SMPS Inductor Protection

trad3mark · Post by **trad3mark** » 15 Feb 2010, 23:34

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

earthtonesaudio · Post by **earthtonesaudio** » 16 Feb 2010, 02:32

No. The reverse biased diode is there to protect the switching transistor, NOT the inductor.

To keep from burning out the inductor, use a lower supply voltage, higher current rated inductor, and/or reduce the on time.

trad3mark · Post by **trad3mark** » 16 Feb 2010, 12:34

http://www.coilgun.info/theoryinductors ... ckback.htm

this suggests otherwise, and would make perfect sense as to why i lost 4 inductors... they're a much higher rating than the voltage and current going into it, so the theory on that page would make sense of what's going wrong...

earthtonesaudio · Post by **earthtonesaudio** » 17 Feb 2010, 13:42

I'm pretty sure the page you linked to is saying the exact same thing I was saying...

Electronic devices controlling inductive loads can be easily damaged, especially the component that "breaks down" in order to satisfy the inductor's craving for continuity of current.

Consider the circuit shown at right. The switch is initially closed, and current I is flowing through the inductor. When the switch is opened, the inductor "tries" to keep current flowing from A to B, as it had been. That means that terminal B goes positive relative to terminal A. In a case like this it may go 1000 volts positive before the switch contact "blows over". This shortens the life of the switch and also generates impulsive interference that may affect other circuits nearby. If the switch happens to be a trasnsistor, it would be an understatement to say that its life is shortened; its life is ended!

The diode is to protect the switching transistor, not the inductor. Are you sure you have calculated the voltage/current requirements of your inductor correctly?

trad3mark · Post by **trad3mark** » 17 Feb 2010, 17:06

ok so i'm getting somewhere.

the input into the SMPS is 9.4V @ 300ma (it's the normal boss regulated PA). So now with a new inductor, the SMPS is outputting 120V. grand. IF i wanted to say, lower the current in from 300ma to something a little lower, just to protect the inductor, how would i do that easily?

trad3mark · Post by **trad3mark** » 17 Feb 2010, 17:41

actually, it's 500ma. So a small resistor, say 200R between the 9V and the inductor should do it, right?