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Formula for minimum drive current for mosfet

J

Jon Slaughter

Jan 1, 1970
0
Ok, I finally found the equations. All my calculations say that the
switching power is negliable compared to the power dissipated by R_ON. (Only
at high frequencies does it start becoming significant)

P = I^2*R + 2*a*V*I*t*F

which can be rewritten as

P = I^2*R + 2*a*V*Q*F

and a depends on the load type(1/2 for inductive and 1/6 for resistive)

So infact for my case I need only like 200mA to switch the load fast enough
for 50khz(so like 99% of the square waveform is constant).
 
Did you use a resistors in each base, or just tie them together?



I just tied them together. It's basically two emitter followers. They
should never be both on at the same time, but if one is slower than the
other, I guess it can happen, and did. The simulation looked OK.

Paul

There is charge storage in the base, so you get cross conduction if
switched fast.
 
P

Paul E. Schoen

Jan 1, 1970
0
Jon Slaughter said:
Ok, I finally found the equations. All my calculations say that the
switching power is negliable compared to the power dissipated by R_ON.
(Only at high frequencies does it start becoming significant)

P = I^2*R + 2*a*V*I*t*F

which can be rewritten as

P = I^2*R + 2*a*V*Q*F

and a depends on the load type(1/2 for inductive and 1/6 for resistive)

So infact for my case I need only like 200mA to switch the load fast
enough for 50khz(so like 99% of the square waveform is constant).

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.

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
 
G

gearhead

Jan 1, 1970
0
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.
 
P

Paul E. Schoen

Jan 1, 1970
0
----------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
 
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