Sizing Mosfet Gate Resistors

From a previous post by Terry Given in "Resistor Current Handling

Capability" circa Thurs, Feb 26 2004 :

"This is a common reliability problem with power electronics - "designers" often neglect peak pulse power in MOSFET gate resistors - I saw one design with 4.7Ohm Rg, +12V Vg i.e. 31W peak, using 0805 resistors that failed

after a short time, causing catastrophic failure; so much damage was done each time that the root cause was totally obscured, and the "designer" could not stop it happening. oops. "

I am the lowly test engineer with the 10 fets in parallel test question from several months ago. I the process of getting my head around how hard I can pulse my 10 ohm gate resistors when one side itied to ground. I read Terry's above note.

Anyone care to elaborate or point me to a good app note regarding sizing Mosfet gate resistors?

The board I am testing (yes "they" decided the mosfets need to be individually turned on. . .I knew "they" would) looks suspicously like "oops".

Thanks much, Ed V.

Reply to
EdV
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31 watts peak won't hurt an 0805 if the duration and rep-rate are low.

One side to ground? Are the resistors connected gate-to-source? That's sort of strange. "Mosfet gate resistors" are usually in series with the gate.

John

Reply to
John Larkin

10 ohms works well
Reply to
Mr. J D

The gate resistors are in series with the gates but they are all driven from a single source.

My test stratgey is:

1 - ground the common signal source 2 - apply an appropriate pulse at the test point beween one Fet and its gate resistor 3 - acquire the Vds data (gppib data from oscioscope) before during and after the "appropriate gate pulse" 4 - crunch Vds to see that the correct "dip" is there for a Fet sinking some current.

repeat steps 1-4 for fets through 20

The fets turn on in about 35nsec so I am guessing I can get something constructive done with a 1 usec pulse. Determining the voltage is problematic. I am waiting for the designer to tell me which resistor is being designed in which will have some effect on how hard I can pulse. I should probably consider a worst case scenario since my purchasing dept. will gladly "redesign " this circuit for me.

Thanks for being interested in my evolving dilemnas : - >

Ed V.

John Lark>

Reply to
EdV

Hello Ed,

App notes won't help too much I am afraid. The proper strategy in cases like this would be:

a. Find out what the highest (worst case) pulse current would be, via the slope of the drive signal and the gate capacitance.

b. Get pulse load data from the resistor manufacturer, in writing. If they can't or won't furnish it change to another manufacturer that can.

Should be done by the design engineer but it never hurts to double-check. We all make mistakes.

BTW, test engineers are not "lowly". I value their work, expertise and production-related experience a lot.

Did you see failed resistors?

--
Regards, Joerg

http://www.analogconsultants.com
Reply to
Joerg

Please don't top post.

The average power dissipated in a series gate resistor is about

P = C * F * V^2

where

C is total effective gate capacitance, Miller included

F = drive frequency

V = gate drive voltage.

So calculate that and see if it's within the resistor's power dissipation spec.

John

Reply to
John Larkin

Hello John,

Don't discount pulse load limits. Metal film types exhibit certain pathologies related to that. That is a serious concern with most of my ultrasound pulser designs. I used to prefer Beyschlag because of their excellent specsmanship. They were bought by Vishay but much of the info is now online:

formatting link

They are also very helpful on the phone but if it's still like it used to be, ideally you'd have to be able to speak German. Preferably with a northern accent ;-)

--
Regards, Joerg

http://www.analogconsultants.com
Reply to
Joerg

A fet gate will charge up in nanoseconds if you drive it hard, and the thermal impulse into the resistor will be tiny. If you drive it slow, the resistor peak power will be low, and that's OK too.

The energy dumped into the resistor on each edge is about equal to the gate charge energy [1], which will be 10s of mJ for a big fet (several millicoulombs of gate charge driven to, say, 10 volts.) I've never seen even an 0805 gate resistor fail from the transient charge current. At very high rep-rates, the average power could cook the resistor.

John

[1] assuming the gate driver is infinitely fast, which it usually isn't. The slower the driver, the less power is dissipated in the series gate resistor.
Reply to
John Larkin

Hello John,

That's exactly what we did, drive a gate from a push-pull stage. It's been a while but IIRC they told me that it'll start eating away right under the metallization area.

I did (0805). We replaced them with larger ones. Most of the failures were in the T/R switches though where pulse peaks were higher but still well under the thermal limit. None of the failed resistors ever got hot and there were no scorch marks on the boards. Yet fail they did.

BTW the most strange failure I had on a pulser (but this did exceed the thermal) was when part of a resistor flew off with a loud bang and the remnants had turned into milky green glass.

--
Regards, Joerg

http://www.analogconsultants.com
Reply to
Joerg

I've seen this with 0603 and 0805 gate resistors. Its worse with IGBTs than FETs, for 2 reasons:

- IGBTs tend to be bigger (well at least the big ones are), and have more gate capacitance. 100nF is nothing!

- IGBTs often have bipolar gate drive (to reduce evil effects of Cm),

+/-15V is pretty typical, which quadruples the resistor peak power.

and of course in most smps the rep-rate is not small (although really big IGBT drivers rarely switch above 10kHz)

The high rep-rate drives up the average power dissipation (100nF 30Vpp at 1kHz = 90mW into 10R), and therefore both the average and peak temperatures.

And I should clarify "seen": I havent actually measured a dodgy smt gate resistor, but rather on several occassions surmised this as the cause of clusters of field failures - the numbers certainly supported the hypothesis, and suitable component choices made the problems disappear completely. because the units had already failed, Rg was always toast.

I have however destroyed a number of resistors in this manner. It all started when I turned on a drive in the test lab and noticed a flash, many years ago. I banged the juice off PDQ, and along with another engineer we tested it carefully, but found no "fault" - it still ran. But there was definitely a flash, so we dismembered the drive and lo and behold there it was - a dead PR02 10R resistor, used as damping in an EMI filter. It wasnt our "design", it was in common usage there. some calcs showed that at 415Vac a 10R sees 34.5kW peak pulse, which was clearly a bit much for the PR02. So we went walkabout thru the service area, and lo and behold, there was many an EMI filter whose caps were disconnected.

similar issues apply with DC bus soft-charging resistors; a cheap 10W w/w wont have the mandrel completely surrounded by ceramic material, so there is an air gap and hence a hot spot; hit it with a sizeable thump and you get a pretty red flash from that spot, it work hardens and eventually breaks.

Likewise for Dynamic Braking resistors. Now that is fun; on a cold windy night a 250kW DB resistor is great!

As an interesting aside, in addition to the substrate solder joint, IGBT bond wires are vulnerable to thermal cycling.

Cheers Terry

Reply to
Terry Given

Hello Terry,

During a vacation job back in the college days I build many of these. About the size of a Westinghouse fridge when complete. I guess they are still doing their job in locomotives (in Brazil).

--
Regards, Joerg

http://www.analogconsultants.com
Reply to
Joerg

Q^2 . R . rep rate surely ? Where Q = Vdrive / Ciss

Graham

Reply to
Eeyore

What's your design experience Mr J.D. ?

Graham

Reply to
Eeyore

Sounds sensible.

Which voltage ?

This all sounds a bit arse about face to me.

Graham

Reply to
Eeyore

Correction !!!! But still not right. This assumes the gate drive current is a constant.

The 2 comes from the charge and discharge of the gate capacitance..

But the energy in the series R is created mainly in short pulses therefore increasing the power dissipated in the series R.

If you can solve it for a single pulse then it just needs multiplying by the rep rate. It's doable but I'm not in the mood for integrals involving e right now.

Graham

Reply to
Eeyore

Where did R go ?

Graham

Reply to
Eeyore

rep

dont bother, its wrong. The basic maths is:

E = 0.5*Q*V = 0.5*C*V^2

P = E*f

do a dimensional analysis (IOW look at the units) of your equation, they dont match.

When you put 0.5*C*V^2 J into a cap thru a resistor, the resistor also dissipates 0.5*C*V^2 J. When you then discharge the cap into the same resistor, again it dissipates 0.5CV^2 J, leading to a total resistor energy loss of CV^2 J to charge C to V Volts then discharge C. Repeat at some frequency F, and voila: P = CV^2F

The resistor doesnt appear in the equation because it is not controlling the voltage to which the cap gets charged.

The above assumes the cap is both fully charged and discharged. If it is not, *then* R comes into play. In the case of a FET driver, this is an extremely good assumption; if the gate isnt charging and discharging fully, there is most likely a problem.

The same maths usually applies to snubbers, too, but not to RCD voltage clamps, as the clamp capacitor generally does not discharge between successive clamping operations.

Cheers Terry

Reply to
Terry Given

see my post below.

Cheers Terry

Reply to
Terry Given

rep

QED: Eeyore has been proven incompetent ;-)

...Jim Thompson

--
|  James E.Thompson, P.E.                           |    mens     |
|  Analog Innovations, Inc.                         |     et      |
|  Analog/Mixed-Signal ASIC\'s and Discrete Systems  |    manus    |
|  Phoenix, Arizona            Voice:(480)460-2350  |             |
|  E-mail Address at Website     Fax:(480)460-2142  |  Brass Rat  |
|       http://www.analog-innovations.com           |    1962     |
             
I love to cook with wine.      Sometimes I even put it in the food.
Reply to
Jim Thompson

rep

now.

Eh ? Q . t^-1 = I

OK. Neat trick. Hadn't figured that one.

Yup.

Just the rate. Of course it it was very slow the gate would never charge to V.

Yes.

Sure.

Interesting.

Graham

Reply to
Eeyore

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