A diode pushed backwards
The chapter on semiconductors and the p-n junction showed what a diode does in reverse bias. The p side is joined to the negative terminal, and the depletion layer grows wider. The majority carriers are held back, and only the few minority carriers cross. The reverse current is tiny, a few microamperes or less, and it hardly changes as the reverse voltage grows.
Now keep increasing the reverse voltage. At one particular value the picture changes suddenly. The reverse current, which was almost zero, rises very steeply. A little more voltage gives a great deal more current. This is reverse breakdown, and the voltage at which it starts is the breakdown voltage.
Reverse breakdown is the sudden, steep rise of the reverse current of a p-n junction when the reverse voltage reaches the breakdown voltage. Beyond it the current is set by the rest of the circuit, while the voltage across the junction stays almost constant.
A vertical line holds a voltage
Read the reverse part of Fig 1.1 from left to right. Along the vertical line, the current changes from \(-40\,\text{mA}\) to almost zero. The voltage stays at \(-5\,\text{V}\) the whole way. So once the diode is in breakdown, the voltage across it is fixed at the breakdown value, whatever current the circuit sends through it. That single property is what the next lesson uses to hold a voltage steady.
Breakdown is not damage
The word breakdown sounds like destruction, but the junction is not harmed by it. Harm comes only from heat. The electrical power turned into heat in a component is \(P = VI\), the voltage across it times the current through it. The unit on electricity develops this. A diode in breakdown survives as long as \(V_Z I\) stays below the power its makers allow.
An ordinary rectifier diode has a breakdown voltage of hundreds of volts, and it is never meant to reach it. A Zener diode is made to work in breakdown. Its doping is chosen so that it breaks down at a sharp, known voltage \(V_Z\), anywhere from about \(2\,\text{V}\) to \(200\,\text{V}\). Its symbol, in the figures of this chapter, is a diode whose bar has bent ends.
A Zener diode is a p-n junction diode made to operate in reverse breakdown. Connected in reverse bias, it keeps the voltage across it at its Zener voltage \(V_Z\) over a wide range of currents.
A Zener diode with \(V_Z = 6.2\,\text{V}\) may dissipate at most \(0.5\,\text{W}\). What is the largest current it can safely carry in breakdown, in mA?
A Zener diode is in breakdown. The current through it rises from \(10\,\text{mA}\) to \(30\,\text{mA}\). What happens to the voltage across it?
The field inside the depletion layer
Why does the current suddenly grow? The answer lies in the electric field across the depletion layer. A reverse voltage \(V\) spread across a layer of width \(d\) makes a field of about \(E = V/d\), measured in volts per metre. The unit on electricity develops this link between field and voltage. How thick the layer is depends on the doping, and that decides which of two mechanisms acts.
Zener breakdown: electrons torn from their bonds
In a heavily doped junction the depletion layer is very thin, about \(0.1\,\mu\text{m}\) or less. Then even \(5\,\text{V}\) across it gives a field of \(5 \times 10^{7}\,\text{V/m}\). A field this strong pulls valence electrons straight out of their covalent bonds, without any collision. Each freed electron leaves a hole behind, and the field sweeps both away. This is Zener breakdown. In the language of bands, the electron crosses the gap directly from the valence band to the conduction band, a quantum effect called tunnelling.
Avalanche breakdown: one carrier makes many
In a lightly doped junction the depletion layer is wide, a few micrometres, and the field is weaker. It cannot pull electrons out of bonds directly. But a stray minority carrier crossing the wide layer is accelerated by the field between one collision and the next. If it gains more energy than the band gap before it collides, the collision can free an electron from a bond. This is impact ionisation.
Now there are two carriers, and both are accelerated. After the next collision there are four, then eight. The number doubles again and again, like snow gathering into an avalanche on a mountain slope. After twenty such steps one carrier has become \(2^{20}\), about a million. This is avalanche breakdown.
In the depletion layer of a silicon diode the field is \(3 \times 10^{7}\,\text{V/m}\). An electron travels \(0.05\,\mu\text{m}\) between collisions. How much energy does it gain, in eV? Can the collision free a new electron-hole pair?
Across a distance \(s\) in a field \(E\), the potential difference is \(Es = 3 \times 10^{7} \times 0.05 \times 10^{-6} = 1.5\,\text{V}\).
An electron crossing \(1.5\,\text{V}\) gains \(1.5\,\text{eV}\), by the definition of the electron volt.
Silicon's band gap is \(1.1\,\text{eV}\). The electron has more than enough energy to lift a valence electron across the gap, so it can make a new pair.
Values used: the band gap of silicon, \(1.1\,\text{eV}\); an electron crossing a potential difference of \(V\) volts gains \(V\) electron volts.
In the same field of \(3 \times 10^{7}\,\text{V/m}\), a hotter crystal makes the electrons collide after only \(0.02\,\mu\text{m}\). How much energy, in eV, does an electron now gain between collisions?
In an avalanche, each generation of collisions doubles the number of carriers. About how many generations turn a single carrier into a million?
Which one: the voltage tells
Heavy doping makes a thin layer, so the field is strong at a low voltage. Zener breakdown therefore begins first, below about \(5\,\text{V}\). Light doping makes a wide layer, and the field stays weak until the voltage is higher. Avalanche breakdown therefore dominates above about \(6\,\text{V}\). Between \(5\) and \(6\,\text{V}\) both mechanisms act together.
How each one responds to heat
The two mechanisms respond to temperature in opposite ways. In a hotter crystal the valence electrons already vibrate with more energy, so a slightly weaker field frees them. The Zener voltage therefore falls as the temperature rises, a negative temperature coefficient.
In avalanche breakdown heat works the other way. The atoms of a hotter crystal vibrate more, so a moving carrier collides sooner and gains less energy between collisions. A stronger field, and so a higher voltage, is needed for impact ionisation. The avalanche voltage rises with temperature, a positive coefficient.
A \(12\,\text{V}\) diode breaks down by avalanche. Its breakdown voltage changes by \(0.08\,\%\) for each degree Celsius. Inside a closed box it warms by \(50\,°\text{C}\). What is its breakdown voltage now?
The change is \(12 \times 0.0008 \times 50 = 0.48\,\text{V}\).
Avalanche breakdown has a positive coefficient, so the voltage rises: \(12 + 0.48 = 12.48\,\text{V}\).
A diode near \(5.6\,\text{V}\), where the two effects cancel, would hardly change at all. That is why such diodes are chosen for steady reference voltages.
A \(3.3\,\text{V}\) Zener diode is warmed. What happens to its breakdown voltage, and why?
Which of these are true of avalanche breakdown?
Shops and data sheets call every breakdown diode a "Zener diode", including a \(12\,\text{V}\) or a \(30\,\text{V}\) one. Such a diode actually breaks down by avalanche. The name belongs to the device; the mechanism depends on its voltage.
Why is a Zener diode meant for a low breakdown voltage, such as \(4.7\,\text{V}\), made from heavily doped material?
Heavy doping makes the depletion layer very thin. The field across it is about \(V/d\), so a thin layer gives a very strong field at a small voltage. That field can pull electrons out of their bonds at a low, sharp voltage.
In 1934 the American physicist Clarence Zener asked why a strong electric field makes an insulator suddenly conduct. Using the new quantum theory, he showed that a very strong field lets an electron pass straight from the valence band into the conduction band. In 1951 a team at Bell Laboratories measured this current in thin germanium junctions and named it after him. The diodes built to use the effect have carried his name ever since.
- Thinking that breakdown always destroys a diode.Breakdown is harmless while the heat \(V_Z I\) stays below the rated power. A Zener diode lives there.
- Mixing up the two mechanisms.Zener: heavy doping, thin layer, electrons pulled out by the field, below about \(5\,\text{V}\). Avalanche: light doping, wide layer, carriers multiply, above about \(6\,\text{V}\).
- Giving both mechanisms the same temperature behaviour.The Zener voltage falls with heat; the avalanche voltage rises with heat.