Wiring 120 LED in parallel.It is in square shape.Each side contain of 40 LED.How many power supply i need.?

I'm going to build up a multi-touch. Now I'm facing the wiring problem. The screen is square in shape. Each side contains of 40 LED.Now i need to know how many power supply that i needed..and also the wiring diagram since that i have no knowledge on it.

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frollard8 years ago

With the inputs of 12VDC
1.5v typical for an IR led
15ma current each
120 total LEDS

it suggests: 15 strings of 8 leds in series, with a 1 ohm resistor on each

It will consume 225mA at 12volts.
Go to the link, and input the numbers I did- it will draw you a pretty diagram so you can see how to wire them.
Tzoul frollard5 years ago

Thank You! Would you solve my problem? I am using a 14.8v power source to light 16 bi-colored leds in a parallel. I understand I need 32 Resistors, but what strength? My led specs are like this: 2.9v draw, 25mA, Please help. When I enter this into the many resistor calculators Ive found on the web, they give me such incredible resistor suggestions, however I wonder if this accounts for each led. Another thing I can not rap my mind around is that: occasionally I will have both colors on at the same time. How do I figure that into all this resistor calculatiing.
three pronged 5mm leds... Yehaa!
frollard Tzoul5 years ago
Those are a little more complex to drive if you want some level of control over them -- because you can't run them in series like you would a regular led. The shared 'common' anode or cathode would mean they all share a ground or source...so yes, they do need individual resistors on each one.

Unfortunately the 'system' is much less efficient running single 3v leds on a nearly 15v source. It'll work but most of the power gets burned off in the resistors

negative grounds on the outside, positive tap in the middle

- ---/\/\/---- |<|--+--|>|---/\/\/---- - R = 560 ohms (assuming that's positive common)
+ = positive

+---/\/\/---- |<|--T--|>|---/\/\/----+ R = 560 ohms (assuming negative common)
T = negative
Tzoul frollard5 years ago
Frollard, you rock so hard!

I want to be efficient, you think 32 - (560 ohms, 1/4 watt, 5%) will protect the leds but jeopardize my badass 14.8v battery pack, thats very good to know. I am very thankfull that you helped me in this way because I am early in this faze of the project, I can still change things, Word! thanks for helping in a big way. If you dont hear from me again, know that I am in bliss and I wish the same for you.

However, if you have another solution, now that you know its a "three lead led" with 2 anodes and 1 cathode, Im all eyes.

frollard Tzoul5 years ago

The only improvement I'd use for efficiency, is a lower voltage battery pack. Since you're stuck unable to make series strips of leds, the closer the led voltage matches the battery voltage the less work the resistors have to do to protect the leds.

Another option is called a 'buck' converter. It actively converts DC voltage to a lower DC voltage, and they are 85-95% efficient, which is fantastic. So if you could find a 3-5v buck converter ("variable input") on ebay or electronics sites, it would save you at least half your battery life.
oops, forgot to reverse the diodes. 2nd one should be:

+---/\/\/---- |>|--T--|<|---/\/\/----+ R = 560 ohms (assuming negative common)
T = negative