Resistors
Heres a little info for the guys who play with leds.
http://physics.bu.edu/py106/notes/Circuits.html http://www.1728.com/resistrs.htm -------------- As your led setups expand, you may run into the need to have a high wattage resistor, but with the standard supply for resistance. Your average resistor is 1/4W or 1/2W. An average led runs 0.072W @ 3.6V. If you wanted to run more leds in parallel off one resistor, you would need a higher wattage resistor After fiddling with Ohm's law a little... I came to this answer. Problem: You need X Ohms @ Y Wattage Explanation: Wiring resistors in parallel will cause the resistance to be 1/R (x being your resistance) [use the calculator above] So? for every pair you have in parallel , put a pair in parallel behind it, and you'll stay at the same resistance. Matching the number in series, with how many resistors you have in parallel, and you'll always end up at the same resistance. Examples: using 470ohm 0.5W resistors 1/470+1/470 = 235ohm @ 1W ...combined with another in series... 235Ohm + 235Ohm = 470Ohm @ 1W 1/470+1/470+1/470 = 156.67ohm @ 1.5W ...combined with 2 more in series.... 156.67Ohm+156.67Ohm+156.67Ohm = 470ohm @ 1.5W |
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neat! thanks!
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nice Ill check it out later..
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i personally use 1 resistor wired directly to each led instead of multiple leds of of on resistor. simplifies things for me when it comes to expand. pretty neat calcs u found btw
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this would benefit if you had... say... 8 wired in parallel to a circuit board, and wanted to replace them.
:ninja: |
ok. while my logic does kinda sorta work... it doesn't apply for the situation.
several leds in series need to have the current taken into account. CORRECTLYusing the VIR equation yields... 63.75 ohm @ 1.63 amp. so... using a 270ohm 1/2W x4 (in parallel) will give 67.5ohm @ 2W. I'll test this idea soon... but not terribly soon. Gotta find time/parts |
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oh i definitely have that one bookmarked too.
but it doesn't help with the more advanced stuff. |
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