Mechanism of destruction of a reversely polarized diode

Thread Starter

Marcelo Besch

Joined Mar 25, 2015
10
Hello all!

I've been curious about diodes recently, trying to understand them more completely. This site was very useful already, clarifying to me exactly what happens when a diode is forward polarized by the knee voltage (the depletion region collapses completely).

Now, I have been curious about exactly what happens when the breakdown voltage is applied to a diode. It is said everywhere that this damages the component, but nowhere I have seen it explained why.

Is there some phenomenon that disrupts the structure? Does it simply burn because of the excessive current?
 

nsaspook

Joined Aug 27, 2009
13,272
Is there some phenomenon that disrupts the structure? Does it simply burn because of the excessive current?
Mainly yes, the destruction is caused by heating. Avalanche (kinetic ionization) and tunneling "Zener" (quantum ionization) are usually not destruction (within reasonable limits) from the internal electric field.

Breakdown IR heating can lead to permanent disruption but is not inherently destructive unless the device power dissipation is exceed.
 
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studiot

Joined Nov 9, 2007
4,998
It is said everywhere that this damages the component, but nowhere I have seen it explained why.
Hopefully not by any reputable source.

Beyond the breakdown reverse blocking voltage the effective resistance of the semiconductor is so low that an excessive current would flow (if unlimited) destroying the component by IR heating.
If that current is limited to a safe value then no damage will ensue.
That is the basis of breakdown testers and some relaxation oscillators.
 

MikeML

Joined Oct 2, 2009
5,444
P=IE, Take a 1n4004, with a PIV of ~400V. Now apply a reverse voltage of 405V from a power supply capable of supplying 1A. As the diode breaks down, the instantaneous power going into it is 405W. How long do you think that it can stand that? 1us? 100us? 10ms?
 

cmartinez

Joined Jan 17, 2007
8,253
P=IE, Take a 1n4004, with a PIV of ~400V. Now apply a reverse voltage of 405V from a power supply capable of supplying 1A. As the diode breaks down, the instantaneous power going into it is 405W. How long do you think that it can stand that? 1us? 100us? 10ms?
That's very interesting what you and @studiot just mentioned... what's a relaxation oscillator?
 

studiot

Joined Nov 9, 2007
4,998
A relaxation oscillator has two parts.

A triggered switch between two states eg a bistable
Timing components on which the charge or valtage changes with time eg a capacitor fed current by a resistor.

The output voltage from the capacitor (as it charges from the resistor) is monitored by the bistable trigger.
When it reaches the trigger threshold the bistable flips over, causing some thing to discharge the capacitor, dropping its voltage.

This causes the trigger to fire again setting the bistable to the other state, which allows the charging current to flow once again into the capacitor until........

Gosh I've just described a 555

It's easier to show with a real circuit - Google is your friend - There are lots of types.
 

ian field

Joined Oct 27, 2012
6,536
Hopefully not by any reputable source.

Beyond the breakdown reverse blocking voltage the effective resistance of the semiconductor is so low that an excessive current would flow (if unlimited) destroying the component by IR heating.
If that current is limited to a safe value then no damage will ensue.
That is the basis of breakdown testers and some relaxation oscillators.
I always understood it to be caused by the avalanch effect - increasing the reverse voltage stress dislodges any loose electrons from the crystal lattice, increasing the stress further causes a greater number of electrons to loose their grip - the avalanch bit is when there's enough falling electrons to dislodge others that were trying harder not to let go.

I2R effect is probably ultimately the failure mechanism.
 

takao21203

Joined Apr 28, 2012
3,702
sure its the current / dissipated power, whilqe some semiconductors also are sensible for too fast rises as they cant transport it fast enough so its exceeded locally.

Some are highly structured internally, its not just a flat PN junction.

I made a circuit which can light up small neons even the blue ones so its at least 100v.
It connects directly to the base of a transistor, with no resistor.

You can hold any LED in your hand, touch with one pin, it lights up.

Because its a current source. the voltage will break down and there is only little stored energy.

Even a small tiny rectangle clear red LED will just light up normally. No damage or heating.
But the same, it will light up a neon. It doesnt conduct at first so the voltage is rising, then ignites,
and takes more than 10 volts until its extinguishing.

Voltage is secondary what counts is current, dissipated power and stored electrical charge as such. It could be a magnetic flux which just becomes electrical current of several kinds.
 
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