Semiconductor Electronics

Written and reviewed by Akanksha Tiwari, Physics & Chemistry faculty at UniDrill (M.Sc. Physics, Banaras Hindu University · 8 years teaching Class XI–XII science). Based on NCERT Physics Class XII, Chapter 14.

How can a crystal act as a one-way valve?

Before the late 1940s, controlling an electric current meant a vacuum tube: a glass bulb with a heated filament that boiled electrons off into empty space. Tubes were bulky, fragile, power hungry and needed about 100 V. Then physicists found that a carefully prepared crystal could do the same job with the charge carriers moving inside the solid. A modern phone chip packs billions of such devices. Everything starts with one structure, the p–n junction, which is a crystal that lets current through in one direction only.

Metals, insulators and semiconductors

Solids are first sorted by how well they conduct. The resistivity ranges tell the story:

The reason lies in energy bands. In an isolated atom electrons sit at sharp energy levels. In a crystal, the levels of billions of neighbouring atoms merge into broad bands. The outermost filled band is the valence band (top edge $E_v$). The next one up, normally empty at low temperature, is the conduction band (bottom edge $E_c$). The energy gap $E_g = E_c - E_v$ is the energy an electron needs to jump from one to the other and become free to carry current.

In a metal there is no gap to cross, because the bands overlap or the conduction band is only partly full. In an insulator the gap is so wide that heat cannot lift electrons across it. A semiconductor lies between: at 0 K it is an insulator, but at room temperature a few electrons gain enough thermal energy to jump the small gap. Tin is an interesting contrast. Grey tin has $E_g = 0$ and behaves like a metal, while silicon and germanium, in the same group as carbon with the same crystal structure as diamond, have gaps of 1.1 eV and 0.7 eV, and diamond itself has 5.4 eV. That ordering, $E_g(C) > E_g(Si) > E_g(Ge) > E_g(Sn)$, is all about how tightly bound the outer electrons are.

Worked example: which light can free an electron? What is the longest wavelength of light that can lift an electron across the band gap of Si (1.1 eV), Ge (0.7 eV) and diamond (5.4 eV)? Use $hc = 1240 \text{ eV nm}$.

Visible light (400 to 700 nm) can therefore free electrons in silicon and germanium, but passes right through diamond, which is why it is transparent. The common wrong step: thinking longer wavelengths are more effective. A longer wavelength means a lower photon energy, so only wavelengths up to $\lambda_{\max}$ can free an electron.

Intrinsic semiconductors: pure silicon and germanium

Each Si or Ge atom has four valence electrons and shares one with each of four neighbours, so every bond is a pair of shared electrons in a diamond-like lattice. At 0 K all the bonds are intact and nothing can move: the crystal is an insulator.

At a higher temperature, heat breaks a few bonds. An electron breaks free and wanders through the crystal (it is now in the conduction band), and the broken bond leaves a vacancy called a hole. A neighbouring bond electron can hop into the hole, which moves the vacancy the other way. The hole therefore acts like a particle of charge $+e$ that drifts along the field, in the direction opposite to the electrons. Every broken bond makes one electron and one hole, so in a pure crystal

$$n_e = n_h = n_i, \qquad \text{and the total current is } I = I_e + I_h$$

where $n_i$ is the intrinsic carrier concentration. Electrons and holes also meet and recombine. At any temperature the rate of generation equals the rate of recombination, so $n_i$ is steady. Because raising the temperature breaks more bonds, $n_i$ rises quickly with $T$, which is why a semiconductor's resistivity falls as it heats up. A metal behaves in the opposite way: it has a fixed number of free electrons, and heating only increases collisions, so its resistivity rises.

Pause and predict: you heat a copper wire and a silicon rod from 300 K to 400 K. Which one conducts better afterwards?

Check your answer: the silicon rod conducts better, because heat creates more electron-hole pairs. The copper wire conducts worse, because its number of carriers stays the same while collisions increase.

Doping: the pinch of impurity that changes everything

Pure silicon has so few carriers that it conducts poorly. The cure is doping: replacing about one atom in a million with an atom of similar size from group 13 or 15. Doped crystals are extrinsic semiconductors and come in two flavours.

The loose fifth electron of a donor needs only about 0.05 eV to be freed in Si (about 0.01 eV in Ge), far less than the 1.1 eV gap. At room temperature the thermal energy $kT$ is about 0.026 eV, so nearly every donor releases its electron, and one dopant atom produces one conduction electron. Silicon alone would need 1.1 eV. This is why a few parts per million of phosphorus raises the conductivity thousands of times.

Neither type is charged. A donor atom that gives up an electron becomes a fixed positive ion, so for each free electron there is a fixed positive core, and the crystal as a whole stays neutral. The same balance holds for p-type with fixed negative acceptor ions.

The law of mass action. At a given temperature, whatever the doping, the product of the electron and hole concentrations is fixed:

$$n_e n_h = n_i^2$$

Add donors and $n_e$ shoots up, but extra electrons meet and recombine with holes, so $n_h$ goes down to keep the product constant.

Worked example: minority carriers in n-type silicon. Pure Si has $n_i = 1.5 \times 10^{16} \text{ m}^{-3}$. It is doped with arsenic so that $n_e \approx 5 \times 10^{22} \text{ m}^{-3}$ (about 1 part in a million of the $5 \times 10^{28}$ atoms per m³). Find $n_h$.

Electrons outnumber holes by roughly $10^{13}$ to 1. The common wrong step: writing $n_h = n_i$ because the crystal was originally intrinsic. Doping changes both numbers, and only the product stays at $n_i^2$.

Beyond the textbook. For the exam: a hole is a vacancy that behaves like a particle of charge $+e$. In reality: this is a better description than it sounds. Hall effect measurements on p-type samples show positive charge carriers, just as the model predicts, and in the band picture the hole is the absence of an electron near the top of the valence band. Calling it a particle is a mathematical simplification of what a very large number of electrons do together. Why it matters: it is why a p-type crystal can be treated exactly like a crystal with positive carriers.

The p–n junction

Take a single crystal and dope one half p-type and the other half n-type. The place where they meet is a p–n junction. Two things happen immediately.

Holes are plentiful on the p side and scarce on the n side, and electrons the other way round. Carriers therefore diffuse, holes from p to n and electrons from n to p, simply because they move from a crowded region to an empty one. This diffusion current is made up of majority carriers.

But each carrier that crosses leaves a fixed ion behind. Electrons leaving the n side expose positive donor ions there, and holes leaving the p side expose negative acceptor ions. A thin layer on each side of the junction, about 0.1 µm thick, ends up with no mobile carriers at all, just fixed ions. This is the depletion region. The positive ions on the n side and negative ions on the p side set up an electric field pointing from n to p, which opposes further diffusion and pushes minority carriers back across the other way (the drift current). The potential difference across the layer is the barrier potential $V_0$, about 0.7 V for Si.

The process stops when drift balances diffusion, so at equilibrium the net current is zero. The field is not small: about 0.7 V across 0.1 µm is a field of the order of $10^6 \text{ V m}^{-1}$.

Pause and predict: you connect a voltmeter across an unbiased p–n junction diode. Does it read the barrier potential of 0.7 V?

Check your answer: no, it reads zero. The metal contacts at the two ends of the diode set up their own potential differences that exactly cancel $V_0$. If the voltmeter could read it, the diode would be a free battery.

Forward and reverse bias

Connecting a battery across the junction changes the barrier.

In forward bias the lowered barrier lets many majority carriers cross, and they become minority carriers on the other side. In reverse bias, majority carriers are driven away from the junction, and only the few minority carriers that the field sweeps across contribute. This reverse saturation current does not depend on the applied voltage, because the supply of minority carriers is limited. It does depend on temperature, since heat makes more minority carriers.

The V–I curve in words. In forward bias the current is almost zero until the voltage reaches the cut-in (threshold) voltage, about 0.7 V for Si and 0.2 V for Ge, and then rises steeply. In reverse bias the current is a tiny constant (microamperes) until, at a large enough reverse voltage called the breakdown voltage, it suddenly increases. The curve is not a straight line, so a diode does not obey Ohm's law. Its resistance at a point is described by the dynamic resistance

$$r_d = \dfrac{\Delta V}{\Delta I}$$

which is small in forward bias (a few ohms) and very large in reverse bias.

Worked example: a diode in a circuit. An ideal diode is connected in series with a 6 V battery and a 3 kΩ resistor. Find the current when the diode is forward biased, and when the battery is reversed. Then repeat the forward case for a real silicon diode.

The common wrong step: applying $V = IR$ across the diode with a fixed resistance. A diode's resistance changes along the curve, so you treat it as a switch (or a 0.7 V drop) and apply Ohm's law only to the resistor.

Worked example: dynamic resistance. A silicon diode's forward voltage changes from 0.70 V to 0.72 V, and the current changes from 10 mA to 20 mA. Find $r_d$, and compare it with $V/I$ at the first point.

The common wrong step: dividing $V$ by $I$ and calling it the dynamic resistance. That ratio, 70 Ω, is the resistance at that point. The dynamic resistance is the slope of the curve, 2 Ω, and for a diode the two are very different.

Beyond the textbook. For the exam: a Si diode starts to conduct at about 0.7 V and a Ge diode at about 0.2 V. In reality: the forward current grows exponentially with voltage, by roughly a factor of ten for every 60 to 100 mV, so there is no sharp "switch-on" point. The cut-in voltage is where the current becomes large enough to notice on a graph. Why it matters: it is why the voltage drop across a conducting silicon diode stays close to 0.7 V even as the current changes by a large factor.

Rectifiers: turning ac into dc

A rectifier uses a diode's one-way behaviour to change alternating current into current that flows in one direction.

Half-wave rectifier. One diode in series with the load RL. During the half cycle when the diode is forward biased, the load carries current. During the other half cycle the diode is reverse biased and the output is zero. The output is a series of pulses, one per input cycle, so the output frequency equals the input frequency (50 Hz in, 50 Hz out).

Full-wave rectifier (centre-tap). A transformer with a centre-tapped secondary feeds two diodes $D_1$ and $D_2$. When end A is positive with respect to the centre tap, $D_1$ conducts and $D_2$ is reverse biased. In the next half cycle B is positive and $D_2$ conducts. Current passes through the load in the same direction both times, so there are two pulses per input cycle and the output frequency is twice the input (50 Hz in, 100 Hz out).

The capacitor filter. The output of either rectifier is pulsating dc, not steady dc. Connect a capacitor C in parallel with the load. During each pulse the capacitor charges to the peak voltage. Between pulses it supplies the load and discharges slowly, with a time constant $\tau = R_L C$. The larger $R_L C$ compared with the gap between pulses, the smaller the ripple. Filters smooth the ripple but do not change its frequency.

Worked example: ripple. A full-wave rectifier (input 50 Hz) feeds a 1 kΩ load with a 100 µF capacitor. Estimate the fall in voltage between pulses, and compare it with a half-wave rectifier.

The full-wave circuit gives roughly half the ripple of the half-wave circuit for the same filter, because its pulses come twice as often. The common wrong step: saying the capacitor doubles the frequency. The ripple frequency is fixed by the rectifier type, not by the filter.

Exam traps

Formula sheet

60-second revision

Where this chapter connects

• Previous: Nuclei closes the modern physics block. This chapter turns from the nucleus to the electrons in a solid, where quantised energy levels again decide what the system can do.
• Related: Alternating Current supplies the ac input, the transformer and the 50 Hz mains that a rectifier works on. Current Electricity gives resistivity, conductivity and the temperature dependence that separate metals from semiconductors.
• Link back: Dual Nature of Radiation and Matter shows how photon energy $h\nu$ compares with a band gap, the idea behind light creating electron-hole pairs.