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MOS FET with built-in TVS diode.

Steam
Associate II

I want help to understand this.

The STM MOS FET STP6NK60ZFP has a MAX gate-source voltage (Vgs) at 30V. In order to protect it from ESD it has a built-in TVS diode. Over 30V it can fail. But the bulit-in TVS has breakdown voltage (Vbr) set at 30V. It's when it begins to conduct. The clamping voltage (Vcl) is a couple of volts more. It's when it conducts fully. But then it's over the limit 30V. More voltage than that can make the MOS FET fail. Shouldn't it be that the Vcl should be 30V so an ESD can't exceed 30V? Or is it really enough with Vbr at 30V and Vcl some volts more? Se table below.

 

ScreenShot_20250620014544.jpeg

1 ACCEPTED SOLUTION

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I'm not an expert. You are.

So your answer make sence and I now understand thanks to your explaining and calculation. My question is now answered.

Thank you again for your professional help.

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6 REPLIES 6
Peter BENSCH
ST Employee

Welcome @Steam, to the community!

Since you can and must rely on the information in the data sheet, your conclusion is absolutely correct. In most cases, a TVS with a lower voltage than the typical 30V is provided externally at the gate if protection is actually required.

Does it answer your question?

Regards
/Peter

In order to give better visibility on the answered topics, please click on Accept as Solution on the reply which solved your issue or answered your question.
Not really the answer I was looking for. My concern was if the built in TVS protection really works since the Vbr is 30V and that is the max voltage for the gate.
The Vgs is then some volts more (unclear how much, not described) above the limit voltage for the gate. So it feels that there is a risk of destroying the MOS FET since the TVS conducts to its fully ABOVE the gates limit. Or is there a certain tolerance above the 30V so the MOS FET can endure this anyway?

>Or is there a certain tolerance above the 30V so the MOS FET can endure this anyway?

Sure, there is some margin , otherwise the internal 30V zener would be 100% useless.

But its not intended to "use" this 30V limiter for a circuit, its just some protection for handling the device, 

until its soldered in the target circuit, where its gate drive voltage anyway is lower, in +/- 12 V range (about, typical).

So this kind of gate protection is just to get no damaged mosfets by someone touching the mosfet with some charge, until the final circuit is built.

Its some built in "insurance" to have a still working mosfet in the target circuit.

If you feel a post has answered your question, please click "Accept as Solution".

Really?

I get another impression when I read the data sheet:

The built-in back-to-back Zener diodes have specifically been designed to enhance not only the device’s 
ESD capability, but also to make them safely absorb possible voltage transients that may occasionally be 
applied from gate to source. In this respect the Zener voltage is appropriate to achieve an efficient and 
cost-effective intervention to protect the device’s integrity. These integrated Zener diodes thus avoid the 
usage of external components.

 

What gate driver you use ? supply ? 

So mosfet+driver at max. 15V driver supply : how to get any voltage over 15V to the gate ?

Only at hard switching "off" the mos, you can get a spike from drain via Cgd ;  25 pF from ds STP6NK60ZFP .

Driver going to 0V , to switch mosfet off, at a typical gate resistor (from ds) 4.7 ohms, it would need (30V/5 ohm) about 6A through the 25pF , to come close to the critical voltage at the gate; right ?

Now calculate the current at the given speed of this STP6NK60ZFP, 19ns at 300V, 3A : or - if you cannot calculate -

just think: how to get 6A ever, if the mosfet running at 3A ?? never .

All it could get is a 3A peak to the gate with a bad driver/design, this would switch partially on the mos and give some oscillation (the end of the mosfet, most times). But never you could get 30V at the gate in a circuit with a gate driver.

So this 30V protection is for safer handling the non-installed part - thats it.

If you feel a post has answered your question, please click "Accept as Solution".

I'm not an expert. You are.

So your answer make sence and I now understand thanks to your explaining and calculation. My question is now answered.

Thank you again for your professional help.