just a less mobile one. ✓
Find: the electron share of the current.
Setup: Drift current is \(J=en\mu E\); with equal populations the shares are in the ratio of mobilities, so the electron fraction is \(\mu_e/(\mu_e+\mu_h)\).
Solve: \[ \frac{\mu_e}{\mu_e+\mu_h}=\frac{0.135}{0.135+0.048}=\frac{0.135}{0.183}=0.738. \] Answer: \(\boxed{\text{electrons }73.8\%,\ \text{holes }26.2\%}\).
Check: Equal numbers do not mean equal currents: the nimbler electrons carry the larger share, and \(73.8\%+26.2\%=100\%\) as it must. The hole, though it is the absence of an electron, drifts and conducts like a genuine positive carrier — just a less mobile one. ✓
Example 4 · How rare is a thermal pair?
A \(1\,\mathrm{cm^3}\) block of pure silicon at \(300\,\mathrm{K}\) has \(n_i=1.5\times10^{16}\,\mathrm{m^{-3}}\). How many electron–hole pairs does it hold, and what fraction of its \(\sim5\times10^{28}\,\mathrm{m^{-3}}\) atoms are ionised?
Solution.
Given: \(n_i=1.5\times10^{16}\,\mathrm{m^{-3}}\), \(V=1\,\mathrm{cm^3}=10^{-6}\,\mathrm{m^3}\), atom density \(5\times10^{28}\,\mathrm{m^{-3}}\).
Find: pair count and ionised fraction.
Setup: Pairs \(=n_iV\); ionised fraction \(=n_i/n_{\mathrm{Si}}\).
Solve: \[ n_iV=(1.5\times10^{16})(10^{-6})=1.5\times10^{10}\ \text{pairs},\qquad \frac{n_i}{n_{\mathrm{Si}}}=\frac{1.5\times10^{16}}{5\times10^{28}}=3\times10^{-13}. \] Answer: \(\boxed{1.5\times10^{10}\ \text{pairs};\ 3\times10^{-13}\ \text{ionised}}\).
Check: Only three atoms in ten trillion contribute a carrier — pure silicon is almost an insulator. This vanishing number is the backdrop against which the next section's doping produces its startling effect. ✓
Find: pair count and ionised fraction.
Setup: Pairs \(=n_iV\); ionised fraction \(=n_i/n_{\mathrm{Si}}\).
Solve: \[ n_iV=(1.5\times10^{16})(10^{-6})=1.5\times10^{10}\ \text{pairs},\qquad \frac{n_i}{n_{\mathrm{Si}}}=\frac{1.5\times10^{16}}{5\times10^{28}}=3\times10^{-13}. \] Answer: \(\boxed{1.5\times10^{10}\ \text{pairs};\ 3\times10^{-13}\ \text{ionised}}\).
Check: Only three atoms in ten trillion contribute a carrier — pure silicon is almost an insulator. This vanishing number is the backdrop against which the next section's doping produces its startling effect. ✓
Extrinsic Semiconductors: The Power of Doping
Intrinsic silicon is too resistive and far too temperature-sensitive to build reliable devices. The fix—doping—is the central trick of the whole industry: replace a vanishingly small fraction of silicon atoms with impurity atoms of a neighbouring group, and the carrier population is transformed.
n-type: Donors
Silicon sits in group 14 with four valence electrons. Substitute a pentavalent atom—phosphorus, arsenic, antimony (group 15)—into a lattice site. Four of its five valence electrons form the usual covalent bonds; the fifth is left over, bound only weakly to its parent and easily freed (binding energy \(\sim 0.05\,\text{eV}\), comparable to \(k_BT\)). At room temperature essentially every such impurity donates its spare electron to the conduction band.
The pentavalent impurity is a donor; the material is n-type (negative majority carriers). If \(N_D\) donors per unit volume are added, with \(N_D \gg n_i\), then almost every conduction electron comes from a donor: \[ n_e \approx N_D. \] Electrons are the majority carriers; the few thermally-generated holes are the minority carriers.
p-type: Acceptors
Now substitute a trivalent atom—boron, aluminium, indium (group 13). It has only three valence electrons, so one of the four bonds to its silicon neighbours is left short of an electron: a ready-made hole. A neighbouring bound electron readily fills it, propagating a hole through the lattice. The trivalent impurity accepts an electron and is called an acceptor; the material is p-type (positive majority carriers). With \(N_A\) acceptors per unit volume, \[ n_h \approx N_A, \] holes are the majority carriers, and the sparse electrons are the minority.
Birth of the formula: The Mass-Action Law
Doping changes the balance of electrons and holes, but a beautiful
constraint links them. In thermal equilibrium, electron–hole pairs are
generated (rate \(g\), set only by temperature and \(E_g\)) and recombine (rate
\(\propto n_e n_h\), since an electron must meet a hole). At equilibrium
generation equals recombination:
\[
g = r\,n_e n_h
\quad\Longrightarrow\quad
n_e n_h = \frac{g}{r} = \text{const}(T).
\]
The product depends only on temperature, not on doping. Evaluating it for the
intrinsic case (\(n_e = n_h = n_i\)) fixes the constant:
\[
\boxed{\;n_e\,n_h = n_i^{2}\;}
\]
the mass-action law. It holds for any doping. Raise the
electron population by doping n-type and the hole population must fall
in exact proportion—added electrons recombine away the holes until the
product is restored.
Minority-carrier corollary. In n-type material with \(n_e \approx
N_D\),
\[
n_h = \frac{n_i^{2}}{N_D}.
\]
Worked anchor (Si, \(300\,\)K, \(N_D = 10^{22}\,\text{m}^{-3}\)):
\[
n_h = \frac{(1.5\times10^{16})^{2}}{10^{22}}
= 2.25\times10^{10}\,\text{m}^{-3}.
\]
The doping raised electrons from \(1.5\times10^{16}\) to \(10^{22}\) (a factor of
\(\sim 7\times10^{5}\)) and suppressed holes to
\(2.25\times10^{10}\)—now outnumbered by electrons more than \(10^{11}\) to one.
That is how one impurity atom in ten million multiplies conductivity a
millionfold: it is the answer to the chapter's first Perplexing Question.
Core intuition: Doped Silicon Is Still Electrically Neutral
Doping does not charge the crystal. A donor atom contributes one extra
mobile electron, but the donor nucleus carries one extra proton: the
atom as a whole is neutral, and so is the crystal. The mobile electron is free
to wander, but it leaves behind a fixed, positively-charged donor ion
locked in the lattice. n-type silicon is therefore neutral overall—a sea of
mobile electrons over a fixed background of positive donor ions. Symmetrically,
p-type silicon is neutral: mobile holes over fixed negative acceptor ions. We
will need this fixed-ion background in the very next section to understand the
junction.
Misconception alert: n-type Silicon Is Not Negatively Charged
The single most common slip in this chapter. “n-type” refers to the sign of
the mobile majority carriers (negative electrons), not to a net
charge on the material. A block of n-type silicon, sitting on the bench, is
electrically neutral: every mobile electron is balanced by a fixed donor ion.
Likewise p-type silicon is neutral, not positive. If you could net-charge a
semiconductor merely by doping it, doping would be impossible—the Coulomb
energy would forbid it. Keep “type” (a statement about which carrier
dominates) rigorously separate from “charge” (a statement about an
imbalance of total charge, which doping never creates).
◇ Figure 2
Fig 2. Doping. A pentavalent donor (left) contributes a loosely-bound extra
electron; a trivalent acceptor (right) leaves a bond short of an electron — a
mobile hole. Each impurity stays fixed and, with its core charge, keeps the
crystal neutral.
✓ Checkpoint · Intrinsic and Extrinsic Carriers
- In n-type silicon, are electrons or holes the majority carriers, and what valency is the dopant?Electrons are the majority carriers; the dopant is pentavalent (a donor).
- n-type silicon has \(n_e \approx N_D = 1\times10^{22}\,\mathrm{m^{-3}}\) (\(n_i = 1.5\times10^{16}\)). Find the hole concentration.Mass action: \(n_h = n_i^2/N_D = (1.5\times10^{16})^2/10^{22} = 2.25\times10^{10}\,\mathrm{m^{-3}}\).
- Is a block of n-type silicon negatively charged? Explain.No — it is electrically neutral. Each mobile electron is balanced by a fixed positive donor ion; “n-type” names the carrier sign, not a net charge.
Worked Examples
Example 5 · Majority and minority carriers in p-type silicon
Silicon is doped with \(N_A=2\times10^{22}\,\mathrm{m^{-3}}\) acceptors (\(n_i=1.5\times10^{16}\,\mathrm{m^{-3}}\)). Find the hole and electron concentrations and their ratio.
Solution.
Given: \(N_A=2\times10^{22}\), \(n_i=1.5\times10^{16}\,\mathrm{m^{-3}}\).
Find: \(n_h\), \(n_e\), \(n_h/n_e\).
Setup: An acceptor is trivalent — it grabs one bonding electron, leaving a hole. With \(N_A\gg n_i\), \(n_h\approx N_A\); electrons follow from mass action, \(n_e=n_i^2/n_h\).
Solve: \[ n_h\approx2\times10^{22}\,\mathrm{m^{-3}},\quad n_e=\frac{(1.5\times10^{16})^2}{2\times10^{22}}=1.1\times10^{10}\,\mathrm{m^{-3}},\quad \frac{n_h}{n_e}\approx1.8\times10^{12}. \] Answer: \(\boxed{n_h=2\times10^{22},\ n_e=1.1\times10^{10}\,\mathrm{m^{-3}}}\).
Check: The roles simply swap from the n-type case: holes are now the majority, electrons the vanishing minority. The product \(n_en_h=(1.1\times10^{10})(2\times10^{22})=2.25\times10^{32}=n_i^2\) is preserved, exactly as mass action demands. ✓
Find: \(n_h\), \(n_e\), \(n_h/n_e\).
Setup: An acceptor is trivalent — it grabs one bonding electron, leaving a hole. With \(N_A\gg n_i\), \(n_h\approx N_A\); electrons follow from mass action, \(n_e=n_i^2/n_h\).
Solve: \[ n_h\approx2\times10^{22}\,\mathrm{m^{-3}},\quad n_e=\frac{(1.5\times10^{16})^2}{2\times10^{22}}=1.1\times10^{10}\,\mathrm{m^{-3}},\quad \frac{n_h}{n_e}\approx1.8\times10^{12}. \] Answer: \(\boxed{n_h=2\times10^{22},\ n_e=1.1\times10^{10}\,\mathrm{m^{-3}}}\).
Check: The roles simply swap from the n-type case: holes are now the majority, electrons the vanishing minority. The product \(n_en_h=(1.1\times10^{10})(2\times10^{22})=2.25\times10^{32}=n_i^2\) is preserved, exactly as mass action demands. ✓
Example 6 · The impurity paradox (Perplexing Question 1)
Silicon (\(n_i=1.5\times10^{16}\,\mathrm{m^{-3}}\)) is doped with \(N_D=1.5\times10^{22}\,\mathrm{m^{-3}}\) — about one donor per three million silicon atoms. Estimate the factor by which its conductivity rises over the intrinsic value. Take \(\mu_e=0.135\), \(\mu_h=0.048\,\mathrm{m^2/(V\,s)}\).
Solution.
Given: \(N_D=1.5\times10^{22}\), \(n_i=1.5\times10^{16}\,\mathrm{m^{-3}}\), mobilities as above.
Find: \(\sigma_{\mathrm{doped}}/\sigma_{\mathrm{intrinsic}}\).
Setup: Intrinsic \(\sigma_i=en_i(\mu_e+\mu_h)\); doped \(\sigma\approx eN_D\mu_e\) (electrons dominate). The ratio is \(N_D\mu_e/[n_i(\mu_e+\mu_h)]\).
Solve: \[ \frac{\sigma_{\mathrm{doped}}}{\sigma_i}=\frac{N_D\mu_e}{n_i(\mu_e+\mu_h)} =\frac{1.5\times10^{22}}{1.5\times10^{16}}\cdot\frac{0.135}{0.183} =10^{6}\times0.74\approx7.4\times10^{5}. \] Answer: \(\boxed{\sim10^{6}}\) — a millionfold rise.
Check: One foreign atom in three million lifts the conductivity a million-fold. The paradox dissolves once you see that each donor contributes a mobile electron, whereas the pure crystal's atoms are all locked in bonds: a doping fraction of \(3\times10^{-7}\) buys a carrier increase of \(\sim10^{6}\). ✓
Find: \(\sigma_{\mathrm{doped}}/\sigma_{\mathrm{intrinsic}}\).
Setup: Intrinsic \(\sigma_i=en_i(\mu_e+\mu_h)\); doped \(\sigma\approx eN_D\mu_e\) (electrons dominate). The ratio is \(N_D\mu_e/[n_i(\mu_e+\mu_h)]\).
Solve: \[ \frac{\sigma_{\mathrm{doped}}}{\sigma_i}=\frac{N_D\mu_e}{n_i(\mu_e+\mu_h)} =\frac{1.5\times10^{22}}{1.5\times10^{16}}\cdot\frac{0.135}{0.183} =10^{6}\times0.74\approx7.4\times10^{5}. \] Answer: \(\boxed{\sim10^{6}}\) — a millionfold rise.
Check: One foreign atom in three million lifts the conductivity a million-fold. The paradox dissolves once you see that each donor contributes a mobile electron, whereas the pure crystal's atoms are all locked in bonds: a doping fraction of \(3\times10^{-7}\) buys a carrier increase of \(\sim10^{6}\). ✓
Example 7 · Compensated doping
A silicon sample contains both \(N_D=1\times10^{22}\,\mathrm{m^{-3}}\) donors and \(N_A=6\times10^{21}\,\mathrm{m^{-3}}\) acceptors. Is it n- or p-type, and what are the carrier concentrations?
Solution.
Given: \(N_D=1\times10^{22}\), \(N_A=6\times10^{21}\,\mathrm{m^{-3}}\), \(n_i=1.5\times10^{16}\).
Find: type and carrier concentrations.
Setup: Donors and acceptors cancel; the net dopant is \(N_D-N_A\). Here \(N_D\gt N_A\), so the sample is n-type with \(n_e\approx N_D-N_A\).
Solve: \[ n_e\approx N_D-N_A=4\times10^{21}\,\mathrm{m^{-3}},\qquad n_h=\frac{n_i^2}{n_e}=\frac{(1.5\times10^{16})^2}{4\times10^{21}}=5.6\times10^{10}\,\mathrm{m^{-3}}. \] Answer: \(\boxed{\text{n-type};\ n_e\approx4\times10^{21}\,\mathrm{m^{-3}}}\).
Check: Acceptors “mop up” six of every ten donor electrons; only the net excess conducts. Had \(N_A\) exceeded \(N_D\), the same bookkeeping would give a p-type sample — compensation, the very idea behind forming a junction inside one crystal. ✓
Find: type and carrier concentrations.
Setup: Donors and acceptors cancel; the net dopant is \(N_D-N_A\). Here \(N_D\gt N_A\), so the sample is n-type with \(n_e\approx N_D-N_A\).
Solve: \[ n_e\approx N_D-N_A=4\times10^{21}\,\mathrm{m^{-3}},\qquad n_h=\frac{n_i^2}{n_e}=\frac{(1.5\times10^{16})^2}{4\times10^{21}}=5.6\times10^{10}\,\mathrm{m^{-3}}. \] Answer: \(\boxed{\text{n-type};\ n_e\approx4\times10^{21}\,\mathrm{m^{-3}}}\).
Check: Acceptors “mop up” six of every ten donor electrons; only the net excess conducts. Had \(N_A\) exceeded \(N_D\), the same bookkeeping would give a p-type sample — compensation, the very idea behind forming a junction inside one crystal. ✓
Solved examples
Five fully-worked problems from this chapter, free — solution and answer shown in full. The complete set of worked examples is in the full book.
Example 1 · Naming the majority carrier
A silicon crystal is doped with phosphorus to a concentration
\(N_D = 5\times10^{22}\,\text{m}^{-3}\). Name the majority and minority
carriers and state the approximate majority-carrier concentration.
Solution.
Given: Si doped with phosphorus (pentavalent), \(N_D =
5\times10^{22}\,\text{m}^{-3}\).
Find: carrier types and majority concentration.
Setup: A pentavalent donor gives up one electron each; with \(N_D \gg n_i\) almost every conduction electron comes from a donor.
Solve: The material is n-type. Majority carriers are electrons, \(n_e \approx N_D = 5\times10^{22}\,\text{m}^{-3}\); minority carriers are holes.
Answer: \(\boxed{\text{n-type; electrons majority, } n_e \approx 5\times10^{22}\,\text{m}^{-3}}\)
Check: Only the minority holes are \(\sim10^{13}\) times rarer, yet they never vanish—and each fixed donor ion keeps the crystal electrically neutral overall. ✓
Find: carrier types and majority concentration.
Setup: A pentavalent donor gives up one electron each; with \(N_D \gg n_i\) almost every conduction electron comes from a donor.
Solve: The material is n-type. Majority carriers are electrons, \(n_e \approx N_D = 5\times10^{22}\,\text{m}^{-3}\); minority carriers are holes.
Answer: \(\boxed{\text{n-type; electrons majority, } n_e \approx 5\times10^{22}\,\text{m}^{-3}}\)
Check: Only the minority holes are \(\sim10^{13}\) times rarer, yet they never vanish—and each fixed donor ion keeps the crystal electrically neutral overall. ✓
Example 2 · Minority-carrier concentration
For the crystal of the previous example (\(n_i = 1.5\times10^{16}\,
\text{m}^{-3}\) at \(300\,\)K), find the equilibrium hole concentration.
Solution.
Given: \(n_e \approx N_D = 5\times10^{22}\,\text{m}^{-3}\),
\(n_i = 1.5\times10^{16}\,\text{m}^{-3}\).
Find: minority hole concentration \(n_h\).
Setup: mass-action law \(n_e n_h = n_i^2\).
Solve: \[ n_h = \frac{n_i^2}{n_e} = \frac{(1.5\times10^{16})^2}{5\times10^{22}} = 4.5\times10^{9}\,\text{m}^{-3}. \] Answer: \(\boxed{n_h = 4.5\times10^{9}\,\text{m}^{-3}}\)
Check: The product \(n_e n_h=(5\times10^{22})(4.5\times10^{9})=2.25\times10^{32}=n_i^2\) is preserved: doping raised electrons and suppressed holes by the same factor. ✓
Find: minority hole concentration \(n_h\).
Setup: mass-action law \(n_e n_h = n_i^2\).
Solve: \[ n_h = \frac{n_i^2}{n_e} = \frac{(1.5\times10^{16})^2}{5\times10^{22}} = 4.5\times10^{9}\,\text{m}^{-3}. \] Answer: \(\boxed{n_h = 4.5\times10^{9}\,\text{m}^{-3}}\)
Check: The product \(n_e n_h=(5\times10^{22})(4.5\times10^{9})=2.25\times10^{32}=n_i^2\) is preserved: doping raised electrons and suppressed holes by the same factor. ✓
Example 3 · Classifying solids by the gap
Three solids have energy gaps \(E_g = 0\) (bands overlap),
\(1.1\,\text{eV}\), and \(6\,\text{eV}\). Classify each, given
\(k_BT \approx 0.026\,\text{eV}\) at room temperature.
Solution.
Given: \(E_g = 0,\ 1.1,\ 6\,\text{eV}\); \(k_BT = 0.026\,\text{eV}\).
Find: conductor / semiconductor / insulator.
Setup: carriers crossing a gap scale as \(e^{-E_g/2k_BT}\); compare \(E_g\) with \(k_BT\).
Solve: \(E_g=0\): no gap, abundant carriers \(\Rightarrow\) conductor. \(E_g=1.1\,\text{eV}\): \(E_g/k_BT \approx 42\), a small but non-negligible thermal population \(\Rightarrow\) semiconductor. \(E_g=6\,\text{eV}\): \(E_g/k_BT \approx 230\), the exponential is astronomically small \(\Rightarrow\) insulator.
Answer: \(\boxed{\text{conductor, semiconductor, insulator}}\)
Check: Only the middle gap gives a carrier population that is both small and tunable by temperature—the defining trait of a semiconductor, absent in both the conductor and the insulator. ✓
Find: conductor / semiconductor / insulator.
Setup: carriers crossing a gap scale as \(e^{-E_g/2k_BT}\); compare \(E_g\) with \(k_BT\).
Solve: \(E_g=0\): no gap, abundant carriers \(\Rightarrow\) conductor. \(E_g=1.1\,\text{eV}\): \(E_g/k_BT \approx 42\), a small but non-negligible thermal population \(\Rightarrow\) semiconductor. \(E_g=6\,\text{eV}\): \(E_g/k_BT \approx 230\), the exponential is astronomically small \(\Rightarrow\) insulator.
Answer: \(\boxed{\text{conductor, semiconductor, insulator}}\)
Check: Only the middle gap gives a carrier population that is both small and tunable by temperature—the defining trait of a semiconductor, absent in both the conductor and the insulator. ✓
Example 4 · Reading a bias
A p–n junction has its p-side connected to the positive terminal of a battery
and its n-side to the negative terminal. State the bias, and what happens to
the depletion width and the barrier.
Solution.
Given: p-side at \(+\), n-side at \(-\).
Find: bias type and effect.
Setup: an applied field opposing the built-in field is forward bias.
Solve: This is forward bias. The applied field opposes the built-in field, so the barrier lowers and the depletion region narrows; once the applied voltage nears the barrier (\(\sim0.7\,\)V for Si) a large current flows.
Answer: \(\boxed{\text{forward bias; barrier lowers, depletion narrows, large current}}\)
Check: Reverse the battery and the reasoning runs backward: the applied field now aids the built-in field, widening \(W\) and throttling the current to a trickle. ✓
Find: bias type and effect.
Setup: an applied field opposing the built-in field is forward bias.
Solve: This is forward bias. The applied field opposes the built-in field, so the barrier lowers and the depletion region narrows; once the applied voltage nears the barrier (\(\sim0.7\,\)V for Si) a large current flows.
Answer: \(\boxed{\text{forward bias; barrier lowers, depletion narrows, large current}}\)
Check: Reverse the battery and the reasoning runs backward: the applied field now aids the built-in field, widening \(W\) and throttling the current to a trickle. ✓
Example 5 · Diode in series with a resistor
An ideal silicon diode (knee voltage \(0.7\,\)V, zero forward resistance) is in
series with \(R = 100\,\Omega\) across a \(5\,\)V supply, forward-biased. Find the
current.
Solution.
Given: \(V = 5\,\)V, knee \(0.7\,\)V, \(R = 100\,\Omega\).
Find: current \(I\).
Setup: once conducting, the silicon diode holds \(0.7\,\)V; the rest falls across \(R\).
Solve: \[ I = \frac{V - 0.7}{R} = \frac{5 - 0.7}{100} = \frac{4.3}{100} = 43\,\text{mA}. \] Answer: \(\boxed{I = 43\,\text{mA}}\)
Check: The resistor drops \(IR=(43\,\text{mA})(100\,\Omega)=4.3\,\)V, and \(0.7+4.3=5\,\)V closes the loop. ✓
Find: current \(I\).
Setup: once conducting, the silicon diode holds \(0.7\,\)V; the rest falls across \(R\).
Solve: \[ I = \frac{V - 0.7}{R} = \frac{5 - 0.7}{100} = \frac{4.3}{100} = 43\,\text{mA}. \] Answer: \(\boxed{I = 43\,\text{mA}}\)
Check: The resistor drops \(IR=(43\,\text{mA})(100\,\Omega)=4.3\,\)V, and \(0.7+4.3=5\,\)V closes the loop. ✓
Problem bank
Five questions from this chapter’s 50-question bank, free — attempt each one before you reveal the answer. The rest of the bank, and the timed test that draws on all of it, are in the full book.
✎ Free practice questions
- Carrier type:
Silicon is doped with arsenic (pentavalent). Name the resulting type and the majority carrier.n-type; majority carriers are electrons (the donor's fifth electron is freed). Minority carriers are holes. - Minority concentration:
n-type silicon has \(n_e \approx N_D = 1.0\times10^{22}\,\mathrm{m^{-3}}\). Find the hole concentration.Mass action: \(n_h = n_i^2/n_e = (1.5\times10^{16})^2/10^{22} = 2.25\times10^{10}\,\mathrm{m^{-3}}\). - LED colour:
An LED has gap \(E_g = 2.0\,\mathrm{eV}\). Find its emission wavelength.\(\lambda = hc/E_g = 1240/2.0 = 620\,\mathrm{nm}\) (orange-red). - Two diodes in series:
A silicon and a germanium diode, both forward, are in series with \(R = 330\,\Omega\) across \(6\,\)V. Find the current.\(I = (6 - 0.7 - 0.3)/330 = 5.0/330 = 15\,\mathrm{mA}\). - Suppressed minority carriers:
In n-type silicon with \(N_D = 1\times10^{21}\,\mathrm{m^{-3}}\), find the hole concentration and explain why it is far below \(n_i\).\(n_h = n_i^2/N_D = (1.5\times10^{16})^2/10^{21} = 2.25\times10^{11}\, \mathrm{m^{-3}}\), well below \(n_i=1.5\times10^{16}\). The extra electrons from doping recombine away holes until \(n_en_h=n_i^2\) is restored, so the more you raise \(n_e\), the more you suppress \(n_h\).
Chapter test
A paper drawn at random from this chapter's bank. Choose the exam you are training for — the marking scheme, pace and difficulty mix follow the real pattern. Work on paper; when you finish (or the clock runs out), the answers are revealed and you mark yourself honestly.
Pattern
Questions
Clock
§ 4 – 10 · The rest of the chapter
The chapter continues.
You’ve read the opening, the first three theory sections, the opening run of worked examples and five bank questions — all free, with no account. The rest of the chapter is behind the pass.
Still ahead in this chapter
- The p–n Junction
- The Junction Diode
- The Diode as a Rectifier
- Special-Purpose Diodes
- Logic Gates: The Digital Layer
- Beyond the Syllabus: The Transistor
- Common Pitfalls and Exam Strategy
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