Semiconductor Electronics

Warm a copper wire from $27^{\circ }$C to $127^{\circ }$C and its resistance rises by about $40\%$. Warm a silicon wafer over the same range and its resistance falls by a factor of a few hundred. Why do the two go opposite ways?

Because they change different things.

In a metal the number of free electrons is fixed by the chemistry and cannot change. Heating only makes the lattice vibrate harder, so the electrons scatter more often, the relaxation time falls and the resistance climbs. Modest, linear, unsurprising.

In a semiconductor the carriers must first be created, by lifting an electron across the energy gap. The number available follows a Boltzmann factor:

$n_i\propto T^{3/2}{e}^{-E_g/2kT}$

For silicon, $E_g=1.1$ eV, and $kT$ is $25.9$ meV at $300$ K but $34.5$ meV at $400$ K. The exponent falls from $21.3$ to $16.0$, so the carrier concentration is multiplied by

$e^{5.3}\times {\left(\frac{400}{300}\right)}^{3/2}\approx 200\times 1.54\approx 300$

Three hundred times more carriers. Increased scattering is utterly swamped, and the resistance collapses. Everything else in this chapter is built on the same exponential.

1. Bands, gaps and carrier concentration

Bringing atoms together spreads their discrete levels into bands. What matters is the gap between the highest filled band (valence) and the lowest empty one (conduction).

A conductor has overlapping bands, so carriers exist at any temperature. An insulator has a gap of several electronvolts, which thermal energy of $0.026$ eV cannot bridge. A semiconductor's gap of about $1$ eV is small enough that a tiny but useful fraction of electrons make it across.

Conductivity counts both carriers, and their mobilities differ:

$\sigma =e\left(n_e{\mu }_e+n_h{\mu }_h\right)$

with ${\mu }_e$ roughly two to three times ${\mu }_h$, because a hole moves only by successive electron hops.

Illustration 1

Intrinsic silicon has $n_i=1.5\times 10^{16}$ m${}^{-3}$ at $300$ K, with ${\mu }_e=0.135$ and ${\mu }_h=0.048$ m${}^2$ V${}^{-1}$ s${}^{-1}$. Find its conductivity and resistivity.

$\sigma =\left(1.6\times 10^{-19}\right)\left(1.5\times 10^{16}\right)\left(0.135+0.048\right)$

$=\left(2.4\times 10^{-3}\right)\left(0.183\right)=4.39\times 10^{-4}$ S m${}^{-1}$

$\rho =\frac{1}{\sigma }=2278\Omega$ m

Compare copper's $1.7\times 10^{-8}\Omega$ m. Intrinsic silicon is over a hundred billion times worse — which is exactly why it is useless until it is doped.

2. Doping and the mass-action law

Adding a pentavalent atom to silicon contributes one loosely bound extra electron: n-type. A trivalent atom leaves a vacancy: p-type. One dopant atom in $10^6$ raises the conductivity by orders of magnitude.

Two points are constantly examined. First, the crystal stays electrically neutral — a donor atom supplies an electron but also a fixed positive ion core, so no net charge appears anywhere. Second, doping does not merely add majority carriers; it suppresses the minority ones, because the extra electrons increase the recombination rate. The product is conserved:

$n_e{n}_h=n_i^2$

which holds for intrinsic and doped material alike, and is the fastest route to any minority-carrier question.

Illustration 2

Silicon with $n_i=1.5\times 10^{16}$ m${}^{-3}$ is doped to give $n_e=5\times 10^{22}$ m${}^{-3}$. Find the hole concentration.

$n_h=\frac{n_i^2}{n_e}=\frac{{\left(1.5\times 10^{16}\right)}^2}{5\times 10^{22}}=\frac{2.25\times 10^{32}}{5\times 10^{22}}$

$n_h=4.5\times 10^9$ m${}^{-3}$

Holes have been suppressed by a factor of three million. Doping is not simply addition — the majority carriers actively mop up the minority ones, which is why a doped semiconductor conducts by essentially one carrier type.

3. The p-n junction and its barrier

Join p and n material and electrons diffuse across into the p side while holes diffuse the other way. Each departure leaves behind an immobile ion, so a layer of fixed charge builds up: negative on the p side, positive on the n side.

That charge produces a field opposing further diffusion, and equilibrium arrives when the drift current of minority carriers exactly cancels the diffusion current of majority carriers. The region is now stripped of mobile carriers — the depletion layer — and carries a barrier potential of about $0.3$ V in germanium and $0.7$ V in silicon.

Illustration 3

Explain why the reverse saturation current of a diode is almost independent of the applied voltage but strongly dependent on temperature.

Reverse current is carried by minority carriers, which are swept across the junction by the field. Every minority carrier that reaches the depletion layer is collected, so raising the voltage cannot increase the current — it is limited by how many carriers exist, not by how hard they are pushed.

Their number, however, comes from thermal generation across the gap, and so follows $e^{-E_g/2kT}$.

A rule of thumb is that reverse current roughly doubles every $10$ K. This is why silicon, with its larger gap, is preferred over germanium for anything that must work warm.

4. Biasing, the diode equation and dynamic resistance

Forward bias lowers the barrier and narrows the depletion layer, so current rises steeply once the applied voltage approaches the barrier height. Reverse bias does the opposite. Quantitatively,

$I=I_0\left(e^{eV/\eta kT}-1\right)$

with $kT/e=26$ mV at room temperature. Because the current is exponential, a diode has no single resistance. Two must be distinguished:

$R_{static}=\frac{V}{I},r_{dynamic}=\frac{dV}{dI}\approx \frac{26\text{mV}}{I\left(\text{mA}\right)}\Omega$

The dynamic resistance is what matters for small signals, and it falls as the current rises.

Illustration 4

A silicon diode carries $2$ mA at $0.70$ V. Find its static and dynamic resistances.

$R_{static}=\frac{0.70}{2\times 10^{-3}}=350\Omega$

$r_{dynamic}\approx \frac{26}{2}=13\Omega$

A factor of nearly thirty between them. Quoting a single resistance for a diode is meaningless, and which one a question wants is decided by whether the signal is large or small.

5. Breakdown: Zener against avalanche

Push the reverse bias far enough and the current rises abruptly. Two quite different mechanisms produce that, and Advanced questions distinguish them.

Zener breakdown happens in heavily doped junctions, where the depletion layer is only about $10$ nm wide. Even a few volts across such a thin layer gives a field near $10^8$ V m${}^{-1}$, strong enough to tear electrons directly out of covalent bonds. It dominates below about $5$ V, and its breakdown voltage falls as temperature rises, because warm bonds break more easily.

Avalanche breakdown happens in lightly doped junctions with a wide depletion layer. A minority carrier accelerates over a long distance, gains enough energy to knock another electron free, and the two then liberate four, and so on. It dominates above about $6$ V, and its breakdown voltage rises with temperature, because increased lattice vibration shortens the mean free path so a stronger field is needed.

Between $5$ and $6$ V both operate at once and their opposite temperature coefficients partly cancel. That is why diodes rated near $5.6$ V are the most thermally stable and are the standard choice for voltage references.

Neither mechanism damages the diode. What destroys it is heat, so the only real design constraint is the power rating.

Illustration 5

A diode breaks down at $3.3$ V. Identify the mechanism, and state what happens to the breakdown voltage when the device warms up.

At $3.3$ V, well below $5$ V, the mechanism is Zener breakdown by direct field ionisation.

Its temperature coefficient is negative, so warming the device lowers the breakdown voltage slightly.

A $12$ V device would be avalanche instead, with the opposite sign — which is why a designer needing thermal stability chooses neither extreme but a device near $5.6$ V.

6. Rectification and Zener regulation

A single diode gives half-wave rectification: one half-cycle passes, the other is blocked. Two diodes with a centre-tapped transformer, or four in a bridge, give full-wave.

Full-wave doubles the ripple frequency, which is what makes it so much easier to smooth: the filter capacitor has half as long to discharge between peaks.

A Zener diode is deliberately operated in reverse breakdown, where the voltage across it is almost constant over a wide range of current. As a regulator with a series resistance,

$R_S=\frac{V_{in}-V_Z}{I_Z+I_L}$

and the design must keep $I_Z$ above a minimum for regulation and below the power limit.

Illustration 6

A $9$ V supply is to be regulated to $5$ V for a load drawing $20$ mA, using a Zener that needs at least $5$ mA. Find the series resistance and the Zener power dissipation when the load is disconnected.

$R_S=\frac{9-5}{25\times 10^{-3}}=160\Omega$

With the load removed, all the current goes through the Zener:

$I_Z=\frac{4}{160}=25$ mA, so $P=5\times 0.025=0.125$ W

The worst case for the Zener is no load at all, which is the scenario a regulator must be rated for and the one students most often forget to check.

7. Optoelectronic devices

Three devices use the junction with light, and the direction of bias differs in each.

An LED is forward biased. Electrons and holes recombine across the junction and emit a photon of energy roughly $E_g$, so

$\lambda \approx \frac{1240}{E_g\left(\text{eV}\right)}\text{nm}$

Visible light needs $E_g$ between about $1.8$ and $3.1$ eV, which is why silicon at $1.1$ eV cannot make an LED and gallium compounds are used instead.

A photodiode is reverse biased, which seems perverse until you see why: the dark current is tiny, so the fractional change caused by illumination is large and easily measured. In forward bias the photo-generated current would be lost in a much larger dark current.

A solar cell has no external bias at all. The junction's own field separates the photo-generated pairs and drives them through the load.

Illustration 7

Find the band gap needed for an LED emitting at $650$ nm, and explain why silicon cannot be used.

$E_g=\frac{1240}{650}=1.91$ eV

Silicon's gap is $1.1$ eV, which would emit at $1127$ nm — deep in the infrared and invisible.

Silicon is also an indirect-gap material, so recombination there tends to release heat rather than light. Both reasons rule it out for light emission, though it remains ideal for detection.

8. The transistor as an amplifier

A junction transistor has a thin, lightly doped base between a heavily doped emitter and a moderately doped collector. That asymmetry is the whole design: most carriers injected from the emitter cross the base without recombining and are collected.

$\alpha =\frac{I_C}{I_E},\beta =\frac{I_C}{I_B},\beta =\frac{\alpha }{1-\alpha }$

Since $\alpha$ is typically $0.99$, $\beta$ is around $100$. In the common-emitter configuration the voltage gain is

$A_V=-\beta \frac{R_L}{r_i}$

the minus sign indicating the $180^{\circ }$ phase inversion between input and output.

The circuit imposes $I_C=\frac{V_{CC}-V_{CE}}{R_C}$, a straight load line drawn across the family of output characteristics. Where it crosses the curve for the chosen base current is the operating point. Bias it near the middle of the line for an amplifier; drive it between the two ends, saturation and cut-off, to use the transistor as a switch.

Illustration 8

A common-emitter stage has $V_{CC}=10$ V, $R_C=2\text{k}\Omega$ and $\beta =100$. Find the operating point for a base current of $20\mu$A, and the base current that would saturate the transistor.

$I_C=\beta I_B=100\times 20\times 10^{-6}=2$ mA

$V_{CE}=10-\left(2\times 10^{-3}\right)\left(2000\right)=6$ V

Comfortably in the active region. Saturation means $V_{CE}\approx 0$, needing $I_C=\frac{10}{2000}=5$ mA, so

$I_B=\frac{5}{100}=50\mu$A

Beyond $50\mu$A nothing further happens. The transistor is saturated and the extra base current is wasted, which is exactly the regime a switching circuit is designed to sit in.

Illustration 9

A transistor has $\alpha =0.98$. Find $\beta$, and the collector and base currents for an emitter current of $5$ mA.

$\beta =\frac{0.98}{0.02}=49$

$I_C=0.98\times 5=4.9$ mA

$I_B=5-4.9=0.1$ mA

Note how sensitive $\beta$ is to $\alpha$. Raising $\alpha$ from $0.98$ to $0.99$ doubles $\beta$ to $99$, which is why base thickness is controlled so tightly in manufacture.

Illustration 10

A common-emitter amplifier has $\beta =100$, load resistance $2\text{k}\Omega$ and input resistance $1\text{k}\Omega$. Find the voltage and power gains.

$A_V=-100\times \frac{2000}{1000}=-200$

Current gain is $\beta =100$, so power gain $=\left|A_V\right|\times \beta =200\times 100=2\times 10^4$

The output is inverted. A rising input current increases the collector current, which increases the drop across the load and therefore lowers the collector voltage.

9. Logic gates

The three basic gates are OR, AND and NOT. Two composite gates matter more in practice:

NAND and NOR are each universal — any logic function whatever can be built from copies of either one alone. A NOT is a NAND with its inputs tied together; an AND is a NAND followed by that NOT; an OR follows from De Morgan's relations.

$\overline{A\cdot B}=\bar{A}+\bar{B},\overline{A+B}=\bar{A}\cdot \bar{B}$

This is why integrated circuits are manufactured from a single repeated gate type rather than from an assortment.

Illustration 11

Show how to build an OR gate using only NAND gates.

Invert each input with a NAND used as a NOT: $\bar{A}$ and $\bar{B}$.

Feed both into a third NAND: the output is $\overline{\bar{A}\cdot \bar{B}}$.

By De Morgan's relation, $\overline{\bar{A}\cdot \bar{B}}=A+B$, which is OR.

Three NAND gates suffice. The same construction underlies every logic family, and it is why the NAND gate is the single most manufactured component in history.

Illustration 12

Two inputs $A$ and $B$ are fed to a NAND gate whose output goes to a NOT gate. Write the truth table and identify the function.

NAND gives $0$ only when both inputs are $1$. Inverting that gives $1$ only when both are $1$.

$A=0,B=0\to 0$; $A=0,B=1\to 0$; $A=1,B=0\to 0$; $A=1,B=1\to 1$

This is an AND gate, which is exactly how AND is constructed in practice — as a NAND followed by an inverter, since NAND is the cheaper primitive.

Summary

• Metals worsen on heating (fixed carriers, more scattering); semiconductors improve exponentially, since carriers must first be created.
• $n_i\propto T^{3/2}{e}^{-E_g/2kT}$: silicon gains about $300$ times more carriers between $300$ K and $400$ K.
• Gaps: conductor overlapping, semiconductor about $1$ eV, insulator above $3$ eV. $\sigma =e\left(n_e{\mu }_e+n_h{\mu }_h\right)$, with ${\mu }_e>{\mu }_h$.
• Doping keeps the crystal neutral and suppresses minority carriers; $n_e{n}_h=n_i^2$ always.
• The junction barrier comes from immobile ions left behind by diffusion; $0.3$ V in Ge, $0.7$ V in Si.
• Equilibrium is drift cancelling diffusion, not an absence of current.
• Reverse saturation current is set by minority-carrier supply, so it is voltage-independent but roughly doubles every $10$ K.
• $I=I_0\left(e^{eV/\eta kT}-1\right)$; $kT/e=26$ mV; $r_{dynamic}\approx \frac{26}{I\left(\text{mA}\right)}\Omega$, quite unlike $V/I$.
• Zener breakdown below $5$ V (thin layer, field ionisation, negative coefficient); avalanche above $6$ V (wide layer, impact ionisation, positive coefficient).
• Devices near $5.6$ V are the most thermally stable, because the two coefficients cancel.
• Half-wave: ripple at $f$, $I_{dc}=I_m/\pi$, efficiency $40.6\%$. Full-wave: ripple at $2f$, $I_{dc}=2I_m/\pi$, efficiency $81.2\%$.
• Zener regulator $R_S=\frac{V_{in}-V_Z}{I_Z+I_L}$; the worst case for the Zener is no load.
• LED forward biased with $\lambda \approx 1240/E_g$ nm; photodiode reverse biased for a large fractional change; solar cell unbiased.
• Load line $I_C=\frac{V_{CC}-V_{CE}}{R_C}$ sets the operating point: mid-line for an amplifier, saturation or cut-off for a switch.
• $\beta =\frac{\alpha }{1-\alpha }$, so $\alpha =0.99$ gives $\beta =99$; $A_V=-\beta R_L/r_i$, with the sign meaning phase inversion.
• NAND and NOR are each universal; De Morgan's relations convert between them.