running led project


Thread Starter

ravi chandran

Dear sirs,

Could anybody help me to overcome this problem.
I have made a chaser light project. the leds were mounted at 5 words.
Total of LED used were 3800pc. The problem is i cannot use the transistor to drive the leds. In this case i use relay to switch on the leds. I have purchased a 6V, 120 A transfomer for this project.
My worries is that how could i limit the current to all the leds. A word have appr. 750 led and were connected parallel. I am worried if i use the powersupply directly the led, the led might burn. how could i overcome this.Please advise me.

Thanks in advance,
Ravi Chandran

David Wooden


Add a resistor in series with each LED. Assuming that your LEDs draw 20mA each, and with a 6V source, and assuming that LEDs (diodes in general) always present a .7V drop, your resistance should be: Ohm's law: Resistance = voltage / current Voltage = 6 - .7 = 5.3 Current = 20ma = .02A Resistance = 5.3 / .02 = 265 ohms
The closest standard value is 270 ohms (Way close enough; the value isn't very critical.) Power requirements: P = IE (Power in Watts = Current(Amps) * Voltage 6V * .02 = .12W = 120mW, or
approximately 1/8W. Good design calls for 100% overload, so choose 1/4W resistors ( They're probably cheaper anyway.)

Hope this helps.


Johan Bengtsson

This is the way to go, but LED:s do generally not have the 0.7V drop a normal diode have but rather something around 1.6-2.4V depending on type (you must check this if you want to be sure of the current).
Generally always round the calculated resistor value upwards, LED:s break easily when the current is too high.

The more current the brighter light, if 270ohms give you light enough use that, you won't have 20mA but rather something around 13-16mA. This also depends on how stable your DC voltage is.

You cannot use the transistor, I suppose that means you turn on/off several LED:s at the same time. If that is the case can you connect some LED:s in series instead of parallel? if you for example coul pair the LED:s two and two in series you would not need more than one resistor for each pair instead of one resistor per LED.
In that case calculate (6-2*Vf)/If to get the resistor value. Be sure to have at least 1-1.5V for the resistor or the LED:s will probably break at too small variations in the DC power source.
Pairing the LED:s would also bring down power needed and need a smaller relay. If you could have a higher DC voltage and put more LED:s in series you can get down the number of resistors even further, for example by feeding 10 diodes in series with a Vf of 2.1 V from a 24V source would give you:
R=(24-10*2.1)/If, R=3/If, for If=20mA this means R=150ohm
The DC source have to be more stable the less voltage you put at the resistor.

/Johan Bengtsson

P&L, the Academy of Automation
Box 252, S-281 23 H{ssleholm SWEDEN
Tel: +46 451 49 460, Fax: +46 451 89 833
E-mail: [email protected]

If you are making a display, and you don't want to have a discrete driver for each LED, couldn't you multiplex the display? If you are, you should be able to do current control through duty cycle. But I don't think this is feasible with relays!

It's difficult to answer your question without knowing a little more. There have been several good posts, so perhaps your question is already answered.

I did build something like this with several thousand LED's in a matrix, and I didn't need current limiting resistors because of the multiplexing arrangement; the drop of the LEDs plus the switching transistor were enough to limit the current to an acceptable level. The LEDs are able to take a lot more current, to bring the apparent bightness up to a bright level
during the OFF time. Of course, if the multiplexing got stuck, LEDs popped! Perhaps you could use some combination of current limit plus multiplexing.

Hope this helps a little.


Willy Smith
Numatics, Inc.
Costa Rica

Jeff Frischman

have you thought about using 7 segment display drivers instead of relays to drive your leds ? generally they can sink or source about 25ma per segment, using the display driver you won't need any series resistors either.

Jeff Frischman
Algood Casters Limited
605 Fenmar Drive, Toronto, Ontario, Canada, M9L 2R6
Tel: 416/749-7743 * 800/254-6633 (800/ALGOOD3)
Fax: 416/747-5671
email: mailto:[email protected]

Gregor Slemensek

Instead of directly connecting LEDs to power supply use voltage regulator that lowers voltage to let's say 2V.

Best regards
Gregor Slemensek