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From: email@example.com (Bill Allison)
Subject: Re: Help - Power mosfets - difficult load
Date: 25 Oct 2002 14:31:02 -0700
References: <firstname.lastname@example.org> <email@example.com> <firstname.lastname@example.org> <email@example.com> <firstname.lastname@example.org> <3DB81206.E28CA99F@rica.net> <email@example.com>
NNTP-Posting-Date: 25 Oct 2002 21:31:02 GMT
Thanks for that really lucid explanation and for your time and
trouble. And I got your post by E-mail first - this reply is the later
and slightly fuller.
One or three comments…
1) Considering fet temperature rise, I have in my favour the fact that
the motor will be powered up for around 10 seconds at most, every 5
minutes or so at worst.
2) Your calculations are based on the motor's winding resistance i.e.
on the current that would flow into a stalled motor, but the average
current will surely be less when the motor is rotating? The motor's
spec says 380A stalled, 70 to 80A with rated load. And stall current
will only flow for the briefest of moments at start-up (the motor will
always be started with no load other than brush and bearing friction).
3) Your calculations are also based on switching the fet as fast as it
is capable of. Is there a possibility that I might need to use slower
rise and fall times to prevent the amplitude of transients (due to
load and stray inductances) exceeding drain to source limits? That
would be at the expense of switching losses and might make switching
frequency a consideration? The idea is from IRF app note AN-936
Or am I worrying about nothing, because those massive fets' ratings
are in all circumstances going to be well in excess of the stresses
I'm presenting them with…?
Winfield Hill wrote in message news:...
> Bill Allison wrote...
> > FYI and FWIW, I now know that my motor has a DC resistance of
> > 32 millohms (not nearly as low as I feared) and the battery 10
> > when new so I'm feeling more encouraged that this can work...
> Good, some numbers we can sink our teeth into. Let's assume
> a low 2.5 milliohms of cable and connector resistance each way.
> That's 10+32+5+5 = 52 millihoms (I've used the Rds(on) for a
> pair of warmed-up FB180SA10). This means when the FETs are
> on you can expect 230A of current. The I^2*R loss in the
> 5-milli-ohm FET pair will be 266 watts. If you run them at
> 50% duty cycle, that's 133 watts in the FETs and another 133W
> in the leads. You should consider the 266W total, because
> heavy copper wire conducts heat very well.
> In contrast, if you used a high frequency to take advantage
> of the motor's inductance, at the same 50% duty cycle you'd
> have a steady current of 115A flowing. The FET dissipation
> is reduced to 66W during the on time and averages only 33W.
> That's a full factor of 4x less, and nothing to sneeze at.
> Another bonus, the wire loss will also drop by 4x, which is
> another help to the expensive closely-connected FETs.
> You've expressed concern with the switching loss. Let's do
> the calculation. We'll assume a modest 3A FET gate driver
> chip for each FET (e.g. a TC4403 or TC4424), and we'll add
> a 3.9 ohm series resistor to reduce the gate current to 1.5A.
> The FB180SA10 data sheet tells us the Gate-to-Drain ("Miller")
> Charge, Qgd = 110nC. With 1.5A gate drive the switching-
> transition that you're worried about will be very fast, only
> t = q/I = 75ns. The FET carries the full 115A for the entire
> time, so the dissipation will be E = 75ns * 115A * 12V/2 =
> 52 uJ, and if you switch at say 20kHz your switching losses
> will be only P = f * E = 1 watt. (Note, serious designers
> would use much higher gate drive to get faster switching
> for such a big FET, but because that means a much tougher
> job controlling lead-inductance spikes I won't suggest it.)
> So in conclusion, you'll pay a considerable penalty in I^2R
> power heating in your FET by switching at only 50Hz. You
> can do it if you like, but don't do so under the assumption
> you'll be making a big savings in switching losses.
> - Win
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