Working LED's from 54Volt

Status
This old topic is closed. If you want to reopen this topic, contact a moderator using the "Report Post" button.
Administrator
Joined 2007
Paid Member
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.
 
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
 
Administrator
Joined 2007
Paid Member
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.
 
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.
 
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

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)
 
Last edited:
Robert Kesh Said

“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) “

Thank you for the correction, I should have said.
Your power supply will have to be able to deliver .63W x 5LEDS = 3.15 Watts.

With a total current of.
5 LED x .012A = .06A

Also as Robert Kesh Said, it is difficult to determine LED brightness with current. First buy the LEDS you want to use then determine the resistor by using less than specified current and checking brightness with different values of resistor.
 
Status
This old topic is closed. If you want to reopen this topic, contact a moderator using the "Report Post" button.