Intruductions to (mos)FET

Specifically, what you are describing is a "high side switch".

Chris

"jodleren" schreef in bericht news: snipped-for-privacy@l30g2000vbn.googlegroups.com...

For instance:

BTW. A BUZ72 is an N-channel FET. To switch the high side, a P-channel FET will be more suitable.

petrus bitbyter

If it's an N-channel FET you need to hook the drain to your power supply, the source to your equipment, and raise the gate voltage above the source voltage by 10V (5V if it's a "logic level" FET). That means you need a driver that can bring the gate voltage above your power supply voltage.

If it's a P-channel FET you need to hook the source to your power supply, the drain to your equipment, and pull the gate voltage below the source voltage by 10V (5V if 'logic level'). That's a lot easier to do, but P-channel FETs are inherently less conductive than N-channel ones, so a FET that'll handle the current will tend to be more expensive.

Thank you for your answer. Would you consider a PNP darlington rahther and a P-channel FET for turning on/off power? Or pulsing it?

I wonder, that with my BUZ72, that I can swicth 0W - let me guess: Source to 0V, Drain to load (new 0V for remaining system) and on the other end to +V, and gate by resistor to +V To turn it off, a transistor between gate and 0V. Do I get it right? Turning off 0V, as it all happens in my system. It will not be affected by the outer world (the load is the rest of the world).

How do switch mode supplies work, they "have to" swicth the +V, right? P channel? FET or NPN ?

I dont like the 5V drop, as it would fry off 5W @ 1A, right?

Sorry for asking,

WBR Sonnich

I mean DC-DC converters, such as buck, and and buck boost.

WBR Sonnich

Darlingtons are slow, and they have a considerable voltage drop from collector to emitter -- usually at least 0.9V. So they burn up power that you'd like to use in your load, they're really not much easier to drive than a MOSFET, and they're sloooooooow. Other than that, they're fine devices.

Mostly. The transistor will only pull the gate down to 0.2V, which may or may not turn the FET off completely. To be on the safe side you'd do something like this:

The diode drop between the NPN and the N-FET gate makes sure the resistor can pull the gate all the way to ground. Or just use a little N-FET instead of the NPN.

It depends on the topology of the supply. Boost converters use low-side switches, which makes life easy. Buck converters and H-bridge drivers usually work by generating a separate power rail for the upper side gate, although some lower-power circuits use P-FETs.

What 5V drop? Across the FET gate? No current flows there in static operation -- it 'looks' like a capacitor to the driving circuit.

That approach could work fine in many instances, but the gate bleed resistor will have to be quite large, so the FET won't turn off very fast due to Miller capacitance. I'd want another diode in antiparallel if any significant power is being switched.

Cheers

Phil Hobbs

Good point. Or a cap in parallel with the diode. Or something.

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I don't think a cap does it, you've gotta get 'real' charge off of there.

But thanks for the circuit fragment!

George H.

There's no 5V drop. P-channel MOSFET switched +ve rail to load with negative on the gate, wrt the +ve rail. A switching circuit can run an N-channel MOSFET by bootstrapping the gate driver to supply +ve gate drive above the +ve rail.

Depends what you're making, if you can bootstrap, lots of answers :)

Grant.