#### vincerules

##### Full Time elf

- Joined
- Aug 3, 2012

- Messages
- 100

- Thread starter
- #1

OK, I am trying to get a handle on the power/wiring. I will use my biggest load(most lights) as the example.

I am outling my rooflines using 2811 10 sections/meter, which is .64 amps/meter according to this thread.

The power supply I am using is this one from Ray Wu which is 12v/350W and from the sticker it says 29A.

80% max load = 23.2A

85% max load = 24.65

For my rooflines, I have 420 total pixels.

So I divide that by 10 to get meters = 42.

42 * .64 = 26.88

So if my math is correct, I am overloaded and should split these into two power supplies, correct?

Now onto the cabling. I am using this 18 gauge speaker wire from Monoprice.

I inject power every 50 pixels(or standard 5M length), so the most 1 cable run would need to power is 100 pixels (since I would inject between 2 sets of 50). Conductor resistance is .023

The longest run is probably ~20 feet away, but lets use 10 meters for ease of math and extra length room.

10 * .023 = .23ohms * 2 = .46ohms total return circuit resistance

100 pixels = 10 meters of lights * .64/ampspmeter = 6.4 amps

6.4 * .46 = 2.944v

12v - 2.944v = 9.056

24.6% voltage drop, I'm assuming that is not acceptable.

Or should I be doing it for only 50 pixels since the circuit does not actually go through all 100 pixels and back but instead gets split into 2 sets of 50.

50 pixels = 5 meters * .64 = 3.2

3.2 * .46 = 1.472v

12 - 1.472 = 10.528

12.3% voltage drop. Is that accetable?

Do I need to use thicker gauge wire for the longer runs?

Is my math correct? I am doing this while reading through the ACL manual (for the 1000th time) and trying to make sense of it all.

Thank you for taking the time to read this and any help you can give.

I am outling my rooflines using 2811 10 sections/meter, which is .64 amps/meter according to this thread.

The power supply I am using is this one from Ray Wu which is 12v/350W and from the sticker it says 29A.

80% max load = 23.2A

85% max load = 24.65

For my rooflines, I have 420 total pixels.

So I divide that by 10 to get meters = 42.

42 * .64 = 26.88

So if my math is correct, I am overloaded and should split these into two power supplies, correct?

Now onto the cabling. I am using this 18 gauge speaker wire from Monoprice.

I inject power every 50 pixels(or standard 5M length), so the most 1 cable run would need to power is 100 pixels (since I would inject between 2 sets of 50). Conductor resistance is .023

The longest run is probably ~20 feet away, but lets use 10 meters for ease of math and extra length room.

10 * .023 = .23ohms * 2 = .46ohms total return circuit resistance

100 pixels = 10 meters of lights * .64/ampspmeter = 6.4 amps

6.4 * .46 = 2.944v

12v - 2.944v = 9.056

24.6% voltage drop, I'm assuming that is not acceptable.

Or should I be doing it for only 50 pixels since the circuit does not actually go through all 100 pixels and back but instead gets split into 2 sets of 50.

50 pixels = 5 meters * .64 = 3.2

3.2 * .46 = 1.472v

12 - 1.472 = 10.528

12.3% voltage drop. Is that accetable?

Do I need to use thicker gauge wire for the longer runs?

Is my math correct? I am doing this while reading through the ACL manual (for the 1000th time) and trying to make sense of it all.

Thank you for taking the time to read this and any help you can give.