Re: Formula for minimum drive current for mosfet

Jon Slaughter wrote:

[snip]

I think you need to work it out. You don't seem to have any clue what your
talking about any just about any google search can prove you wrong.

You might be right in that the maxmimum instantanous power dissipated in
charging the capacitances is when V is half way min and max but what you
fail to realize is that this is independent on frequency. (I might linger
around Vpp/2 longer when switching slower but thats only if you take into
account a single switch)

The power loss isn't about charging the gate capacitance. This is tiny
compared to the losses due to Idrain current and Vds.

Draw a load line through the Id, Vds curves for a MOSFET. Draw equal
power curves on top of that.

For an ideal device, power dissipation is zero at either end of the load
line (where Id=0 or
Vds=0). In the middle of the load line, the MOSFET power dissipation is
maximum. The longer the MOSFET takes to travel through this region, the
more power it dissipates.

I don't give a crap about what happens instantanous but average. Sure
switching slower might mean that I'm at Vpp/2 longer but on average its the
same if I switch as a frequency twice as fast. Why? Doubt you even read this
but your at the Vpp/2 twice as many times... so it doesn't matter how fast
you switch w.r.t to this power dissipation. Although you end up dissipating
more power for other reasons when you switch faster.

Each switching transition will result in a fixed amount of energy (ideal
case) dissipated in the MOSFET. For the same switching frequency, the
faster you make the transition, the lower this is. Assuming a square
wave, of course.

So, point being, If I switch at 2x the frequency I might only be at Vpp/2
half as long but it occurs twice as often and hence cancels. THIS means that
it doesn't matter which frequency I switch at and your logic of "Switching
slower cause more power dissipation" is simply wrong.

By slower, we aren't talking about the switching frequency. We are
talking about how fast each transition (the slope of the sides of the
pulses) occurs.

In fact almost every site I have seen gives a formula where the power
dissipation is proportional to F and not inverse as you have said.

Yes, but it is inversely proportional to the speed of each transition.
To speed up the switching transition, you have to drive charge into, or
suck it out of the gate as quickly as possible. The average gate current
is proportional to the switching frequency, but the instantaneous gate
current determines the transition speed and therefore the MOSFET power
dissipation. You will run into the limitations of the uC (and associated
circuitry) instantaneous current before you have problems with average
current.

--
Paul Hovnanian paul@xxxxxxxxxxxxx
-----------------------------------------------------------------------
Procrastinators: The leaders for tomorrow.
.

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