Electronics > Beginners
Does break down voltage actually damage the diode?
Cerebus:
--- Quote from: Synthtech on September 18, 2018, 10:26:19 am ---A lot of vintage gear that I work on use reversed biased base-emitter transistor junctions as a white noise source using standard transistors such as a 2SC945. They have been working fine in their avalanche region for decades.
Whether this messes up the transistor for use as a transistor I don’t know.
--- End quote ---
It's generally accepted that 'zenering' BE junctions is a Bad ThingTM that will generally alter the characteristics of the transistor even if it doesn't destroy it.
Probably worth a word of two about the damage mechanisms in play when you 'zener' a junction that wasn't explicitly designed for it.
Gross damage can be caused by the simple mechanism of dumping too much power into the device. A diode that is designed to handle, say, 1W will generally produce that with a voltage drop of, say, 0.75 and so at a current of 1300 mA. Stick 1300mA through the same device wired as a zener at, say, 6V would produced a dissipation of 8W. Obviously at an 8 times higher thermal load something is quite likely to give, perhaps even melt.
When you're not exceeding the power rating of the device the mechanism of damage is more subtle. Avalanche "zenering", such as in a transistor BE junction, inherently involves electrons that are moving fast enough to ionize the atoms that they hit going across the junction (impact ionization). If you get electrons that are moving fast enough they will have enough energy to not merely ionize atoms, but (simplistically) to start introducing faults in the crystal lattice. That's much more subtle and can alter the characteristics of the junction without destroying it.
james_s:
I reverse engineered a commercial HeNe laser power supply brick that had two 2N3904 transistors wired with only the B-E pins as a zener. They were made like this for decades, I think the dead one I de-potted was about 30 years old and the failure wasn't related to those.
T3sl4co1l:
Yes, it can. Back in the bad old days of early silicon diodes, avalanche wasn't typical, and diodes were rated by maximum voltage without breakdown (as opposed to minimum breakdown voltage at some bias current).
Processes weren't quite well enough controlled, and what happens is a microscopic defect breaks down, concentrating the entire applied current (~mA?) into a tiny area, which burns through and shorts the diode.
Anywhere you see a rating that applies current, you are guaranteed to have at least that much avalanche capability.
Although not necessarily more current than that. The defects might be less dramatic than the early diodes, but not so good as to avalanche the whole diode junction evenly. For that, there's actual avalanche rated parts. TVSs are made to break down over the whole die, and handle considerable energy.
Some rectifier diodes do carry an energy rating, 1-20mJ being fairly common. In the case of schottky diodes, this is absorbed by the guard ring, a PN junction that surrounds the schottky junction. The same may be true of other types, I don't know. This is why the energy rating might be quite small relative to the overall die size.
There are still plenty of diodes and transistors that can't handle avalanche. In the transistor case, you see V_CES ("collector-emitter sustain voltage"), rather than V_CEO (collector-emitter-open voltage, i.e., voltage applied C-E, with B open/unconnected).
I don't know of any IGBTs rated for avalanche. Probably for good reason, like parasitic SCR activation from putting free charges everywhere.
InGaN LEDs (i.e., blue, white, high efficiency green, etc.) don't seem to handle any avalanche, they die suddenly at 20-30V even with very low bias current. Older LED types (GaP, GaAs and alloys -- IR to green) usually handle a few mA, at random voltages (20 to >200V).
Tim
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