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From: Winfield Hill
Subject: Re: Help - Power mosfets - difficult load
Date: 25 Oct 2002 06:26:10 -0700
Organization: Rowland Institute
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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.
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