BetterSense

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Can i use a 120v panel indicator light on 240v if i put a diode in series?

I can't imagine why not but i've been striking out lately with AC circuits.

FastAndy

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Not sure what the diode would do for you in this situation. My guess is that the indicator would have a significantly shorter life if it is run on 240 instead of its intense 120.

MrNatural

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You may be thinking of resistor rather than diode?
If you have neutral leg (versus ground leg) you can use one leg of the 240V with it to run your 120V stuff.

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AllanMar

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Depends on the circuit of the LED.

In most of the cheap ones (which are just a resistor and a capacitor) you can't do this. Even though adding a resistor would limit the current but not the reverse voltage which is likely to burn out the LED, you would have to adjust the capacitance to reduce the voltage that the LED sees.

BetterSense

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I never said anything about capacitance? I never said LED? I am confused. My indicator is incandescent. What kind of LED comes with a capacitor?

The diode is to convert 240V into 120V.


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I tried to do that but the indicator light stays on even with my SSR is off.

AllanMar

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Oops, I assumed it was an LED, LED panel lights have become the norm. Inside most cheap LED pilot lights is a resistor/capacitor circuit (there are other ways also) to allow then to run off higher voltage AC. An incandescent MAY survive, but will get brighter and hotter.

A diode alone is not going to convert 240V AC to 120V AC. Since it is an incandescent, a resistor as stated above would work as long as it was of suitable wattage. You'll need to double the resistance of the bulb. This isn't a great way to do this though, for what its worth your probably better off getting a new indicator.

ajdelange

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For an incandescent lamp all you need is a series resistor. The resistor should have the same value as the lamp when it is on. To find this you would have to measure the current through the lamp when it is on (connected to 120). Divide the voltage (120) by the current and get a resistor of that size. If, for example, the lamp draws 10 ma the resistance is 120/0.01 = 12000 Ω. Multiply the voltage by the current: P = 120*.01 = 1.2 W is the power dissipated by the lamp. The same power will be dissipated by the resistor which should be sized to be twice the power dissipated i.e. in this case you would want a 12K 2W resistor.

If you connect a 120V bulb across 240 V it will burn 4 times as brightly as it would connected to 120V and its life will be short indeed. If you connect a diode in series it will only conduct on alternate half cycles so that it will burn twice as brightly as normal. The life will still be dramatically shortened.

For a LED lamp find out what size resistor is in there (or connect to a power source and measure the current calculating the resistance as for the bulb) and add another one of the same size in series. Adding another diode in series will insure that the PIV of the LED is not exceeded but you will have to put large resistors in parallel with the diodes in order to share the reverse voltage across them.

AllanMar

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As I said before, your best bet for LED's is just to change the resistor & capacitor. Adding a diode will cause them to flicker. Here is a pic of a cheap 120V pilot LED disassembled (1k res and 0.1uF cap).
And the bottom one is a typical schematic (they usually cheap out and use reverse LED's instead of a diode, also makes flicker less noticable)

ajdelange

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That could be a real problem for you if you can see 60 Hz flicker.

No need to fiddle with the cap. Just measure the AC current through the assembly, as I said above, and calculate the impedance. Then add another resistor of the same size. The current will stay about the same. In the previous post's diagram the measured current at 110 would be 19 ma from which you would conclude that the impedance is 5700 Ω. So add a 5.7K resistor. The impedance now becomes 8760 and the current at 220 is 25 ma. Close enough for government work. If you really want to limit the current to what it was at 120 then add sqrt(3) times the determined impedance or, in this example, 9.9K, which would result in a current of 18 ma at 220 (if I did all the math right).

ajdelange

Recipes 

Forget what I said in #7 about PIV protection with a series diode. Yes it will work (and no it won't cause flicker) but it's a silly way to do it. All you have to do is put another diode (LED or conventional) across the led in the opposite polarity (anode to cathode and cathode to anode). The led will never see more than a couple volts in the reverse direction.