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Vrystaat 13th November 2012 06:39 PM

Working LED's from 54Volt
 
Gooday all,
Can somebody please help with the easiest way to connect 5 LED's to work from 54 V DC. The LED's will be in parallel.

Regards
Jan

Elvee 13th November 2012 06:51 PM

Running LEDs from 54V is bad enough, paralleling them will make things worse. You should at least connect them in series, and use a 4.7K dropping resistor

Mooly 13th November 2012 07:02 PM

Will just add that many modern LED's are really efficient and the 10 ma or so that Elvees' 4K7 will give may be way to high for visual comfort.

Vrystaat 13th November 2012 07:05 PM

Thanks for the reply Elvee,
They need to work in parallel, I think I can manage a 12 V AC transformer, after rectification it is about 16V DC. Would this be better ? I can use a LM317 voltage regulator but is there an easier way ?

Thanks
Jan

Mooly 13th November 2012 07:10 PM

Why parallel ? 10 Led's (for example) at 10ma each is 100 ma in parallel but still only 10ma in series. And you can't really parallel them directly without equalising resistors so thats another 10 components needed. You have enough voltage available so why not use it to best advantage.

Perhaps if you can tell us what they are used as we could advise better.

Vrystaat 13th November 2012 07:24 PM

The five led's are connected to five relay's normal closed contacts, so all leds are working when the relay's are switched and are only an indication that the relay's are switched. This is a circuit for testing relay modules. The contacts are voltage free, that's why i want to power up the leds with a common supply.

Thanks
Jan

Mooly 13th November 2012 07:45 PM

It would probably help to see it drawn out but saying that if you just want an LED to light when the relay is active then an LED and resistor across the relay will suffice, one for each relay. If you wanted the LED to work the other way around and be lit when the relay was off then it can still be done using a transistor to sense the relay voltage and "short" the LED, or you can use contacts on the relay itself.... lots of ways to do it.

For simple on/off indication with a modern high brightness LED around 1ma or less will be plenty.

Vrystaat 13th November 2012 07:51 PM

Thanks Mooly, I'l just use a LM317 voltage regulator with a 12v Transformer to sort it out,
Thanks for all the help !

Regards
Jan

powerbob 13th November 2012 11:29 PM

Here is the design procedure for a 54VDC supply. I used some assumed vales to work it out.

Use .5 to .75 current rating of LED for long life and still good visibility.

Assume you are using a 20ma LED, run at 12ma.

Get LED voltage from spec sheet. Assume 1.5vdc
54vdc - LED voltage 1.5vdc = 52.5VDC

Each LED must have a resistor in series with it.

R = E/I

R = 52.5vdc / .012A = 4,375 ohm, use nearest standard value.

Power rating of resistor.
P = I x E
P= .012A x 52.5VDC = .63W
To be safe, use a 1 watt resistor.

Your power supply will have to be able to deliver .63W x 54VDC x 5LEDS = 170 Watts!!!
That is A LOT.

You can recalculate this again if you can get another DC power supply of a lower voltage. I would say no lower than 5VDC and no higher than 12VDC.

Robert Kesh 14th November 2012 12:17 AM

1 resistor per diode at 5K6 for about 10mA current.

you'll need 1W resistors

Or use 10K half watt resistors for slightly less brightness (brightness isn't linear with current, so this could be OK)

Or use 22K resistors of normal quarter watt type. Some super bright LEDs give a decent brightness at a few mA

Quote:

Originally Posted by powerbob (Post 3240555)
Your power supply will have to be able to deliver .63W x 54VDC x 5LEDS = 170 Watts!!!
That is A LOT.

Total wattage will be about half a watt per diode in first instance, 1/4 in second, 1/8 watt in last. (powerbob was thinking amps when he multiplied watts by volts)


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