Mosfet Vds Rating

[PICMICRO]

In mosfet datasheet the maximum Drain to Source Voltage is Listed. But I never find any time specification for it, unlike the Maximum Current rating, for which time is clearly stated.
Does this means, Vds rating shouldn't be exceeded even for very very brief moment, like, 1 nanoseconds or 1 picoseconds?
I tried some simulations and liberally added parasaitic inductances in every branch.
No matter what snubbers I design, I can't get rid of very brief overvoltages exceeding its Vds rating due to various parasaitic inductances, notably of the snubber branch itself.
Should I worry?

 

[crutschow]

If you have a small ceramic capacitor with short leads mounted close to the MOSFET then the stored inductive energy in the parasitic inductance is very small and should be absorbed by the MOSFET parasitic drain capacitance.

What values are you using for your parasitic inductances?

 

[ronsimpson]

Unlike junction transistors, MOSFETs function more like a zener diode when facing too much Vds.
With over voltage conditions a MOSFET starts leaking current from D to S, in a non destructive fashion, as long as the silicon does not get too hot. A 100V MOSFET at 110V and 1A will get hot very fast! For uS it is fine.

See safe operating area fig. 8 in the data sheet attached.

 

[PICMICRO]

Unlike junction transistors, MOSFETs function more like a zener diode when facing too much Vds.
With over voltage conditions a MOSFET starts leaking current from D to S, in a non destructive fashion, as long as the silicon does not get too hot. A 100V MOSFET at 110V and 1A will get hot very fast! For uS it is fine.

See safe operating area fig. 8 in the data sheet attached.

In the datasheet you linked and other datasheets I have seen, No region above Maximum Vds (in this case 100V) is shown to be safe, no matter how brief is the time.

So, for 100V mosfter 110 V for a 1usec is fine. What about 1100 Volts for 1 usec? It could be great if I could find Vds overvoltage Vs time period graph.

Oh! I forgot to incorporate Vds Capacitance in my simulations. <I have been simulating mosftets with ideal switch>. After including them, the short Inductive surge is will absorbed.

So, I think I can answer my own question, the Voltage Vs Time graph is nothing but the 'Charging' time for the Vds Capacitor before it crosses Vds_max value. Am I right ?

 

[johansen]

In the datasheet you linked and other datasheets I have seen, No region above Maximum Vds (in this case 100V) is shown to be safe, no matter how brief is the time.

So, for 100V mosfter 110 V for a 1usec is fine. What about 1100 Volts for 1 usec? It could be great if I could find Vds overvoltage Vs time period graph.

Oh! I forgot to incorporate Vds Capacitance in my simulations. <I have been simulating mosftets with ideal switch>. After including them, the short Inductive surge is will absorbed.

So, I think I can answer my own question, the Voltage Vs Time graph is nothing but the 'Charging' time for the Vds Capacitor before it crosses Vds_max value. Am I right ?

its in there but its a bit cryptic unless you know what to look for: avalanche energy and avalanche current.
you can see the test circuit in the last page, where it says unclamped inductive load.
it also shows you a graph of the thermal impedance.

basically, the DS junction breaks down at say 110 volts for a 100 volt mosfet, and if the current passed through this junction is less than the avalanche current listed in the datasheet then the fet should survive. however the energy dissipated must be less than the energy listed, most of the time it is in millijoules. however, the energy able to be safely dumped has to be interpreted via the thermal impedance chart.

 

[ronsimpson]

Many MOSFET spice models do not have breakdown behave included. In spice it is OK to have 1000 volts on a 100 volt part. In the real world the MOSFET will draw current at 110 volts. (using our example)

For a small amount of inductive kick back the MOSFET capacitance and the MOSFET pulling current will eat up the energy form your inductor.

The reason why there is no "over voltage" line on the graph is that you can't have over voltage. I need to say that another way. Using out 100V MOSFET, connect a 10k resistor from D to 1000 volts. At some where from 100V to 120V the MOSFET will conduct current and hold the drain at about 110 volts. This voltage is not states but is ">100 volts", and probably not much larger.

You can't get 1100 volts for 1uS. It will conduct 100Amps, over heat and explode.

I may not have answered your questions.

There is also a D-G capacitor. It you pull up hard on the D (very fast) the D-G capacitor will pull up the gate and turn on the MOSFET thus limiting the rise time.

 

[Bob Scott]

Ron is right. The FET will act like a Zener.

What current is involved? How much current flows through your inductive load? The FET Drain will automatically sink as much current as required to keep the voltage no higher that its Vds. The FET is its own protection.

 

[PICMICRO]

Ron is right. The FET will act like a Zener.

What current is involved? How much current flows through your inductive load? The FET Drain will automatically sink as much current as required to keep the voltage no higher that its Vds. The FET is its own protection.

Oh! it sounds quite impressive.
Almost the FET's Rated current is involved and that is what flows through the inductor.
How much current the FET Zener can support?

 

[johansen]

Oh! it sounds quite impressive.
Almost the FET's Rated current is involved and that is what flows through the inductor.
How much current the FET Zener can support?

that is in the datasheet where it says Avalanche current.
note however that it cannot pass that current for very long. microseconds.

google "avalanche rated mosfets" for more

 

[PICMICRO]

that is in the datasheet where it says Avalanche current.
note however that it cannot pass that current for very long. microseconds.

google "avalanche rated mosfets" for more

Thanks.
From the datasheet of IRFP460 Mosfet, which is a 500V 20A mosfet,

Avalanche Current
IAR    Avalanche Current, Repetitive or Not-Repetitive (pulse width limited by Tj max)    20 A
EAS   Single Pulse Avalanche Energy (starting Tj = 25 °C, ID = IAR, VDD = 50 V)        960 mJ

I can't fully grasp the concept. From the IAR rating, I guess that it can support 20A avalanche current for until the Mosfet's temperature is within limit.
That means 20A*500Volts = 10,000 watts.
Now from the second EAS rating, it says it can't give more than 960 mJ single Avalanche energy. 10,000 watts for 100 microseconds will give 1000 mj.
So, Does this means, the avalanche breakdown shouldn't last for more than ~100 useconds?
If yes, then to be consistent with IAR rating, does this also means, if it lasts more than 100 usec, the Junction Temperature will rise more than Tj Max?
Why is Vdd = 50V specified for EAS rating. Shouldn't it be Vdd = 500V?, since we are talking about avalanche current.
I have a record where I have managed to damage a IRFP460 due to overvoltages (The avalanche current was well under 20A), without its body temperature rising significantly. Can you explain this?

 

[ronsimpson]

Using your numbers: 20A * 500V for 100uS = too hot. OR 5A * 500V for 400uS=too hot. OR 1A * 500V for 2000uS

If the part is already hot them the time must be shorter.

 

[ronsimpson]

MOSFET over voltage is modeled by placing a Zener (500V in your case) from D to G. When the D-S voltage is too high the MOSFET turns on and heat is generated.

In a junction transistor when the voltage is too high the weakest part of the die conducts. (maybe 0.1% of the die conducts and gets very hot very fast because of the small area) With a micro scope I have seen holes in the die where a small piece has exploded.

The time you last mentioned (100uS) is the time for the die to go from 25C to 150C. If the the die started out at 87C it will take only 50uS to reach 150C. I know there are some inaccuracy in my numbers...I am trying to make the point that a hot part has little room for stress.

 

[PICMICRO]

MOSFET over voltage is modeled by placing a Zener (500V in your case) from D to G. When the D-S voltage is too high the MOSFET turns on and heat is generated.

In a junction transistor when the voltage is too high the weakest part of the die conducts. (maybe 0.1% of the die conducts and gets very hot very fast because of the small area) With a micro scope I have seen holes in the die where a small piece has exploded.
So,

The time you last mentioned (100uS) is the time for the die to go from 25C to 150C. If the the die started out at 87C it will take only 50uS to reach 150C. I know there are some inaccuracy in my numbers...I am trying to make the point that a hot part has little room for stress.

Thanks for your explanation.
Just for confirmation,
5A * 500V for 400uS=too hot. It means, the die is too hot, but not necessarily the whole body (that you can touch with your finger) is too hot. Am I right?
I ask because, when I damaged that Mosfet, it wan't untouchably hot, just mildly hot.

 

[ronsimpson]

We are talking die temp.
When we talk 100uS the heat takes time to make it to the case and to the heat sink.
The data sheet talks about the thermal resistance from die to case. With a little math you can get to the die temp from the case temp.