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Author Topic: fpevul88's build log  (Read 30318 times)
Drewid
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« Reply #60 on: January 25, 2007, 10:53:02 AM »

Indeed
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fpevul88
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« Reply #61 on: January 25, 2007, 07:58:14 PM »

so ive got my leds and are on their way. Next, I found numerous 12v power sources but there are variations in amps. there are 1.5, 5 etc. Anyone know which one i should get?
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nathansmith
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« Reply #62 on: January 25, 2007, 10:01:38 PM »

That's really going to be a function of the current draw of your LEDs. Also, if you plan on running your monitor and LEDs off the same power supply, then you need to sum up the current draw of both the monitor and the LEDs.

For instance, if you are using 10 LEDs that draw 25 mA each, then you need a power supply that can provide at least 10*25 = 250 mA. If your monitor says that it needs 3A, then you need to add that in to show that you need a total 3.25 A power supply. It's always nice to have a larger power supply than you think you need in order to handle peak current draw as well. If you only need 3.25 A, I'd shoot for a 3.5 A power supply or bigger.

The amperage doesn't matter so long as it is MORE than the sum of the current requirements of all your devices.

If you're current draw is low enough, you could even draw the power from USB.
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fpevul88
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« Reply #63 on: January 26, 2007, 03:25:25 AM »

ok so going with your suggestion for the usb thing which i want to try, i was wondering, cause i cut open my twin usb cable and this is what i saw:



what lines do i connect, the red and the black right? red being positive?
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nathansmith
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« Reply #64 on: January 26, 2007, 03:34:57 PM »

Yup, red + and black - (http://en.wikipedia.org/wiki/USB#USB_signaling). USB provides as much as 500 mA at 5 volts, though it is only required to provide 100 mA unless the device requests more current.
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fpevul88
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« Reply #65 on: January 26, 2007, 05:48:45 PM »

the previous LED array layout assumed that i am going to be having a twelve volt power source. It seems that when I use a 5 volt power source, the layout will be like this:

Solution 0: 1 x 18 array uses 18 LEDs exactly
    +----|>|---/\/\/----+  R = 100 ohms
    +----|>|---/\/\/----+  R = 100 ohms
    +----|>|---/\/\/----+  R = 100 ohms
    +----|>|---/\/\/----+  R = 100 ohms
    +----|>|---/\/\/----+  R = 100 ohms
    +----|>|---/\/\/----+  R = 100 ohms
    +----|>|---/\/\/----+  R = 100 ohms
    +----|>|---/\/\/----+  R = 100 ohms
    +----|>|---/\/\/----+  R = 100 ohms
    +----|>|---/\/\/----+  R = 100 ohms
    +----|>|---/\/\/----+  R = 100 ohms
    +----|>|---/\/\/----+  R = 100 ohms
    +----|>|---/\/\/----+  R = 100 ohms
    +----|>|---/\/\/----+  R = 100 ohms
    +----|>|---/\/\/----+  R = 100 ohms
    +----|>|---/\/\/----+  R = 100 ohms
    +----|>|---/\/\/----+  R = 100 ohms
    +----|>|---/\/\/----+  R = 100 ohms

does this mean that unlike the 12v array where leds were divided into groups of three on my terminal block (shown above) i am going to have to divide them to groups of one (if there is such a thing) and rewire my block?

it was my assumption also, that my block be rewired as shown:



is this correct?




« Last Edit: January 26, 2007, 05:57:28 PM by fpevul88 » Logged
Drewid
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« Reply #66 on: January 27, 2007, 11:30:22 AM »

So they're all in parallel. Yes that wiring is right.   Do you know how many amps the 2 arrays draw?
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fpevul88
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« Reply #67 on: January 27, 2007, 05:48:28 PM »

i do not know, and since that led array calculator told me to assume that it was 20mA, then i will say 20mA. The leds are 8mm leds.
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nathansmith
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« Reply #68 on: January 28, 2007, 02:08:49 AM »

That's probably 20 mA per LED, so it's likely that you'll draw 360 mA if you are using 18 LEDs (from an earlier post of yours). The resisters will dissipate some current as well, I believe (is that right?). Most LEDs I've seen draw between 20 and 25 mA unless you get some super bright LEDs, so plan a power budget between 360 and 450 mA. A 500 mA power supply would probably be good just to be safe. I wish I knew how much current a backlight inverter drew so that we could try to determine how much current has been freed up by unhooking the backlight.
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Drewid
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« Reply #69 on: January 28, 2007, 11:36:47 AM »

Is that two rows of 18 LEDs or two rows of nine?
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fpevul88
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« Reply #70 on: January 28, 2007, 05:08:25 PM »

i settled for 18 because that is how many leds two teminal blocks can carry... however i bought 20 leds.
 I dont understan your question drewid does this answer your question:

Solution 0: 3 x 6 array uses 18 LEDs exactly
    +----|>|----|>|----|>|---/\/\/----+  R = 120 ohms
    +----|>|----|>|----|>|---/\/\/----+  R = 120 ohms
    +----|>|----|>|----|>|---/\/\/----+  R = 120 ohms
    +----|>|----|>|----|>|---/\/\/----+  R = 120 ohms
    +----|>|----|>|----|>|---/\/\/----+  R = 120 ohms
    +----|>|----|>|----|>|---/\/\/----+  R = 120 ohms

so that is 3 rows of 6...

if you were asking whether ill have a row at the top of my lcd and at the bottom, i am not sure... perhaps i will have 9 uptop and 9 at the bottom. Is this bright enough?
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Drewid
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« Reply #71 on: January 28, 2007, 08:15:54 PM »

Sorry, yeah, I was asking wherther you were going to have 18  top and 18 bottom of your monitor, or nine.

I'm not sure how bright it will be,  I guess you're on point for this one Smiley
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fpevul88
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« Reply #72 on: January 31, 2007, 08:22:37 PM »

so i got the leds and am currently wiring it. I also got a 12v. power adapter with 500mA but the problem is that the wires are not colored. both are with and there is no colored wire underneath the white cover. anyone know how to determine which is positive and negative? one of the wires of the adapter has writing on it while the other doesnt.
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Drewid
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« Reply #73 on: January 31, 2007, 09:55:48 PM »

Do you have access to a multimeter?
If you have a spare LED you could try it.  It will probably kill it if it's the wrong way round though.   Most power supplies seem to have positive in the centre.
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« Reply #74 on: February 24, 2007, 11:22:59 PM »

Yeah, it was a pain finding a negative center for my UD-1212-R

This looks like its gonna be awesome
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