Electronics > Projects, Designs, and Technical Stuff

Getting the correct wire guage for 12V/5A LED extension wires

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cliffyk:

--- Quote from: DW1961 on July 17, 2020, 05:18:15 pm ---
--- Quote from: Siwastaja on July 17, 2020, 07:08:19 am ---
--- Quote from: DW1961 on July 17, 2020, 07:00:20 am ---So if the voltage drop is the same for any wire, why use 24v over 12v?

--- End quote ---

assuming the load is re-designed to do what it is doing at this higher voltage).

--- End quote ---

I think this is what I was missing?

--- End quote ---

Yup, that's it ...

In looking up 2835 series specs I found that 600 LED 16 ft light strips (like this from Amazon) using 12 v 5 A power supplies are common. Is that what you have?

If so, just because it recommends a 5 A supply does not mean the LED strip actually draws 5 A. Do you know what that actual draw is?

Also, many of the power supplies recommended for with SMD LED strips are constant current devices, controlling their output voltage as needed to keep the load current within specs which adds another twist to you situation.

A somewhat exaggerated example is the  30 W UV LED I recentlly used in a UV curing oven I built from a salvaged microwave oven. It uses a constant current 18-39 V power supply and draws 900 mA at 33 V from same. The supply is capable of delivering up to 39 V (if the specs are valid) so I could place the LED 520 feet away using 18 ga. wire. The power supply iwould then deliver it's full 39 volts; the drop across the wire would be 5.96 V; leaving 33.04 V to make the LED happy. It would not be an especially efficient way of doing it, but it would work

Putting these numbers into the online calculator validates this: 



If you are indeed using one of those 16 ft LED strips, you might want to do some empirical testing. Get 15 ft of 14-2 w/g Romex ($7.55 at Home Depot), hook it up (you won't even need to take the wire from it's plastic bag) and see what happens,I suspect it will work fine--you'll lose 0.7 V in the 14 ga. wire but that should not have any major effect on the LED output...

 

DW1961:

--- Quote from: blueskull on July 17, 2020, 05:37:48 pm ---
--- Quote from: DW1961 on July 17, 2020, 05:14:16 pm ---When I input the metrics into this calculator, nothing changed except voltage percent of total volts. Voltage drop in volts stays the same.

--- End quote ---

Did you keep using the same current?

--- End quote ---

I did! 5V for both.

cliffyk:

--- Quote from: DW1961 on July 17, 2020, 09:02:41 pm ---
--- Quote from: blueskull on July 17, 2020, 05:37:48 pm ---
--- Quote from: DW1961 on July 17, 2020, 05:14:16 pm ---When I input the metrics into this calculator, nothing changed except voltage percent of total volts. Voltage drop in volts stays the same.

--- End quote ---

Did you keep using the same current?

--- End quote ---

I did! 5V for both.

--- End quote ---

I assume you mean "5 A", which is of course why nothing but the voltage loss expressed as a percent of total voltage changed.  The voltage loss is dependent on the current flow, not the voltage.

Ohm's law for dummies. Read it...

DW1961:

--- Quote from: cliffyk on July 17, 2020, 06:28:40 pm ---
--- Quote from: DW1961 on July 17, 2020, 05:18:15 pm ---
--- Quote from: Siwastaja on July 17, 2020, 07:08:19 am ---
--- Quote from: DW1961 on July 17, 2020, 07:00:20 am ---So if the voltage drop is the same for any wire, why use 24v over 12v?

--- End quote ---

assuming the load is re-designed to do what it is doing at this higher voltage).

--- End quote ---

I think this is what I was missing?

--- End quote ---

Yup, that's it ...

In looking up 2835 series specs I found that 600 LED 16 ft light strips (like this from Amazon) using 12 v 5 A power supplies are common. Is that what you have?

If so, just because it recommends a 5 A supply does not mean the LED strip actually draws 5 A. Do you know what that actual draw is?

Also, many of the power supplies recommended for with SMD LED strips are constant current devices, controlling their output voltage as needed to keep the load current within specs which adds another twist to you situation.

A somewhat exaggerated example is the  30 W UV LED I recentlly used in a UV curing oven I built from a salvaged microwave oven. It uses a constant current 18-39 V power supply and draws 900 mA at 33 V from same. The supply is capable of delivering up to 39 V (if the specs are valid) so I could place the LED 520 feet away using 18 ga. wire. The power supply iwould then deliver it's full 39 volts; the drop across the wire would be 5.96 V; leaving 33.04 V to make the LED happy. It would not be an especially efficient way of doing it, but it would work

Putting these numbers into the online calculator validates this: 



If you are indeed using one of those 16 ft LED strips, you might want to do some empirical testing. Get 15 ft of 14-2 w/g Romex ($7.55 at Home Depot), hook it up (you won't even need to take the wire from it's plastic bag) and see what happens,I suspect it will work fine--you'll lose 0.7 V in the 14 ga. wire but that should not have any major effect on the LED output...

 

--- End quote ---

That's the strip for which I was calculating, yes. I'm going to test the strip amperage to see what it really consumes also. According to this website, 2835 LEDs are max 20mA and .06 watts. I did the math of 2835 LED max current and it came out to  3Amps. (12v @ 36 watts). So I am calculating for 5Amps, and should be calculating for 3 Amps. Good point.
https://somanytech.com/what-is-smd-led-what-is-led-5050-5630-2835/

But other than the application, I'm just trying to get my head around the voltage drop, whatever it is, at the end of the run when I fill in the calculator blanks, regardless of volts at teh supply, if everything else remains the same, the voltage drop is the same at the end of the run.

DW1961:

--- Quote from: cliffyk on July 17, 2020, 09:10:53 pm ---
--- Quote from: DW1961 on July 17, 2020, 09:02:41 pm ---
--- Quote from: blueskull on July 17, 2020, 05:37:48 pm ---
--- Quote from: DW1961 on July 17, 2020, 05:14:16 pm ---When I input the metrics into this calculator, nothing changed except voltage percent of total volts. Voltage drop in volts stays the same.

--- End quote ---

Did you keep using the same current?

--- End quote ---

I did! 5V for both.

--- End quote ---

I assume you mean "5 A", which is of course why nothing but the voltage loss expressed as a percent of total voltage changed.  The voltage loss is dependent on the current flow, not the voltage.

Ohm's law for dummies. Read it...

--- End quote ---

5 Amps is the "current flow?" That's how much the application needs, in this example we're using. Let's just say it needs 5 Amps to function. Maybe I am misunderstanding "current flow?" I'm understanding that as 5 Amps current.

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