Re: Formula for minimum drive current for mosfet

On 11 apr, 05:56, "Jon Slaughter" <Jon_Slaugh...@xxxxxxxxxxx> wrote:
<bill.slo...@xxxxxxxx> wrote in message

news:20e9d847-ea21-4aa1-bb24-4a2848d7a57a@xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx





On 11 apr, 01:33, "Jon Slaughter" <Jon_Slaugh...@xxxxxxxxxxx> wrote:
Whats the formula?

I = 1/2*F*Q*V?

Trying to figure out if I can drive

http://www.fairchildsemi.com/pf/FD/FDD8424H.html

with a uC directly? (I think it can supply up to 20mA or so)

V = 12V if I use pullup and 5V if not.

Check out the data ***. Figure 7 shows the typical gate charge
required to get the gate voltage up to a level where the part is
turned on - something like 10nC. The worst case total gate charge
listed earlier in the data *** is 24uC.

20mA s going to take 1.2usec to deliver that 24uC of charge - this is
slow switching by MOSFET standards, and you won't want to switch that
slowly very often, because if you do there is a real risk that the
switch will overheat.

Why is slower going to going to cause it to heat up? Its less current so
less heat (same amount of charge). In fact it probably would be better
because its spread out over time. (like, say, charging a battery at 1A for 1
year compared to 365A in one day. Same amount of charge but totally diffrent
results).

The maximum heat dissipation during switching occurs when the voltage
across the switch is half way between the rails. The longer the time
the drain takes to get from one rail to the other, the longer this
dissipation keeps on happening.

Spice can work it out for you if you model your load tolerably
accurately

My switching is at most 100kHz(its for motor control so anything about 20khz
should be ok but I'm going for about 50khz). I figure I need about 5 to 10
times this but really it shouldn't be that important (don't need it to be
exact).

Really though, Can you explain to me why a slower switching speed will cause
it to heat up more? It contradict's everything I know about transister
switches and switching speed.

It ain't what you don't know that screws you up, but what you think
you know that ain't so.

You need to develop a better understanding of what is going on while
the current through the switch moves from off to full on, and the
voltage across the switch moves from the full rail voltage to
practically nothing.

In the middle of this process the instantaneous power dissipation in
the swtich gets pretty high - you can work out exactly how high - and
since the process takes a finite time - 10 to 20nsec if you know what
you are doing, a microsecond or so if you cheapskate on the driver -
each switching opertion dumps a predictable amount of energy into the
switch junction.

Work it out.

--
Bill Sloman, Nijmegen
.

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