On 9 Dec 2015 at 1:35, Neil wrote: > What I'm not sure of is why putting a schottky across the MOSFET would=20 > generate so much heat (as I noticed previously). With this, the energy=20 > should be dissipated across the inductor, the schottky, the sense=20 > resistor and the +12V battery... wouldn't it? Yes, but probably not in the way you're expecting. > If this is so, then the=20 > power dissipated in the schottky should be the same as if it were=20 > directly across the inductor, and actually for a shorter time. I really= =20 > need to try this as a direct comparison to directly across the inductor=20 > to see what really happens. In the case where the Schottky diode is across the inductor, when the MOSFE= T=20 turns off the inductor current continues to flow in the same direction, the= Schottky=20 diode becomes forward biased (minimal voltage drop, minimal dissipation) an= d=20 current circulates in a loop through both components as it decays slowly. In the case where the Schottky diode is across the MOSFET, when the MOSFET= =20 turns off the inductor current continues to flow in the same direction, the= Schottky=20 diode becomes REVERSE biased (avalanche breakdown, like a zener, relatively= =20 high voltage, high power dissipation) and current circulates in a loop thro= ugh all of=20 the components (now including sense resistor and power supply but not inclu= ding=20 the MOSFET) as it decays relatively quickly. Probably not the desired/inten= ded=20 operation you want the Schottky to perform. This is kind of what I was trying to say earlier on, still not sure if I've= explained it any=20 better, hope this helps. The MOSFET or the Schottky could go into avalanche, which one depends on=20 which has the lower breakdown voltage. The one that does will be the one th= at gets=20 hot :-) --=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 .