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Wikipedia says:

The Zener effect is distinct from avalanche breakdown. Avalanche breakdown involves minority carrier electrons in the transition region being accelerated, by the electric field, to energies sufficient for freeing electron-hole pairs via collisions with bound electrons. The Zener and the avalanche effect may occur simultaneously or independently of one another. In general, diode junction breakdowns occurring below 5 volts are caused by the Zener effect, whereas breakdowns occurring above 5 volts are caused by the avalanche effect. Breakdowns occurring at voltages close to 5V are usually caused by some combination of the two effects. Zener breakdown voltage is found to occur at electric field intensity of about 3×107 V/m.[1] Zener breakdown occurs in heavily doped junctions (p-type semiconductor moderately doped and n-type heavily doped), which produces a narrow depletion region.[2] The avalanche breakdown occurs in lightly doped junctions, which produce a wider depletion region. Temperature increase in the junction increases the contribution of the Zener effect to breakdown, and decreases the contribution of the avalanche effect.

My question is can in a diode, primarily under Zener breakdown, increasing the reverse-bias voltage cause avalanche breakdown ?

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Zener breakdown (the breakdown mechanism described by Claude Zener) occurs in practical semiconductor devices at low voltages. Avalanche breakdown dominates at high voltages and includes amplification by carrier-generating processes similar to a small rockfall initiating an avalanche.

Zener breakdown has a negative temperature coefficient of voltage, while avalanche breakdown has a positive temperature coefficient.

So, in practical Si devices see figure #1 here both processes are present, but the Zener process dominates low-voltage breakdown behavior, while avalanche dominates at higher applied voltages.

The temperature coefficient at 5V is very near zero, indicating that both processes contribute nearly equally. This doesn't mean Zener behavior is 'converted', though, it simply means that the flood of extra charge carriers produced by avalanche effects can be more important than the Zener contribution.

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  • $\begingroup$ Finally a good answer. $\endgroup$ – Mockingbird Mar 7 '17 at 3:02
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Once you have any kind of breakdown, it dramatically reduces the differential resistance contributed by the depletion region, so if you double the external voltage, you're not doubling the voltage across the depletion region, but increasing it just a little bit, while an increasing fraction of the voltage drop occurs in the contacts and bulk semiconductor and elsewhere.

So, I'm going to say probably not in practice, or at least not much. But in principle, if you crank up the external voltage enough, you will eventually get enough voltage across the depletion region to induce avalanche current, which would supplementing the (presumably) much much larger amount of zener current. (I could be wrong, this is just a guess.)

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  • $\begingroup$ Would have better if you had shown some refererence. But +1 for answering. $\endgroup$ – Mockingbird Mar 1 '17 at 20:34
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If I'm not wrong, Zener breakdown is characterised by two different processes: Field emission and Avalanche breakdown. As far as I remember, my teacher explained avalanche breakdown as the breaking of bonds. He said both these processes happen simultaneously.

Anyway your source quotes them (Zener breakdown and Avalanche breakdown) as two processes which may occur simultaneously or independently. According to my understanding of your question, I fail to understand how increasing the reverse bias voltage will have any effect on the breakdown. Once breakdown is reached, the voltage across the Zener diode remains constant and one thing reinforces the other leading to a sort of chain reaction.

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  • $\begingroup$ Why the voltage would be constant after breakdown? $\endgroup$ – Mockingbird Mar 4 '17 at 16:22
  • $\begingroup$ I don't know why... But it does doesn't it? Why would it be used as a voltage regulator if it didn't? $\endgroup$ – Kunal Pawar Mar 4 '17 at 16:43
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Yes. If you somehow manage to apply a huge bias on your junction far beyond the Zener breakdown condition, that would mean that much of the current that tunnels across the junction would consist of carriers injected at very high energy. These high energy carriers can then avalanche and beget more.

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