As always, remember to multiply the current by 10 volts to see how much power the diode gets to dissipate. There's no magic in there and it's really doing the same thing the linear regulator would have anyway. Only it won't drop out below its set voltage, so maybe there's an advantage to doing it your way. On Fri, May 24, 2013 at 10:31 AM, Spehro Pefhany wrote= : > At 03:53 AM 24/05/2013, you wrote: > >Hi all, > > > >I'm trying to power some MOSFET drivers (which can be powered from up to > >18V) with a battery pack(s) whose voltage may be anywhere from 16-24V. I > >can opt for the "correct"(?) design and just use a linear regulator, but > >I don't want to add a whole regulator just for the MOSFET driver. Is it > >considered acceptable engineering practice to use a Zener diode in > >series with the power supply rail, reverse biased, to act as a 10V (for > >example) drop to power the IC? That way, the IC supply voltage will be > >from 6-16V, which is acceptable. It seems I'd also need something like a > >10k resistor to GND in parallel with the IC to maintain a high enough > >current through the Zener to keep it in the intended reverse voltage > range. > > > >SPICE simulation says it's okay, but I have a feeling there's something > >I don't know... > > Sounds like you've got the bases covered. A ~10V zener has a reasonably > sharp knee, and your 10K makes sure you're around it. > > Note that using a zener increases the voltage range from to 1.5:1 to 2.7:= 1, > plus tolerances on the zener, but if you're okay with that, it should wor= k. > > The zener will conduct as a diode in the forward direction, so you'll > probably kill the MOSFET driver instantly if the battery gets reversed. > > --sp > > -- > http://www.piclist.com/techref/piclist PIC/SX FAQ & list archive > View/change your membership options at > http://mailman.mit.edu/mailman/listinfo/piclist > --=20 http://www.piclist.com/techref/piclist PIC/SX FAQ & list archive View/change your membership options at http://mailman.mit.edu/mailman/listinfo/piclist .