Re: How to quickly turn on/off N-FET switching high side of 55V?

Michael wrote:
Winfield wrote:
Michael wrote:
Winfield Hill wrote:
Michael wrote:
Hi Winfield - a couple more questions for you: What is the purpose of
the resistor across the bottom optoisolator? Also, you mentioned that
the IRF1407 is a large die mosfet. How can you tell this? The reason I
ask is that I am interested in using a FET with a smaller on
resistance, but I want to make sure I know what to be looking for.

I didn't answer this, sorry. You can estimate relative
MOSFET dies sizes by comparing power dissipation ratings
and gate capacitance Ciss.

Wouldn't power dissipation ratings also be highly dependent on the
package that the FET was in?

No. At least not very much. You'll find that dies mounted
in TO-220 packages have the same thermal resistance as the
same dies mounted in a much larger TO-247 package. And as
the smaller D2-PAK package. It's the die's area footprint
that matters.

Well, even so, are you thinking well about the bigger
picture? I mean, is this just a bunch of batteries to
gang up and help run a motor? What's wrong with just
using a diode-OR on each battery? Is it necessary to
isolate the OFF batteries? If so, then you'll need two
MOSFETs back-back (source-to-source) to insure the ON
battery doesn't work to charge the discharged ones.
What about fuses, current-limiting, status indicators,
etc. Does the motor have a PWM modulator or something,
you don't just always give it the full juice, do you?
What about current monitoring, etc.?

So this board is a power distribution board. It carries power to a
number of other boards that run various motors through all sorts of
current loops. It also powers some digital systems that take in the
high voltage and step it down to something more usable.

The reason I want be able to switch on and off the power sources is
that I want to be able to protect them from undervoltage and
overcurrent situations. They are li-polys, so if I drain them too much
or if I pull too much current they die. If I just used diodes I would
lose that protection. I also want it high side switched as we have had
problems due to low side switching - with fun ground loop problems
like through external VGA cables and stuff like that. Also, the hope
was that using FETs would give me a bit more efficiency, though I
suppose that is not a huge gain. Note that there will be 3 power
sources - two batteries and wall power. The wall power input does not
need undervoltage protection, obviously (except for it being too low
to power anything - at about 9VDC).

I don't understand why I would need two mosfets, or even how that
would work... I mean once the FET is off (VGS < VT)current can't flow
through it, or at least that was my understanding. Errr wait -
intrnisic body diode... Argh. It has a 1.3V forward voltage according
to the IRF1407PbF data***. I see your point now. What happens to a
FET when it is reverse biased? It looks like the body diode turns on
at about -1.3V and then can handle a ton of current, but what about
between VDS = -1.3V - 0? I'm assuming it's just off. That's not really
relevant, but now I'm really curious as I've never really heard the
properties of N-FETs discussed with regards to VDS < 0.

They conduct current either way if the gate voltage is high
enough over the source. If the reverse-direction current is
high enough the drop across Rds(on) will increase to the
point where the anti-parallel intrinsic diode also conducts.

But why put two FETs source to source?

To make a fully-bidirectional four-quadrant switch, that is
completely OFF in one state and completely ON in the other.

To me it'd make more sense to put a schottky diode in series
with the FET. I'd probably put it on the drain side of the
FET just so as to increase VGS a little. I mean - if you put
two FETs source to source wouldn't you see a 1.3V drop across
one of them?

When the FETs are both ON you get 2*Rds(on), not 1.3 volts.

You may be better off with a diode - assuming you want to
switch from one source to the other without a power break,
you'll need a diode-OR action. An ON/OFF two-MOSFET switch
could conduct huge currents from a high-voltage battery into
a low one.

Before leaving the topic, it's worth noting that MOSFETs
as active rectifiers are better and cheaper than Schottky
diodes. For example, a 60CTQ045 dual 45V Schottky will
drop about 500mV at 30A, whereas an IRF1405 55V MOSFET
will drop only 210mV at 30A (both warmed up to Tj = 100C).
The diode costs $2.78 qty 100, vs $2.64 for the MOSFET.

All that said, if your batteries are all tied together
and share the ground system, there's a MUCH simpler
approach that can be used. We can dispense with all
the optically-isolated stuff. Like I said, generally
it's more fun on usenet s.e.d. to talk about whatever
interests us most at the time, as conversation goes,
rather than what may actually be best for a project.

The batteries all share a common ground. What simpler system were you
thinking of? I would really like to keep it high side switched, if
that was your idea.

I'll answer later. Meanwhile look at Fred's ASCII drawing,
as a start. I'd change a few things, but the basic idea
is to use a capacitor to deliver the MOSFET operating
voltages, and transistor level-shift and driver stages.
.

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