Re: Gate drive resistor question.
- From: Terry Given <my_name@xxxxxxxx>
- Date: Thu, 24 Jan 2008 01:09:28 +1300
seegoon99@xxxxxxxxx wrote:
On Jan 23, 1:21 pm, Terry Given <my_n...@xxxxxxxx> wrote:
seegoo...@xxxxxxxxx wrote:
On Jan 23, 12:56 pm, Terry Given <my_n...@xxxxxxxx> wrote:
seegoo...@xxxxxxxxx wrote:
Hi to all.
I have been asked to look into a SMPS that is experiencing a higher
than wanted failure
rate in the field.Something that I've come across is ringing a bell.
The controller chip (NCP1207)
is driving a fet(IRFBE30) via a 47R resistor. The resistor is an 0805
smt package.
Somewhere on this group in the past I've read something about gate
resistors failing because
they are under sized.(power wise)
Is this possible in this case with such a small package. If the gate
resistor goes high impedance
it would obviously cause catastrophic failure of the FET and then
surrounding components.
Which is what we are seeing.
Could this be an issue?
Cheers
Rob
Hi Rob,
it certainly could be. if you look at the peak-pulse-power rating of
0805 resistors, you will see it isnt very good at all - the resistor
comprises a very thin layer of resistive material atop a ceramic
substrate, so there isnt much meat within which to dump the pulse energy.
the IRFBE30 Qg is about 60nC at 12V, so looks like about 5nF. at 47R
thats 235ns time constant, which is a lot slower than the controller
rise time. so the peak current is pretty much +12V/47R = 0.25A, and
12V^2/47R = 3W peak pulse power.
thats not really very much; an 0805 can eat about 1W no problems, so I
wouldnt expect this to be a problem. were the gate resistance more like
4R7, that would definitely ring alarm bells.
part of the problem, of course, is that when the FET goes to the great
silicon plant in the sky, it shorts D-G and then snots Rg and the
controller.....making diagnosis that little bit trickier.
Panasonic make some AWESOME 0603, 0805, 1206 surge rated resistors (they
make MMA0204 and MMB0207 look puny). although they are more expensive,
they fit in the same footprint, and can be used to remove suspicion from Rg
BTW a 235ns gate time constant is pretty slow, and miller capacitance
might be winning the fight against Rg, thereby stretching out the
switching time even further, exacerbating switching losses. look at the
gate waveform, at the plateau that occurs around Vth. This should be
short (< 50ns). If it is not short, the gate drive is too weak. true for
both turn on and turn off (although life is easy at turn-on for DCM).
if its a DCM smps, check it has a soft-start, as DCM supplies ALWAYS
power up in CCM until the output cap is charged. A useful test is to set
up some time delay relays to turn a unit on for, say 10s (enough for it
to start running) then off for long enough for all the caps to
discharge. repeat ad nauseum. this will pretty soon let you know if
there is a start-up issue.
Cheers
Terry- Hide quoted text -
- Show quoted text -
Hi there Terry.
There is something I forgot to add. There is a reverse diode across
the resistor to decreace turn off time. On turn on the Miller
capacitance is definately winning. In fact the gate voltage starts to
fall for about 2V , never mind level out!!
It seems that the gate resistor may well be a potential problem. I'm
going to see if I can increace it to a 2010 footprint.
As you have mentioned , once things have gone "pear shaped" it's
difficult to find the cause of the failure :0(
I'll check out if the supply has a decent soft start as well.
Thanks for the help.
Cheers
Robin
Hi Robin,
the diode halves the number of 3W pulses seen by the resistor, and in so
doing halves the duty cycle, making life a bit easier for the resistor.
in that case, I'd definitely:
a) look at the panasonic parts (digikey has them) - easier than a PCB
re-spin.
(an alternative would be to parallel 5 x 220R 0805, which will happily
suck up 3W PPP ad infinitum.)
b) look carefully at the turn-on switching loss. if the gate drive is
losing the fight with miller, the turn-on loss (if CCM) will be a lot
higher than necessary. if its DCM, then the turn-on loss is
0.5*Cstray*Vpk^2*Fsmps, regardless of switching time (Cstray = Cds +
Ctransformer + Cwhateverelse)
Cheers
Terry- Hide quoted text -
- Show quoted text -
Hi Terry.
I'll see if I can improve the turn on time. The fet is not getting to
hot though. Even at full power (36W about) it's not getting
much above 50deg c , so I'm not sure if it is that serious. There are
some other changes that need to be
made so a board redesign is probably on the cards anyway.
I'm not sure I understand your turn-on loss equasion.
Is (0.5*Cstray*Vpk^2*Fsmps) the power dissapated in the FET during
switch-on? There is no reference to Rds or current. I think I'm
missing the boat here :0(
Cheers
Rob.
Oh, OK.
if its DCM, then the current at turn-on is (by definition) zero. BUT when the FET turns on, the D-S capacitance (and xfmr capacitance and any other strays) have to be discharged, by the FET. so the FET dissipates 0.5*Ctotal*Vpk^2 joules, each turn-on cycle - assuming the D-S cap was charged to Vpk just before the FET was switched on. so the power dissipation in the FET due to this (P = E*f, 1W = 1J/s) is 0.5*Ctotal*Vpk^2*Fsmps. usually (but not always) Vpk = Vin
in CCM, in addition to this capacitive loss, there is 0.5*Vpk*Imin*Tfall energy dumped into FET, where Imin is the CCM current value at turn-on and Tfall is the drain voltage fall time - this assumes the FET current ramps up to Imin before Vds begins to fall. so this switching loss is 0.5*Vpk*Imin*Fsmps. ditto for turn-off, 0.5*Vpk*Imax*Fsmps*Trise (this last one is there in DCM too)
while the FET is on, Rdson comes into play, and you get Irms^2*Rdson conduction loss.
and just because the average FET temperature is 50C doesnt mean the junction is OK - ESPECIALLY during the turn-on (and turn-off) transients. Fairchild give thermal models for their FETs (eg FDP047AN08) use one of those and plug the swiching loss into it, you might be (unpleasantly) surprised at how high Tj can get in a short period of time - during switching edges, you can treat the IGBT die as if it were adiabatic, so all the switching energy causes the junction alone to heat, no heat flows into the tab. see "electro-thermal modelling of multi-megawatt power electronic applications using PSPICE" from www.powerex.com
Cheers
Terry
.
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