LED Sidemarkers (Need opinion)
#21
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Originally Posted by ares
nope with the wiring setup you have you would need the same resistor for a single LED as for all of those; that being about 550ohms.
at 9volts you should have used a 275-300ohm resistor! your at about 400mA at 15ohms on 9volts. double check your resistor they should have surely fried at that current level. within like milliseconds lol. 400mA is WAY out there.
at 9volts you should have used a 275-300ohm resistor! your at about 400mA at 15ohms on 9volts. double check your resistor they should have surely fried at that current level. within like milliseconds lol. 400mA is WAY out there.
Ohm's Law for my case of a 9v battery.
(v1 - v2) / (I*ledcount / 1000)
(9V - 3.5V)/(20mA * 42 / 1000) = 6.55 ohms
So I use a 10 ohm limiting resistor. for the 9 V battery. It lights up well. Thats how i took the pics
I hope i didn't miss anything. In any case. I will attach 275 -300 ohm resistor like it and see what happens. I suppose the worst that can happen is it doesn't turn on. I will write back maybe this weekend with my results.
#22
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well yes... unless I have my definitions wrong(I hope not; but not ruling it out yet) that led count would be for LEDs in a series; not parralel. making your LED count 1. resulting in the aforementioned 275ohm number...
if you had 2 in series; itd be (9-7)/(40/1000)
the 7 being the voltage adding up.
at 3 you would have 10.5 being why you really cant go to 4 in a series because you dont have enough volts. you actually have extremely close to the correct voltage at 4; but I tried to "skip" the resistor and it was really just better to use 3 and keep a resistor.
to be honest; Ive always used an online calculator to do it lol; Im lazy. online calculators only request number of LEDs in series; voltage for each, mA, and supply voltage. they never mention anything about LEDs in parrellel which makes me assume you'd use only those in series for the equation as well.
from what I know; and have done; you could cut that down in that wiring setup and go to 10LEDs, or 1 LED, and the resistor would not change at all. the current still passes from positive-resistor-single LED-negative; it has 42 paths to do so; but that makes no nevermind to the resistor if it has 42 or 1.
if you had 2 in series; itd be (9-7)/(40/1000)
the 7 being the voltage adding up.
at 3 you would have 10.5 being why you really cant go to 4 in a series because you dont have enough volts. you actually have extremely close to the correct voltage at 4; but I tried to "skip" the resistor and it was really just better to use 3 and keep a resistor.
to be honest; Ive always used an online calculator to do it lol; Im lazy. online calculators only request number of LEDs in series; voltage for each, mA, and supply voltage. they never mention anything about LEDs in parrellel which makes me assume you'd use only those in series for the equation as well.
from what I know; and have done; you could cut that down in that wiring setup and go to 10LEDs, or 1 LED, and the resistor would not change at all. the current still passes from positive-resistor-single LED-negative; it has 42 paths to do so; but that makes no nevermind to the resistor if it has 42 or 1.
Last edited by ares; 04-29-2005 at 01:30 PM.
#23
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Well I finished the project pics are attached. I did end up using a 5W 15ohm resistor as a limiting resistor. I taped into the positive end on the parking wire and the negative end i simply attached it to a screw.
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Originally Posted by Acree
those look great man. man that picture sure does look familiar.
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