Could you help me with questions about building an led array?
I have 20 ir leds, http://www.mouser.com/ProductDetail/Vishay/TSAL6400/?qs=oSAwVt7aKTHCOCv1ythi7g%3d%3d
I'm trying to run them in parallel and power it from USB @ 5v and 100ma.
whats the best way to wire them and do I need resistors? can I just use one resistor at the entrance of the parallel wiring. (I assume I could and would need a 1.8 ohm 12 w resistor)
I'm trying to wire them on here http://www.radioshack.com/product/index.jsp?productId=2104052#
should I use this instead for the parallel setup? http://www.radioshack.com/product/index.jsp?productId=2102845#
Additional Details oh and I'm completely clueless about wiring....... schematics, pictures or video examples would really help
billrussell422010-02-25T04:29:18Z
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First, NEVER wire up a bunch of LEDs with only one resistor. You need one resistor per LED (or per series string) to keep the current at the correct value.
Also, the 100 mA rating on the LEDs is the maximum, you might want to keep the current a bit lower for reliability. If you are planning on packing them close together, you may want to reduce the current even more. Best is if you leave a bit of a gap between them for air flow.
USB is not the way to power these. USB is rated at 1 amp usually, not 100mA, but that is not enough to power 22 LEDs, which would need more than 2 amps.
Resistor calculation: do NOT use the 3V number, again that is a max, not the actual voltage. Use the 1.35 volt number. If you pick 80 mA as operating current, then R = (5–1.35) / 0.08 = 46 ohms. I'd use 47, the next standard value. Wattage is 3.6*.08 = 0.29 watts, so use 1/2 watt resistors.
The entire combination of LEDs and resistors will be dissipating 5v x (0.08 x20) = 8 watts. Enough to get quite hot and perhaps too hot. Be sure you don't totally enclose the assembly, leave openings for air flow. Better yet, get one of those tiny fans powered off of the 5 volts.
It is also possible to use two LEDs in series for each string, this will cut the power and current requirements in half. 2x 1.35 = 2.7 volts, and the resistor would be (5–2.7) / 0.08 = 29 ohms, you can use 30 ohms 1/2 watt, total of 10.
Before jumping in the deep end, you could try a simpler approach in order to better your understanding.
I once built a traffic light array using a board, wires, solder and resisters, it was fun too.
You may learn from a simpler approach. Also, understanding it is another thing. I learnt how to produce it but how it actually worked, I still don't know!
I understand computer logic gates, and, nand, or, nor etc, but how they work with electricity, I have no idea.
You would need to search for the most similar schematic on the internet to what you are looking for and you would have to copy it. Understanding it is beyond me.
All I know is resistors, logic gates, wiring and the correct tolerance levels of the resistors are important. Perhaps the most important is finding the correct schematic. Its out there, just need to search! Not an answer, I know but your question is difficult and my knowledge on this is limited with a little experience on a 2 week electronics course about 20 years ago!
First of all. DO NOT try to power that from a USB. you will totally kill the USB power supply. the LEDs each takes 3 volts forward, @ 100 ma max.. that is 2 AMPS of power @ 3 volts. a USB can supply at most 0.5 amps before cutoff.
I did a few calculations. ASSUMING you are using a power supply that can provide MORE than 2 amps of power, your array (assuming all parallel), will need only a 1 ohm resistor rated to 4 watts.
the optimum however, considering the massive current of the LEDs, is by putting it in parallel with a resistor on each LED, as follow:
1 x 20 array uses 20 LEDs exactly +----|>|---/\/\/----+ R = 22 ohms +----|>|---/\/\/----+ R = 22 ohms +----|>|---/\/\/----+ R = 22 ohms +----|>|---/\/\/----+ R = 22 ohms +----|>|---/\/\/----+ R = 22 ohms +----|>|---/\/\/----+ R = 22 ohms +----|>|---/\/\/----+ R = 22 ohms +----|>|---/\/\/----+ R = 22 ohms +----|>|---/\/\/----+ R = 22 ohms +----|>|---/\/\/----+ R = 22 ohms +----|>|---/\/\/----+ R = 22 ohms +----|>|---/\/\/----+ R = 22 ohms +----|>|---/\/\/----+ R = 22 ohms +----|>|---/\/\/----+ R = 22 ohms +----|>|---/\/\/----+ R = 22 ohms +----|>|---/\/\/----+ R = 22 ohms +----|>|---/\/\/----+ R = 22 ohms +----|>|---/\/\/----+ R = 22 ohms +----|>|---/\/\/----+ R = 22 ohms +----|>|---/\/\/----+ R = 22 ohms ' In solution 0: each 22 ohm resistor dissipates 220 mW the wizard thinks ½W resistors are needed for your application together, all resistors dissipate 4400 mW together, the diodes dissipate 6000 mW total power dissipated by the array is 10400 mW the array draws current of 2000 mA from the source.
as you can see from the above, total power is 10 watts. a USB cannot provide more than 2 watts. do not attempt to run it on an USB. I would say you can safely run no more than 5 of the LEDs within the USB's limit (where you will put it all in parallel, then use a 4.7 ohm resistor for them all)
here is a calculator for the resistance you will need: http://led.linear1.org/led.wiz http://led.linear1.org/1led.wiz
Any a million.5 volt alkaline might artwork, you basically get much less existence. Use C cells or maybe AA cells particularly of D. yet, analyzing the tips sheet, you may not basically connect 9 volts to this. that's surely a gaggle of LEDs and the Vf could be between 8.3 and 9.7 volts at a cutting-edge of seven-hundred mA. you will need some style of continuous cutting-edge driving force to ward off the easy being broken by thermal runaway. it somewhat is a capability dissipation of 6.3 watts, as severe as 6.8 watts, so which you may desire to offer a warmth sink. The datasheet has an occasion of heat sink assembly. .