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Aftermarket headlights baked and opened

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Old 08-27-2006, 05:35 AM
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hmmm, not sure. But you can find alot of helpfull info here...
http://hidplanet.com/main.html
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Old 08-27-2006, 03:19 PM
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thanks for the info about the lights
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Old 08-28-2006, 03:21 AM
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still watching....
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Old 08-28-2006, 03:33 AM
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Don't exactly know what you guys are waiting for when you post "waiting". I just order my projector yesterday, so I won't start swaping projector mid next week.
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Old 08-28-2006, 03:35 AM
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which type of projector did you order?
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Old 08-28-2006, 03:38 AM
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watching
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Old 08-28-2006, 03:51 AM
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Originally Posted by Yanki01
which type of projector did you order?
This one:







Here are the approximate dimensions:
width: 2 5/8" Edge to Edge
width: 2 1/4" Lens
Lenth: 5 1/4" Depth without bulb holder

If you look at the two bottom screws, the orientation is exactly like the aftermarket's projector. Hopefully this will fit fine. If not, I'll make some adjustment
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Old 08-28-2006, 03:56 AM
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what brand?
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Old 08-28-2006, 04:05 AM
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Originally Posted by Yanki01
what brand?
BMW E46: Single xenon
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Old 08-28-2006, 04:14 AM
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kool! Hopefully it works out well with it! keep us posted!
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Old 08-28-2006, 05:25 AM
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Well come on man you got to update...Which projectors did you order? Because I had the same idea but glad to see someone that actually did it, I hope everything turns out great man! Keep us posted
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Old 08-28-2006, 05:26 AM
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Well I take back which projects because it clearly says it right there, sorry. But keep us updated like everyone else says...
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Old 08-28-2006, 05:57 AM
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Yah watching keep us updated plz! Much appreciated I can't wait!
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Old 08-28-2006, 11:39 PM
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sounds like he went with e46 projectors. if you need ballast to match let me know. i have a nice set of oem bmw ballast i just got 3 days ago for free and i could cut you a nice price you cant beat .
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Old 08-30-2006, 03:36 PM
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e46's are easy to work with, but the performance lacks. E-code e46's are pretty nice.
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Old 08-31-2006, 02:00 AM
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It looks like I won't get the projector until friday, I'll post more when I get it.

In the meantime, I do have one question for any of electronic literate out there.

I will be using 12 LEDs (each has a forward voltages of 2.3 and reverse voltage of 5) I want to connect these 12 LEDs to the parking light wires (parking light wires have 12.7 volt. These 12 LEDs will be compiled to a stip of leds

From my calculation, I have 12.7V - 2.3V = 10.4V, which approximate 100 Ohm. Should I use 12 "100 Ohm" resistors or should I use 12 "8.3 Ohm" (100/12) resistors?????

Can someone help?
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Old 08-31-2006, 02:47 AM
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^^^ I wouldn't wire all 12 of them in series. Wire 4 in series, and have 3 groups of 4. Use 1 150ohm resistor for each group.
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Old 08-31-2006, 03:10 AM
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Originally Posted by WhiteBlazer01
^^^ I wouldn't wire all 12 of them in series. Wire 4 in series, and have 3 groups of 4. Use 1 150ohm resistor for each group.
My knowledge is very limited pertaining to the electricity flow. Can you explain how you get 150 Ohm for 4 LEDs? and why a series of 4 instead of 12?

thanks
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Old 08-31-2006, 04:43 AM
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Originally Posted by Noodle
Originally Posted by WhiteBlazer01
^^^ I wouldn't wire all 12 of them in series. Wire 4 in series, and have 3 groups of 4. Use 1 150ohm resistor for each group.
My knowledge is very limited pertaining to the electricity flow. Can you explain how you get 150 Ohm for 4 LEDs? and why a series of 4 instead of 12?

thanks

R = (Vs - Vl) / I

Vs = Source Voltage = 12.2V (you should actually test whether it is 12.2 or 14.4)
Vl = Lead Voltage = 2.3V x 4 (4 LED's in the array)
I = Forward Current = 20mA

R = (12.2 - (2.3 x 4)) / .020A (NOTE: You use Amps, not milliamps. 20mA = .020A)
R = (12.2 - 9.2) / .020
R = 3 / .020
R = 150

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Old 08-31-2006, 05:01 AM
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Originally Posted by WhiteBlazer01
Originally Posted by Noodle
Originally Posted by WhiteBlazer01
^^^ I wouldn't wire all 12 of them in series. Wire 4 in series, and have 3 groups of 4. Use 1 150ohm resistor for each group.
My knowledge is very limited pertaining to the electricity flow. Can you explain how you get 150 Ohm for 4 LEDs? and why a series of 4 instead of 12?

thanks

R = (Vs - Vl) / I

Vs = Source Voltage = 12.2V (you should actually test whether it is 12.2 or 14.4)
Vl = Lead Voltage = 2.3V x 4 (4 LED's in the array)
I = Forward Current = 20mA

R = (12.2 - (2.3 x 4)) / .020A (NOTE: You use Amps, not milliamps. 20mA = .020A)
R = (12.2 - 9.2) / .020
R = 3 / .020
R = 150

AHHHHHHHHH

I rely on your source base on your works with the LEDs. But did you concluded that Vs is 12.2 or was that a typo. I tested the wire and it said 12.7. If we use 12.7, then we would get 175 Ohm
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