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View Full Version : my echo hatch photoshoot


taKuto
04-09-2009, 01:28 AM
hope theres some love for the echo's :redface:

btw i heart mismatched lighting

jinxor
04-09-2009, 02:10 AM
The led style fogs are my fav mod by far....

dvlnblkdrs
04-09-2009, 02:17 AM
looks dope! :thumbsup:

yaris-me
04-09-2009, 02:45 AM
Nice!:thumbsup:

taKuto
04-09-2009, 02:46 AM
i'm not fond of the LED fogs at the moment, i'm going to try and make them two parallel lines instead of the honeycomb design whenever I have time to fix it.

evil genius
04-09-2009, 02:47 AM
cool. Love the fogs.

mr.white_hb
04-09-2009, 02:47 AM
Nice Photos, what rims are you using on your echo? 16" or 17"

taKuto
04-09-2009, 03:14 AM
16" enkei rf-1

thebarber
04-09-2009, 10:17 AM
looking good

taKuto
04-09-2009, 10:58 AM
leave the honeycomb, a line will reduce you illumination. with the honeycomb you increase the viewing angle while with the lines you don't. the honeycomb makes the light look more intense. I like them the way you have them. what current are you using for the LEDs.

60mA per series. Not concerned with output as these are angled to the ground for foreground lighting, theres no real significance for these LED's except for a little rice effect, so layout > output performance :biggrin:

cali yaris
04-09-2009, 02:01 PM
^ I think he said "per series" -- that's more than one LED, or maybe I have the terminology incorrect.

marcus
04-09-2009, 02:03 PM
led fogs.. nice.

taKuto
04-09-2009, 03:19 PM
Current is equal within one series circuit. Voltage drops on each LED. For my setup it's 3 LED's per circuit, so 60mA flows through all three LED's. Each LED gets 60mA and drop some voltage. A resistor drops the rest to maintain the current value (i believe i used 18ohms, cant remember) since the power source is a regulated 12V LM7812. These LED's are multichip, china overrates them at 100mA, so I run them at 60mA.

taKuto
04-09-2009, 03:32 PM
series-parallel :tongue:

taKuto
04-09-2009, 03:40 PM
quick sketch of the setup should be easier to understand :)

mrbond
04-09-2009, 06:03 PM
i'm not fond of the LED fogs at the moment, i'm going to try and make them two parallel lines instead of the honeycomb design whenever I have time to fix it.

I agree. But everything else is fantastic! I really wish I could have found one of these for sale before I bought my Yaris. There's just something about that body style...

Kaotic Lazagna
04-09-2009, 06:07 PM
Just plain sweet!

taKuto
04-09-2009, 07:22 PM
yeah thats what i was thinking , however i hope your resistors are 1Watt , you're pulling about 720mW in each branch. Also the fist LEDs to go will be the ones right after you resistor which will disconnect all the other ones.


how did you calculate that? Assuming resistor is 18 ohms (most likely higher),
P = I^2 * R
P = 0.06A * 0.06A * 18 = 0.0648 W = 64.8mW
64.8mW is burned through the resistor itself.

Don't worry about checking my work, i'm not studying electrical engineering for nothing. Everything electrical i've got no problems with, it's when it comes to the mechanical stuff I'm no good with

HTM Yaris
04-09-2009, 10:29 PM
All this means crap to me . Just tell me what the photon output is ...........:smile:

taKuto
04-10-2009, 01:58 AM
well , you want to have a constant 60mA right? Remember Ohms law V=IR right? then 12V divided by 60mA equals 200Ohms. Now for the power you can use like you said I^2R or V X I. 60mA^2 X 200Ohms = 0.72Watt

If you use an 18ohm resistor you only dropping 1.08V

200ohms is calculated for a fixed resistor burning all that current (so no LED's in that circuit, just a resistor alone). The way you can factor in LED's is that you need to subtract all the forward voltage drops and resistor handles the leftover voltage. If you want to look at it that way, assume the LED's were dropping 3.5V at 60mA (these specific ones are not dropping that exact value, but let's assume anyways).

V = 12 - 3.5 - 3.5 - 3.5 = 1.5V. This is the leftover voltage you will use a resistor to deal with. So V/I = R, 1.5V / 0.06A = 25 ohms. So 3 LED's + 25 ohms for that circuit.

Since LED's are a function of current, I usually will use an ammeter and potentiometer to dial in the desired current. China LED's don't have nice and pretty voltage to current charts to work with, and their bins are always different, so it's always a guess and check.

taKuto
04-10-2009, 11:51 AM
The voltage drops are still pretty average for a white LED (3.4-3.8V typical), it's current that usually varies. I got the LED's off some random china supplier from ebay. 10mm 5 chip white LED's i think.

taKuto
04-10-2009, 07:48 PM
yup, but china stuff is still much cheaper :)
only things i've been browsing on digikey are the offerings from cree. I might pickup some P4's to play with. The higher powered offerings from cree are far too expensive off digikey, i'd rather get them from china suppliers such as dealextreme.

taKuto
04-11-2009, 03:30 AM
I was going for digital/analog hardware but changed focus to power systems

taKuto
04-11-2009, 12:53 PM
university of toronto

Doc Zaius
04-16-2009, 01:07 AM
:offtopic:

Hahhaha... j/k. Whatever. Nice pics, taKuto. Nice cutoff too... *is* it a kit? Or a retro?

red_rims
04-16-2009, 01:42 AM
sick

yaris-me
04-16-2009, 03:14 AM
university of toronto

:bellyroll::thumbsup:

mastaofdisasta
04-16-2009, 02:32 PM
Current is equal within one series circuit. Voltage drops on each LED. For my setup it's 3 LED's per circuit, so 60mA flows through all three LED's. Each LED gets 60mA and drop some voltage. A resistor drops the rest to maintain the current value (i believe i used 18ohms, cant remember) since the power source is a regulated 12V LM7812. These LED's are multichip, china overrates them at 100mA, so I run them at 60mA.


Man i work at radioshack and have no idea what the hell yall are talking bout:iono:

taKuto
04-16-2009, 04:23 PM
TL projectors with philips ultinons running off hella gen 3 ballasts. OEM purple power!

mastaofdisasta, just basic circuit theory, time for you to brush up! :)