Board logo

Help! Powering massive LEDs
tegwin - 12/2/10 at 12:47 PM

I am giving myself a headache with this one..

I need to power an array of 9

They are rated at 700ma, 3.2v

With an array of 9..thats just over 20w...


There are plenty of constant current 230vAC modules that can handle these LEDs... but for £60 thats a bit steep for a prototype....

How can I power the LEDs? Its just a simple lab type test for the time being, so I dont want to spend lots until I am satisfied that the LEDs will do the job required of them...

I dont have a suitable DC transfomer... so assume that the supply is 230vAC


Can anyone help?

[Edited on 12/2/10 by tegwin]


flak monkey - 12/2/10 at 12:54 PM

How are they wired? Series or parallel?

You can always just use a 6v lead acid battery with a voltage regulator

http://uk.rs-online.com/web/search/searchBrowseAction.html?method=getProduct&R=5339319

or:

http://uk.rs-online.com/web/search/searchBrowseAction.html?method=getProduct&R=5456391


Doofus - 12/2/10 at 01:05 PM

Not A voltage Reg on diodes

You need a current reg as Tegwin says.

Don't go letting the smoke out

[Edited on 12/2/10 by Doofus]


tegwin - 12/2/10 at 01:05 PM

Hmm...

That might work as a constant voltage for testing...

But ideally I need constant current at 700ma


I have not wired up the LEDs, so could be either series or parallel...


This is a very basic constant current reg... but I dont think it would work for 6000ma..

http://www.discovercircuits.com/PDF-FILES/CONSTANTCURRETLED1.pdf

[Edited on 12/2/10 by tegwin]


afj - 12/2/10 at 01:07 PM

Mobile phone battery 3,7 volts 1000mah same thing as a single cell lipo battery


graememk - 12/2/10 at 01:09 PM

its current opperated, not voltage, going to need something like a resistor in series there chap.

or just leave it how it is and have very bright LED's, for a short period of time.


[Edited on 12/2/10 by graememk]


Doofus - 12/2/10 at 01:10 PM

4 Ohm resistor on a 6v bat (or regulated supply/transformer) would do.
2w though

[Edited on 12/2/10 by Doofus]


tegwin - 12/2/10 at 01:33 PM

Hmmm...

If I use a 12V battery....

Connect the LEDs in three parallel rows of 3 series LEDs... Then each series row would need a 2.2ohm 1W resistor

not an ideal solution... But would be the cheapest way for testing perhaps..

Solution 0: 3 x 3 array uses 9 LEDs exactly
+----|>|----|>|----|>|---///----+ R = 2.2 ohms
+----|>|----|>|----|>|---///----+ R = 2.2 ohms
+----|>|----|>|----|>|---///----+ R = 2.2 ohms

The wizard says: In solution 0:
each 2.2 ohm resistor dissipates 1078 mW
!! the wizard thinks the power dissipated in your resistors is a concern
together, all resistors dissipate 3234 mW
together, the diodes dissipate 22050 mW
total power dissipated by the array is 25284 mW
the array draws current of 2100 mA from the source.


hobbsy - 12/2/10 at 01:34 PM

If you're going to be messing around prototyping stuff just get yourself a bench power supply with current limiting.

Not that cheap but its useful for a load of stuff.

I'm sure there are some second hand units on eBay etc?


02GF74 - 12/2/10 at 03:06 PM

you saying you want the leds powered off the mains?

or the mains is there to provide a power, whcih may need some additional conditioning?

and do you want to power one or how many at the same time?

there are a number of ways to do this.

you can use a drop resistor with the mains - I did this with a teeny 5 mm LED but resistor can get hot.


these 700 mA led sound like LED emitters.

the most efficient way to drive these is by using a constant current driver - read my atcile on DIY LED lamps.

look on ebay for teap0t42 items,

see


these take a input voltage 7-24 V; so if you have an RC toy, use the battery from that or use car battery.

else get a mains power supply for that voltage range.

BTW I have some less 3 W (that one is 5 W) drivers that I don't need - make me an offer .....


02GF74 - 12/2/10 at 03:11 PM

quote:
Originally posted by tegwin
Hmmm...

If I use a 12V battery....

Connect the LEDs in three parallel rows of 3 series LEDs... Then each series row would need a 2.2ohm 1W resistor

blah blah ...


not sure how you get 2.2 ohm

(12 v - 3 x 3.2) / 700 mA = 3.4 ohm

would dissipate 2.3 W .


oh, one more thing, these LEDs are gonna chuck out some heat so will need heat sinking - are they star LEDs perchance?



BTW - 700 mA LEDs are not the birghtest/high poer.

you could, for a couple of years at elast, get 1 A LEDs, seoul P4 emitters - these wold need 5 w Driver; your need 3 W so my post above needs correctiong.

there are even more powerful ones now.

[Edited on 12/2/10 by 02GF74]


MikeRJ - 12/2/10 at 05:38 PM

quote:
Originally posted by tegwin
Hmmm...

If I use a 12V battery....

Connect the LEDs in three parallel rows of 3 series LEDs... Then each series row would need a 2.2ohm 1W resistor



The main drawback with this is although you minimise power dissipation in the resistor, you also make the LED current very sensitive to voltage. This makes sense if you consider an ideal current source has infinite resistance.

You can use a simple voltage regulator such as the LM317T + a suitable resistor as a current source, but you will need a suitable DC supply. Don't even think of connecting stuff like this direct to the mains with simple dropper resistors, the power dissipation will be horrendous

e.g. the best case of all diodes in series gives a forward voltage of 28.8v. With a nominal mains voltage of 230v, that gives you 201.2v at 700mA to dissipate, or around 141 Watts! The traditional approach for these power levels was to use mains light bulbs for the dropper resistors, but the switch on surge current could easily destroy your LEDs.


rf900rush - 12/2/10 at 07:41 PM

Sounds simular to the led drives I made for a brother.

LED Bike lamps using a 3 amp drive IC controlled but a PIC microcontroller.

Will drive any constant current upto 3Amps.

this type of PCB should be able to drive 3 in series (11v approx) and 3 set of these in parallel (2100ma) (9 in total).



input 7-24V.


LEDLAMP1
LEDLAMP1



used for this type of lamp

Lamps
Lamps



For simple tests a power resistor to set the current.

Martin