Semiconductor Electronics

Take a crystal of $N$ silicon atoms. Each atom’s outermost shell ($n=3$ for Si, $n=4$ for Ge) holds 4 electrons, two s and two p, so the crystal has $4N$ valence electrons in all. That same shell has room for 8 electrons per atom (2 s-states plus 6 p-states), so there are $8N$ available states. As the atoms are brought together to the actual spacing of the Si or Ge lattice, these $8N$ states split into two bands of $4N$ states each, separated by a gap:

where $E_C$ is the lowest level of the conduction band and $E_V$ the highest level of the valence band. The counting now does the explaining: $4N$ electrons exactly fill the $4N$ states of the lower band at absolute zero, leaving the upper band completely empty. A full band cannot carry current, because every state in it is occupied and an applied field has no empty states to push electrons into, and an empty band has no carriers at all. So a perfect Si or Ge crystal at 0 K is an insulator. Only once thermal energy lifts some electrons across $E_g$ do both bands start to conduct: the electrons now in the conduction band, and the vacancies they leave behind in the valence band.

The gap also explains a family trend. Carbon (as diamond), silicon and germanium share the same diamond-like lattice and the same four valence electrons, but those electrons sit in the second, third and fourth shells respectively, progressively further from the nucleus and more loosely held. The gap shrinks accordingly: about 5.4 eV for diamond (an insulator, well above the 3 eV line), 1.1 eV for Si and 0.7 eV for Ge. Tin, one step further down the same group, has $E_g = 0$ and is a metal.

Adding a pentavalent impurity (e.g. phosphorus, arsenic, five valence electrons) to silicon (four valence electrons): four of the impurity’s electrons form covalent bonds with neighbouring silicon atoms, and the fifth is left loosely bound, easily freed into the conduction band. These donor atoms contribute mobile electrons as the majority carriers, holes remain the minority carriers.

Adding a trivalent impurity (e.g. boron, aluminium, three valence electrons) leaves one covalent bond incomplete, a hole that readily accepts an electron from a neighbouring bond. These acceptor atoms contribute mobile holes as the majority carriers, electrons are the minority carriers.

Typical doping is tiny, a few parts per million, and the dopant atom must be close in size to Si or Ge so that it can take a lattice site without distorting the crystal, which is why dopants come from the neighbouring groups 13 and 15 of the periodic table. Yet that trace of impurity raises the conductivity many times over, and the reason is energetic. The fifth electron of a pentavalent donor is held so loosely that freeing it costs only about 0.01 eV in germanium and 0.05 eV in silicon, against the roughly 0.7 eV or 1.1 eV needed to break a covalent bond and lift an electron across the full gap. In the band picture, the donor’s electron sits at a donor level $E_D$ just below the bottom of the conduction band, $E_C$. A trivalent acceptor likewise creates an empty acceptor level $E_A$ just above the top of the valence band, $E_V$, which a valence electron can jump into with very little energy, ionising the acceptor and leaving a hole behind in the valence band.

How small is “very little”? The typical thermal energy per particle at room temperature (300 K) is of order $kT$, with $k$ the Boltzmann constant:

the same 0.026 that appears as the thermal voltage $V_T = kT/e$ in the diode equation of §14.5. A donor ionisation energy of 0.05 eV is only about twice $kT$, so at room temperature almost every donor is ionised; silicon’s 1.1 eV gap is more than 40 times $kT$, so only a minute fraction of its own bonds break. The conduction electrons in n-type material therefore come overwhelmingly from the donors, and their number is set by the doping level, not by the temperature; the small intrinsic contribution rises only weakly as the crystal warms.

There is a second, less obvious effect: doping suppresses the minority carriers. Flooding n-type material with electrons gives every thermally generated hole far more chances of meeting an electron and being destroyed, so the hole concentration drops below its intrinsic value; the same happens to electrons in p-type material. Throughout, the crystal stays electrically neutral overall, since every extra mobile carrier is matched by an oppositely charged, immobile ionised dopant core.

That suppression can be made exact using the generation-recombination balance from the start of this section. At a given temperature, bonds break at some rate $g$ per unit volume that depends on the material and the temperature, not on how many free carriers happen to be around. Recombination, on the other hand, needs an electron to meet a hole, so its rate is proportional to the concentration of each, $\kappa\,n_e n_h$, with $\kappa$ a constant for that material at that temperature. In equilibrium the two rates are equal:

The right-hand side depends only on the material and the temperature, so the product $n_e n_h$ is the same for every sample of that material at that temperature, however it is doped. The undoped crystal is one such sample, with $n_e = n_h = n_i$, so the constant must be $n_i^2$:

This is why pushing one carrier up must push the other down. In n-type material with donor concentration $N_D \gg n_i$, nearly every donor is ionised, so $n_e \approx N_D$, and the hole concentration is forced down to $n_h = n_i^2/N_D$, far below $n_i$. p-type material works the same way with acceptor concentration $N_A$: $n_h \approx N_A$ and $n_e = n_i^2/N_A$.

Two currents compete across a forming junction, and equilibrium is reached when they cancel. The first is the diffusion current just described: majority carriers moving down their concentration gradients, holes from p to n and electrons from n to p. Both motions carry conventional current the same way, from p to n. The second is the drift current. The depletion region’s field, pointing from the positive ions on the n-side toward the negative ions on the p-side, catches any minority carriers that wander into it, electrons on the p-side and holes on the n-side, and sweeps them across: electrons to the n-side, holes to the p-side. That is conventional current from n to p, opposite to the diffusion current.

At first diffusion dominates, because the gradient is steep and the depletion region, with its field, has barely begun to form. But every carrier that diffuses across uncovers another fixed ion, widening the space-charge layers and strengthening the field, so drift grows while diffusion is increasingly held back. The process stops when the two currents are exactly equal and opposite. A junction in equilibrium therefore carries no net current, even though carriers are still crossing in both directions.

By then the depletion region is only about a tenth of a micrometre thick. The n-side, having lost electrons, sits at a higher potential than the p-side, which has gained them. This potential difference, $V_0$, opposes any further flow of electrons from n to p (and of holes from p to n), and so is called the barrier potential. One practical consequence: a junction cannot be made by pressing a p-type slab against an n-type slab. Even a polished surface is rough on a scale far larger than the 2 to 3 Å spacing between atoms, so the contact would act as a discontinuity for the carriers rather than a continuous crystal. Real junctions are made inside a single wafer, by doping part of a p-type wafer with a pentavalent impurity to turn that part n-type.

Connecting the p-side to the positive terminal of a battery and the n-side to the negative terminal applies a field that opposes the junction’s own built-in field. This narrows the depletion region, and once the applied voltage is large enough to substantially cancel the built-in potential, majority carriers flood across the junction in large numbers: current rises steeply with voltage.

Quantitatively, the applied voltage subtracts from the built-in one, so the barrier carriers must climb becomes:

With $V_0\approx0.7\text{ V}$ for a silicon junction, a forward bias of 0.5 V leaves an effective barrier of only 0.2 V. Only carriers in the highest energy levels can surmount a barrier, so a slight forward bias lets few through, while a larger one lowers the barrier further and lets many more through. Those that cross land on the far side as minority carriers, electrons arriving in the p-region and holes in the n-region, a process called minority carrier injection. Their concentration is highest right at the junction edge and falls away with distance, so they keep diffusing into the bulk; this hole diffusion current plus the electron diffusion current is the diode’s forward current, typically milliamperes.

Connecting the p-side to the negative terminal and the n-side to the positive terminal applies a field that reinforces the built-in field. This widens the depletion region further, sweeping out what few free carriers were near the junction. Only a tiny trickle of thermally-generated minority carriers, electrons from the p-side, holes from the n-side, can still cross, giving a small, nearly voltage-independent reverse current.

Here the applied voltage adds to the built-in one:

so the same 0.7 V junction under a 5 V reverse bias faces a 5.7 V barrier, and diffusion across it is shut off almost completely. What remains is the drift current of §14.3.1, minority carriers near the junction swept across by its field. Even a small reverse voltage sweeps every available minority carrier across, so raising the voltage further cannot increase the current: it is limited by how many minority carriers exist, not by the applied voltage, and is microamperes or less. That holds up to a critical breakdown voltage $V_{br}$, at which the reverse current suddenly rises steeply. Unless an external resistance keeps that current below the manufacturer’s rated value, the diode overheats and is destroyed, and the same fate awaits a forward-biased diode pushed past its rated forward current.

That steep rise is why a measured characteristic, plotted on ordinary linear axes, looks almost flat up to a certain voltage and then shoots upward. The voltage beyond which the current rises significantly is called the threshold (or cut-in) voltage: about 0.2 V for germanium and 0.7 V for silicon. The two sides of the curve are so different in scale that measuring them takes two different meters, a milliammeter for the forward current and a microammeter for the reverse current.

Because the characteristic is curved, a diode has no single resistance. The useful quantity is how much the voltage must change to produce a small change in current, the dynamic resistance:

which is the inverse slope of the I-V curve at the operating point. The diode equation shows exactly how it varies. Differentiating $I = I_0\left(e^{V/V_T}-1\right)$ with respect to $V$:

and since the diode equation itself says $I_0e^{V/V_T} = I + I_0$:

In forward bias $I \gg I_0$, so:

Dynamic resistance falls in inverse proportion to the current: double the forward current and $r_d$ halves. In reverse bias, $I$ approaches $-I_0$, so the denominator $I+I_0 = I_0e^{V/V_T}$ shrinks toward zero and $r_d$ grows without limit, the mathematical form of the flat reverse characteristic. Low resistance one way and enormous resistance the other is exactly the property §14.6 puts to work.

A full-wave rectifier puts both halves of the AC cycle to use. The transformer’s secondary coil has an extra connection at its midpoint, a centre tap, so its two ends A and B are always at opposite potentials relative to it: when A is positive with respect to the centre tap, B is negative, and vice versa. The p-sides of two diodes, D1 and D2, connect to the ends A and B; their n-sides are joined together, and the load $R_L$ is connected between that common point and the centre tap.

During the half-cycle when A is positive, D1 is forward biased and conducts while D2 is reverse biased and blocks; during the next half-cycle B is positive, so D2 conducts and D1 blocks. Either way, current passes through $R_L$ in the same direction, so every half-cycle produces an output pulse, and the pulses arrive twice per input cycle. The price is that each diode works from only half of the secondary winding, so the peak output is half the peak voltage across the whole secondary. (A bridge rectifier uses four diodes instead and needs no centre tap.) In any rectifier, each diode’s reverse breakdown voltage (§14.4.2) must comfortably exceed the peak reverse voltage it faces, or it would break down on exactly the half-cycles it is meant to block.

Rectified output is one-directional but still pulses between zero and the peak value $V_m$. To steady it, a large capacitor $C$ is connected across the output, in parallel with the load $R_L$. While the rectified voltage is rising, the capacitor charges up to the peak $V_m$. Once the rectified voltage falls below the capacitor’s voltage, the diodes stop conducting (the capacitor now holds their n-sides above their p-sides), and the capacitor alone supplies the load until the next pulse rises to meet it. Because these circuits filter out the AC ripple, they are called filters.

How far does the voltage sag in that gap? While discharging, the capacitor drives a current $I = V/R_L \approx V_m/R_L$ through the load, as long as its voltage stays close to $V_m$. Over a time $\Delta t$ it therefore loses a charge $\Delta Q = I\,\Delta t$, and since $Q = CV$, its voltage drops by:

Taking $\Delta t$ as the full time between pulses gives a slight overestimate, since the next pulse starts recharging the capacitor a little before its peak. The product $R_LC$, the time constant, sits in the denominator: the larger it is compared with the gap between pulses, the smaller the ripple, which is why filter capacitors are large, and why a well-filtered output sits close to the peak voltage $V_m$. A full-wave rectifier, with pulses twice as often, halves $\Delta t$ and so halves the ripple for the same capacitor.