Hey all,
So working off a nixie SMPS, i've burnt out 4 inductors, mainly from switching it on/off to adjust the circuit it's feeding into. Anyway, i read a short article about using a diode to protect your inductor from burning out. In a nutshell, it gave details about how when you switch it on/off, the current can sort of surge it and burn it out. The article suggested putting a diode like a 1N4004 parallel to the inductor in such a way that when power is flowing INTO the inductor, the parallel diode would be in reverse bias. This then means that when power is disconnected, the left over power will go through the diode into a capacitor instead of through the inductor, burining it out.
So, even if i got the theory of the above bit mixed up (cos i can't find the article again lol!), onto my SMPS. On one side of the inductor, there's +9v in, and on the other side, a UF4004, which goes to where the HV (175V) outputs. So, am i correct in saying that if i put a 1N4004 in parallel with the inductor in such a way that the Cathode is at the same point as where the 9V comes in and the anode is on the UF4004 side, i can prevent my inductor from burning out?
OR... is there an easier way to save my precious little inductors?
cheers.
tm