# LED tail light conversion

xetmes
Silver - Posts: 586
Joined: May 18, 2003
Posted: August 10, 2004 at 3:05 PM / IP Logged
yea you can do that, use a larger resistor value (like 470 ohms) its just kinda a waste of resistors and power, but if you did that it would only take 70 leds to produce the same current draw...
lilboi
Copper - Posts: 126
Joined: May 01, 2004
Location: United States
Posted: August 11, 2004 at 4:36 AM / IP Logged

hmm.... i found a way not to hyper blink. It's called a load equalizer.

I'll think about it. What would you suggest?

xetmes
Silver - Posts: 586
Joined: May 18, 2003
Posted: August 11, 2004 at 6:52 AM / IP Logged
yea I think load equalizers are just fancy resistive elements, to waste power and draw as much current as incandecent bulbs
lilboi
Copper - Posts: 126
Joined: May 01, 2004
Location: United States
Posted: August 11, 2004 at 3:24 PM / IP Logged
So single resistor for each led or one for 3 leds.
xetmes
Silver - Posts: 586
Joined: May 18, 2003
Posted: August 11, 2004 at 4:06 PM / IP Logged

ok, ill give you a way to draw the same current using 30 leds...

Use 1 resistor per led, each resistor led branch will draw around 25mA, so 30 will draw 0.75A

since a 21 watt bulb draws 1.75A we are short an amp, so to draw 1 A...

we could put a single resistor in parallel will all the resistor and leds, but that would mean we would need a 12 ohm resistor, further more it would need to be able to handle 12 Watts, quite a bit

to use smaller resistors you could put a bunch in parallel, I will assume you are using 1/2 watt resistors....

since P = V(I) and they are in parallel with 12V (assuming the lowest ~12V), V = 12, P= .5...

I = (0.5) / (12) = 42mA, therefore maximum these resistors can only handle 42mA, we will assume 40mA

so the resistors we need will be 1/2 watt, and must draw 40mA each, since we need to draw 1A we will need..

1A / 40mA = 25 quite a bit of resistors...

now to find the resistance value needed, V = I * R, R = V / I

therefore R = 12 / 40mA = 300, so you will need 25, 300 ohm resistors, you can check since the parallel resistance will be 300 / 12 = 12 ohm, so together it is 12ohm and each resistor will be burning off 0.48 Watts...

If you use 1W resistors you can use fewer (12 @ 150 ohms)

hope that helped...

lilboi
Copper - Posts: 126
Joined: May 01, 2004
Location: United States
Posted: August 12, 2004 at 3:19 PM / IP Logged
Damn that's kinda confusing. Alright so if i do get one resistor per led it would be 1/2 watt, 300ohms.
but if i do it in sets of three i only need 120ohms.
xetmes
Silver - Posts: 586
Joined: May 18, 2003
Posted: August 12, 2004 at 6:48 PM / IP Logged

sorry it was confusing, i am saying hook up the leds with a single 470 ohm resistor. Then take 25, 300 ohm resistors and solder them all in parallel with the leds/resistors... not elegant and a waste of power but hey it will draw as much as the incandecent...

lilboi
Copper - Posts: 126
Joined: May 01, 2004
Location: United States
Posted: August 12, 2004 at 7:57 PM / IP Logged

I rather do it the most efficient way, and buy the load equalizer. So how will i use the least amount of energy for say..... 80 led's (each tail).

xetmes
Silver - Posts: 586
Joined: May 18, 2003
Posted: August 12, 2004 at 8:31 PM / IP Logged

I doubt a load equilizer is more efficient, in fact it most likely is just a resistor like a 20 ohm, 10 watt. Problem is that thermal flashers work off heat produced from current draw, there are not that many ways to draw current without using power...

least amount of energy for 80 leds using the 3 per row would be 25mA * (80 / 3) = .67A

most current for 80 leds using single resistor and led: 25mA * 80 = 2A... too much

lilboi
Copper - Posts: 126
Joined: May 01, 2004
Location: United States
Posted: August 12, 2004 at 8:38 PM / IP Logged
so i should go witht he 3 set idea. with 120 ohm resistor?
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