Mosfet controlling a power supply

I'm building a circuit of which part of the secondary functions is to control a high voltage power supply using a mosfet. I've looked around for anybody who has done a similar sort of thing, but typing 'mosfet power control' into Google yields pages of switch mode supply schematics. I've attached what I think is the correct form of the circuit, but as I'm only using an arduino, I am driving the gate to a high of 5V only, which I don't think will be enough, as the datasheet for the IRFZ44N specifies that 10V is required to have the 17.5 milli-ohm on resistance (but at 10A drain current). I do have access to +12v on the board, so could set up a transistor pre-drive stage for the fet if this becomes a problem. The power supply I will be driving is 5W output, and specifies a 85% efficiency when driven @ 12V, so 6.25W (on the low eff. side (80%)) input power. As I'll be driving less than 2A, will the 5v vs 12v gate voltage make a difference?

Power control.png

Please provide information about the power supply....

You've chosen a low-side switch here - is that OK - the ground side of the
PSU will be at 12V if you do this.

Yes, if you are driving a MOSFET with 5V it must be logic level. That means
it has a rating for the guaranteed Rds(on) for 5V or 4.5V gate drive. Hoping it
just works is foolish as device thresholds vary with temperature age and device,
the datasheet tells you what will work for any device at any temperature across
its whole lifetime.

However if you opt for high-side switching with a p-channel MOSFET you don't
need logic level as you'll be driving its gate with a level-shifting circuit (typically
a small NPN transistor) that is at 12V.

Thanks for your help, I've looked around for some logic level MOSFETs, these ones seem perfectly suitable, and :
FDD24AN06LA0_F085: Intelligent Power and Sensing Technologies | onsemi Vgs=5v, Rds = 20mOhms
or STP55NF06L: http://www.st.com/web/en/resource/technical/document/datasheet/CD00002690.pdf,
I realise now that the L suffix on the end refers to logic level, I take it I'd waste a fair bit of power if I tried to turn on the mosfet rated for Vgs=10V with only 5V.

You've chosen a low-side switch here - is that OK - the ground side of the
PSU will be at 12V if you do this.

What do you mean by this, when the MOSFET is on doesn't the drain get tied to source (albeit with a tiny Rds), and when off isn't the Rds infinite?

edit: The supply which I will switching is this one: http://www.ebay.co.uk/itm/Nixie-tubes-Magic-eye-tubes-power-supply-module-kit-/261169220982?pt=Vintage_Electronics_R2&hash=item3ccee60d76, its a 35kHZ switch mode supply, 180V, 5W output. Input 10-18V, I have it running from 12V at the moment.

I researched and understood what you said about low side switching, so I changed the entire design around to use a p-type (I had to use a transistor drive as I couldn't find any suitable logic level p-type MOSFETs, do you think it will work now?

Power control2.png

Yes, that's it - my point was that the ground of the power supply would (sometimes) be
at +12V, which can be an issue (for instance if the frame is metal and grounded by
other means, deliberate or accidental).

high-side switching is usually prefered because of this.

It would be useful to know what chip is used on the power supply board.

Many switchmode chips already have a "disable" input to switch them on or off.

Many switchmode chips already have a "disable" input to switch them on or off.

That was my initial idea before I started looking at MOSFETs, but the chip is the MC34063 that unusually doesn't have an output enable pin :frowning: , the only time I have actually needed one as well...

edit: Do you think the supply will generate back emf when I unload it, and will I need anything more complex than a diode to cope with/absorb it?

woozle:
Do you think the supply will generate back emf when I unload it, and will I need anything more complex than a diode to cope with/absorb it?

Rather the opposite actually - it will have a (large) capacitor across the input which will present an initial heavy load as the FET switches on.