Formula for minimum drive current for mosfet
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From: gearhead on 14 Apr 2008 11:30 > > The datasheet for the National LM5022 has a good discussion of the various > losses for a switching regulator, which can also be applied to your case. I > modified their formulas for a more generalized case. > > Conduction losses are determined by RdsOn as: > > Pc = D * ( Id^2 * RdsOn * 1.3 ) where D is duty cycle, Id is RMS drain > current during conduction, and the 1.3 factor is for heating effect on > RdsOn. > > Switching losses are: > > Psw = 0.5 * Vg * Id * fsw * (tr + tf) , where Vg is gate voltage, Id is > drain current, tr is rise time, tf is fall time, and fsw is switching > frequency. > ----------You mean drain voltage of course, not gate voltage. > I found that, at 100 kHz, with a 42 watt load, switching loss can be as > high as 2.3 watts even with a 6 amp driver with an FQP90N08, and about 0.43 > watts with a 0.5 amp driver with STP35NF10, and 1.3 watts with a 0.2 amp > driver. A lot depends on the actual switching time of the MOSFET. I was > surprised that the FQP90N08 has 730 nSec rise and 330 nSec fall, so even a > 0.5 amp driver only brings the losses up to 3 watts. The conduction losses > are usually more than Psw, but not always. > > I made an Excel spreadsheet that calculates various losses for a typical > boost regulator. It may not be perfect, but it gives a pretty good idea, I > think. LTspice is probably better, but this is faster. You are welcome to > try it: > > www.smart.net/~pstech/MOSFET_Losses.xls > > Good luck, > > Paul
From: Paul E. Schoen on 14 Apr 2008 13:30
"gearhead" <nospam(a)billburg.com> wrote in message news:e6698ebc-2bcf-4c3d-90c8-01c41db0762a(a)f63g2000hsf.googlegroups.com... > >> >> The datasheet for the National LM5022 has a good discussion of the >> various >> losses for a switching regulator, which can also be applied to your >> case. I >> modified their formulas for a more generalized case. >> >> Conduction losses are determined by RdsOn as: >> >> Pc = D * ( Id^2 * RdsOn * 1.3 ) where D is duty cycle, Id is RMS drain >> current during conduction, and the 1.3 factor is for heating effect on >> RdsOn. >> >> Switching losses are: >> >> Psw = 0.5 * Vg * Id * fsw * (tr + tf) , where Vg is gate voltage, Id is >> drain current, tr is rise time, tf is fall time, and fsw is switching >> frequency. >> > ----------You mean drain voltage of course, not gate voltage. Well, yes, that does make more sense. The gate voltage is only for computing gate drive power. Actually I think the formula should be: Psw = 0.25 * Vd * Id * fsw * (tr + tf) The maximum power will be at 1/2 V and 1/2 I. The power during the transition might be half again that amount. I'll have to look at that and maybe try a simulation. But the above formula gives me values that make sense. In my case, gate voltage and starting drain voltage are both 12 volts, but the final output voltage is about 50 volts. When the MOSFET switches ON, the drain is at that voltage, so maybe it makes more sense to use that value. The LM5022 datasheet uses Vin for the calculation. That will make a big difference! Updated: www.smart.net/~pstech/MOSFET_Losses.xls Paul |