Rectifier help

[johno_71]

Hello all,
I am new to the forum and need a little help. I need to illuminate some leds. The problem is the only power i have available is 3 phase 240 volt with NO neutral. I have tested a polyphase bridge rectifier to convert from a/c to d/c and then doing a big resistor to limit the current to the leds. This method creates a lot of heat. Can I do the same thing and then use a zener diode to drop the voltage down to 12 to 24vdc. I don't know if the zener will work. The other problem is, I am extremely limited on space and heat is an issue. Does anyone know of a compact transformer/rectifer package for an application like this?
Any suggestions would be greatly appreciated.
Thanks,
John

 

[crutschow]

A zener will not reduce the power dissipated.

To limit the current without dissipating power you could use a capacitor in series with the input(s) (not output) of the bridge rectifier. The impedance of the capacitor is 1/(2*pi*f*C) where f is the power line frequency. Choose the cap size to limit the current to the desired value for the LEDs your are using. You should also add a small resistor in series to limit the surge current due to the capacitor charging when you energize the circuit, otherwise you may blow the LEDs.

The capacitor has to be non-polarized (such as a film type) with voltage rating of at least 400V.

Caution: The above circuit exposes you to 240 Volts so observe great care when working on such a circuit.

 

[johno_71]

Could you help me out a little more. My power is 240v 60hz. I am a little unclear about some of the things that need to be plugged into the formula. Here is what I think i understand so far: 1/(2*3.14*60*?) What is C? Have i done this correctly so far? I am trying to give my leds 20 milliamps
Thanks for the education,
John

 

[KeepItSimpleStupid]

Not knowing anything about your LED application I can say the following: You can use the voltage between phases. a resistor and a diode. You have to pay attention to the peak current specs of the LED's. I do believe there are indicators with small transformers attached. LED illumination for industrial switches typically operates on 24 VAC/DC.

 

[johno_71]

can you give me an example of a schematic?
Thanks,
John

 

[KeepItSimpleStupid]

208a---(-)diode---resistor----(-)LED----208b

Diode should be 400 PRV or more e.g. 1N4004
Resistor ~= (208*1.414)/50e-3

Note: I'm sizing it for peak current of 50 mA an neglecting the drop of the LEDS.
W > (I^2)*R; where I is the current (this is non-exact, but will work)

 

[Reloadron]

Given a choice I would isolate from the 240 VAC using a control transformer similar to these. That way you are isolated from the 240. The transformer is connected 240 VAC phase to phase as you seem to have a delta configuration. Rectify and filter the transformer output for your LED power. Another option is a 240 / 120 control transformer to drive a simple wall wart for your DC. This sounds like an industrial application based on the 3 phase 240 VAC delta. Personally (and just my take) I would not want my secondary LEDs or whatever linked directly to mains power. That is why I suggest a control transformer. They are small, cheap and offer mains isolation.

Ron

 

[crutschow]

Could you help me out a little more. My power is 240v 60hz. I am a little unclear about some of the things that need to be plugged into the formula. Here is what I think i understand so far: 1/(2*3.14*60*?) What is C? Have i done this correctly so far? I am trying to give my leds 20 milliamps
Thanks for the education,
John

C is the capacitance.

For 20mA and 240V the capacitor needs an impedance of about 240/20mA = 12k ohm. Plugging into the equation gives C = 1(2*pi*60*12k) = 0.22uF. To limit surge current to 10 times the nominal, add a 1.2k ohm resistor in series with the cap.

But as Ron recommended, it's much better to use a transformer for safety, if at all possible.