4 replies to this topic

[Andykettle]

Im in the process of making my own hi-vis brake light wich involves alot od LED's, 100 to be precise, might be slightly more. I cant run them directly of 12v as they work on 2.6v. I was given a pack of resistors to wire them to 12v individually but beacuse ive connected them all in parrallel theres not enough currnt (i think) and are to dim.
Can anyone tell me which one i need?
Here's the LED stats.


Peak Wave Length (nm)---632

Forward Voltage (V)---< = 2.6

Reverse Current (uA)---< = 30

Absolute Maximum Ratings---( Ta = 25°C )

Max Power Dissipation---PM = 80 mw

Max Continuous Forward Curret---IFM = 30 m

Max Peak Forward Current---IFP =75 mA

Reverse Voltage---5 ~ 6 V

Thanks
Andy

[Guess-Works.com]

Each LED must have it's own resistor, or what you can do is buy LED's which are speciafically designed for a 12v system, and have the resistor built in...

eg... http://www.maplin.co..... 12V&doy=12m5

but if my math is right you need a 470 Ohm resistor on the annode of each LED.

[Bungle]

5 LED's wired in series should be able to run on 12

so wire 5 in seriers in banks of 20 paraell gives you 100 wiered stright in to 12

i think

[dklawson]

Bungle beat me to it.

Series wire the LEDs to equal 12v... (2.6V x 5) = 13V And that's great. Your car's actual battery voltage will probably be about 12.5V and while running, the charging system will likely bring that up to 14.2V (or close to it). 14.2V / 5 = 2.8V which is only about 7% high. Since this is a brake light you're working on it's not going to have a 100% duty cycle and short over voltages will be OK.

If you wire all 100 lamps in parallel, each one draws 75mA for a total of 7.5 Amps (0.075A x 100). Furthermore, the voltage drop on each resistor is dissipated as heat. The basic electrical equation is V = I x R. A single parallel wired LED would need a voltage drop of: 14.2V - 2.6V = 11.6V That is to say, each resistor needs to create an 11.6V drop from the car's operating voltage to the maximum 2.6V the LED wants. The other electrical equation you need to keep in mind is that power (Watts) = V x I. So each resistor creating that 11.6V drop has to carry 0.075Amps or power= 11.6*0.075 = 0.9 Watts. So each resistor would need to be rated for 1 Watt (AND more importantly) with 100 LEDs/resistors, you'd be dissipating 0.9 Watts x 100 = 90 Watts. That's way too much heat.

The advantage of series wiring is that you'll turn on all 5 LEDs for the same total current, specifically, no more than 75mA. For 100 LEDs though, that's still 75ma x 100 LEDs / 5 (per group of LEDs) = 1.5 Amps total. All that current is used to produce light so the only heating is what happens inside the LEDs.

Sorry about the EE lecture.

[Bungle]

i'm glad you under stood me dk as i didnt