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IGBT & MOSFET faults

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kappa_am

kappa_am

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hi all,
I have a question about broken IGBT and MOSFET. Almost all switches that I dealt with faulted short circuit. But I know sometimes these switch faulted open circuit. Here is my question.
When an IGBT/MOSFET break down short circuit and when it break down open circuit?

Thank you for your replies
 

Usually, a MOSFET will fail short first. This is because excessive heat will, by diffusion, mix the dopants enough to create a good conductor instead of the p-n or n-p barriers that were there originally. Often, the gate oxide will be taken into the diffusion, too, causing a short betweem all three terminals.

Only if the short circuit current after this first mode of failure is high enough to blow the bond wires or the entire transistor, there is an open circuit.
Click to expand...

(source: https://electronics.stackexchange.com/questions/27124/do-mosfets-usually-burn-open-or-closed)
 

To assist the statement in the link, after catastrophic failure with sufficient short circuit current applied to an IGBT or MOSFET, there may be no device left at all. Short circuit currents of a shorted bridge leg with sufficient bus capacitors can easily reach 10 kA or more, and power amounts in a high MW range. There's a reason why people use ear protection in a power electronics lab.

Below this "exploded circuit" energy level, failure to short circuit is the usual case.
 

How can I calculate the amount of energy that cause a semiconductor device falls open circuit?

Thank you
 

kappa_am said:
How can I calculate the amount of energy that cause a semiconductor device falls open circuit?

Thank you
Click to expand...
I think that will be near impossible to make a number out of. Because two failed devices (even in same brand, factory and type) will behave different after fail and will reads different resistance between terminals.
 

Thank you for your replies.
Dear Cart, I am afraid, I do not understand your meaning of poor. defect in which characteristics of driver would cause open circuit failure?

I am working on a kind of fault-tolerant inverter, I need a method to estimate he amount of energy that cause a switch fails open. a rough estimation is ok, though. Isn't there any method?

Thank you
 

Its the wrong question.

All this has already been done by the semiconductor manufacturers.

The right question to ask is:
"How can I design a fault tolerant inverter with multiple layers of protection that will keep my switching devices well within the manufacturers published limits".
If you do that, you stand a fair chance of success.

Not a guarantee of absolute certainty, but the odds will be with you.
 

Its the wrong question.
Click to expand...
It's a different question, and not certainly wrong, although probably hard to answer.

Please review the initial post:
When an IGBT/MOSFET break down short circuit and when it break down open circuit?
Click to expand...

I'm basically with Prototyp_V1.0 "near impossible to make a number out of". However, if you can assume that most of the energy goes into the transistor chip (which happens under certain conditions), then an energy amount that vaporizes the chip completely will very likely result in open circuit failure.
 

Warpspeed said:
Its the wrong question.

All this has already been done by the semiconductor manufacturers.

The right question to ask is:
"How can I design a fault tolerant inverter with multiple layers of protection that will keep my switching devices well within the manufacturers published limits".
If you do that, you stand a fair chance of success.

Not a guarantee of absolute certainty, but the odds will be with you.
Click to expand...

A converter is always design in such way the semiconductors stay in their predefined limitations. But this situation cannot be guaranteed. I am going to design a converter that when some of its switches failed, it stays functional. I have proceed well, but in this stage I like to know how to determine the amount of energy that cause a switch to fail open-circuit.

However, if you can assume that most of the energy goes into the transistor chip (which happens under certain conditions), then an energy amount that vaporizes the chip completely will very likely result in open circuit failure.
Click to expand...
definitely, the amount of energy that cause OC failure depends on area of semiconductor, its resistance, type of heatsink, semiconductor technology. is there another factor?
I do not need a number just a rough estimation or relation to semiconductor characteristics.

Thank you
 

I have seen some high current dc/dc power supplies with multiple mosfets each individually fused (in the drains) by low cost fast blow automotive fuses.

While fuse sizing and I^2t calculations are a bit of an art form, it may be one possible solution worth exploring to clear a short circuit.

While there are well defined purpose built semiconductor fuses available, they are often too expensive which makes them uneconomic for production.
 

Anna Conda said:
Yes, really the only way to design fault tolerant inverters is to have fuses in the drain/collector of the devices...
Click to expand...
I am not talking about protection; I am designing a multilevel inverter with fault tolerance capability. It means it can provide desired output, while a few of its switches are destroyed. It is ok. Just to Simplify the control, I just want know with determined what switch level of DC voltage cause switch falls O.S. Is there any method that give a rough estimation or a relation to characteristics?
 

kappa_am said:
I am not talking about protection; I am designing a multilevel inverter with fault tolerance capability. It means it can provide desired output, while a few of its switches are destroyed.
Click to expand...
I doubt if you will ever realize that redundancy in practice.

If the operating conditions are severe enough to kill a proportion of your devices, the survivors are in turn, going to be individually stressed out even more.
 

You are talking about protection, as, if one device fails (say you have 4 devices in parallel, and 4 or six sets of these in your inverter) you need to isolate the failed device in a group in order for the inverter to carry on, a fuse in the collector is commonly used to do this and provide fault tolerance for the power switches...
 

You are talking about protection, as, if one device fails (say you have 4 devices in parallel, and 4 or six sets of these in your inverter) you need to isolate the failed device in a group in order for the inverter to carry on, a fuse in the collector is commonly used to do this and provide fault tolerance for the power switches...
Click to expand...

The switches are not parallel, in a multilevel structure switches are series.

Thank you all guys,
I gather some information from your replies. I am working on it; I will post the outcomes. :)
 

suppose we deal with H-bridge multilevel inverter. If a switch falls open circuit we can bypass the cell that contains this switch by this cell other switches of by a solid-state relay. then by changing other cells' voltage ratio the exact voltage on output can be made.
A simple idea. the software is implemented on dspic it works ok. a few days ahead I am going to check power section and fault detection.
 

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