I have an induced voltage of 9-14VAC from a 200+ foot cable run of 120VAC. In other words, the 7-wire cable has 120VAC present, which is inducing somewhere between 9 & 14V in one of the wires that comes from an open contact. This is enough to barely turn on a 15V LED pilot light. (I know we should have used twisted-pair or shielded cable.)
I'd like to know if a capacitor to ground could be used to bleed off the induced voltage, when the circuit is open. And what would happen when the circuit is closed, providing 120VAC to the LED light fixture. The manufacturer says the fixture does not contain a dropping resistor, but that the 15V LED bulb doesn't care.
If the capacitor would work, how do we figure the size of the capacitor?
Well magnetic induction is a common problem for long wires. You could use bleeding resistor. Do not think that the capacitor will work.
A bleeding resistor is used to discharge a capicitor. I could use a drop-down resistor to reduce the voltage to the bulb, by the amount of known induced voltage. Normally, a swtich supplies 120VAC to turn the 15V LED bulb on. But with the swtich open, the bulb turns on (slightly) with the induced 14VAC. So, given that, what size resitor is put in series with the bulb? Does it depend on the magnitude of the induced voltage?
A capacitor charges up for a direct current, but starts passing current for an alternating current (proportional to frequency). So, if I put a capacitor in parallet with the LED bulb, I think it would pass enough current to bleed off the induced voltage when the 120VAC switch is open. But what happens when the switch is closed, supplying 120VAC across the capacitor?
Hii... Sorry for the bleeding resistor, Yeah you are right, it will be needed a load resistor.
I am not sure about the capcitor. In some cases parallel capacitor works, but it will create a LC filter and the light may start to blynk. You need to reduce the value and see where the bulb actually constant. Try using a 1uF or 2.2uF and reduce the value.
However, For the load resistor it may draw current as aflux source from the actual one that carrying the signal, I am not sure (Think like a poor transformer).
Best way is to check if the conductor having twisted pairs. Use that. Or change the cable.
Sam
PermalinkI was just trying to preview my post, so I'm confused as to why it requires that I comment. Also, there seems to be no way to edit my post before I submit it. I was going to say that I was surprised there is no dropping resistor in the fixture.
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