Getting the correct wire guage for 12V/5A LED extension wires - Page 1
Forum section: electronics / projects

author: DW1961
author_rank: Frequent Contributor
timestamp: July 15, 2020, 05:32:28 am
content: Having a hard time finding this information.This is for a single color 12V white LED strip using 600 2835 LEDs on a 16' strip..I just need to extend the wires between the supply/controller about 14 feet one way. I'll use stranded copper wire.12V max amps will be 5.Thanks again!

author: cliffyk
author_rank: Frequent Contributor
timestamp: July 15, 2020, 06:43:32 am
content: Quote from: DW1961 on July 15, 2020, 05:32:28 amIt is not practical to pull 12 VDC @ 5 A across 145 ft with any reasonable size wire--even let's say 12 ga. (actually a pretty hefty wire)--at that length he wire's resistance (both directions) would 0.4606 Ω.Pulling 5 A across that wire would cause a voltage drop of 2.30 V, leaving just 10.7 V for your load,Using a more manageable wire size, say 18 ga. "doorbell" wire would bump the resistance of the conductors (again in both directions) to 1.85 Ω. The voltage drop at 5 A would be 9.25 V, leaving just 2.75 V for the load.You need to move the power supply closer to the load.

author: DW1961
author_rank: Frequent Contributor
timestamp: July 15, 2020, 07:11:59 am
content: Crap, mistype. I meant 14 feet. Changed in original post!Also updated the information for the strip itself.

author: cliffyk
author_rank: Frequent Contributor
timestamp: July 15, 2020, 07:59:11 am
content: Quote from: DW1961 on July 15, 2020, 07:11:59 amWell that's a horse of a different color!In that case the aforementioned 18 ga. wire would be OK, at 14 ft the resistance (both ways. you have to use "both ways" as the electricity travels both ways) would be just 0.1788 Ω, so the voltage drop at 5 A will be just 0.89 V. Use 16 ga. and it will be just 0.56 V...

author: cliffyk
author_rank: Frequent Contributor
timestamp: July 15, 2020, 08:59:41 am
content: Quote from: blueskull on July 15, 2020, 08:29:36 amMy wife explains to people all the time that when "He says "OK", he means 'marginally acceptable'", not 'Oh boy, it's Miller time!'"...It's kind of like when I say "I don't care.", she explains that I am not being dismissive or petulant but instead "He doesn't really give a crap."

author: DW1961
author_rank: Frequent Contributor
timestamp: July 15, 2020, 05:54:08 pm
content: I see a lot of 22 gauge wire for sale on Amazon for LED strips. I actually bought some to connect strips together for my computer. Total length of the wire was like 1 foot at max 2 Amps using 12V.Like this:https://www.amazon.com/C-able-Extension-Lights-Extend-Connectors/dp/B07FDW4M89/ref=sr_1_2?dchild=1&keywords=12v+rgb+wire&qid=1594835311&sr=8-2I know I researched using 22AWG, but now you have me questioning. What is that stuff anyway? lol Why is there so much of it? I also think most of the RGB extension cables sold are in the 22 AWG range too, but they are mostly not over 3 feet.Thanks also for answering the question didn't ask, which is in this application, do I need to calculate to and from, and yes, you said I do.On another note, the voltage drop calculations you offered are based on the extension wire I will use, but what about total voltage drop for the wire plus the strip? The strip is 16' itself. How do I calculate the drop on a particular light strip? I guess I could DMM it?

author: cliffyk
author_rank: Frequent Contributor
timestamp: July 15, 2020, 07:45:38 pm
content: 28 ft of 22 ga. wire has a resistance of 0.452 Ω (here's a nice online calculator).V = A * Ω; so, the voltage drop across 28 ft of 22 ga. wire carrying 5.0 A will be 2.26 V--leaving 9.74 V for the LEDs. If that is acceptable (I.e. the LEDs remain bright enough for your need) then I guess it would be OK (see the definition of "OK" from the SWMBO dictionary above).However there is no way I'd pass 12 VCD @ 5.0 A through 22 ga. wire. It's not really a fire hazard or anything like that, but it will look like something a teenage kid rigged up.

author: DW1961
author_rank: Frequent Contributor
timestamp: July 16, 2020, 12:22:36 am
content: Quote from: cliffyk on July 15, 2020, 07:45:38 pmBeen using this one. I guess just go with the resistance they input automatically?https://www.rapidtables.com/calc/wire/voltage-drop-calculator.htmlI use all of their calcs.

author: cliffyk
author_rank: Frequent Contributor
timestamp: July 16, 2020, 12:38:24 am
content: Quote from: DW1961 on July 16, 2020, 12:22:36 amYup, that's the resistivity of annealed copper, as "universal" a value as you need for plain ol' copper wire...

author: DW1961
author_rank: Frequent Contributor
timestamp: July 17, 2020, 07:00:20 am
content: So if the voltage drop is the same for any wire, why use 24v over 12v?

author: Siwastaja
author_rank: Super Contributor
timestamp: July 17, 2020, 07:08:19 am
content: Quote from: DW1961 on July 17, 2020, 07:00:20 amBecause given the same current, at 24V, a drop of 1V is just 4.2% of the supply, when it is double that at 12V. This alone means doubling the voltage halves the losses.Add to this that doubling the voltage allows halving the current for same power transfer (assuming the load is re-designed to do what it is doing at this higher voltage). This halves the voltage drop.Add the two effects: the relative voltage drop, and also the power loss, is one quarter!Increasing voltage pays off quickly.

author: DW1961
author_rank: Frequent Contributor
timestamp: July 17, 2020, 05:14:16 pm
content: Also reply to @SiwastajaQuote from: blueskull on July 17, 2020, 07:09:59 amWhen I input the metrics into this calculator, nothing changed except voltage percent of total volts. Voltage drop in volts stays the same.https://www.rapidtables.com/calc/wire/voltage-drop-calculator.htmlSo I'm confused.

author: DW1961
author_rank: Frequent Contributor
timestamp: July 17, 2020, 05:18:15 pm
content: Quote from: Siwastaja on July 17, 2020, 07:08:19 amI think this is what I was missing?

author: Siwastaja
author_rank: Super Contributor
timestamp: July 17, 2020, 05:47:23 pm
content: Well let's give an easy numerical example.#1, let's assume you have a magical constant current load drawing 1A.Input voltage = 10VWire resistance = 1 ohmLoad current = 1AVoltage drop on wire = 1 ohm * 1A = 1VInput power: 10V*1A = 10WPower at load: 9V*1A = 9WPower loss in wire = 1V*1A = 1WEfficiency of transfer = 9W/10W = 90%Then up the voltage to 20V. Still,Wire resistance = 1 ohmLoad current = 1AVoltage drop on wire = 1 ohm * 1A = 1VInput power: 20V*1A = 20WPower at load: 19V*1A = 19WPower loss in wire = 1WEfficiency of transfer = 19W/20W = 95%So already better, you are transferring twice the power with the same loss!But, in reality, if you used a constantpowerload... For example, you used a wide input voltage switch mode converter. Or, if you used a simple resistive load, like a light bulb or a heating element, you would just change the load to one designed for the higher voltage, lower current, same power.Let's recalculate the 20V case:Load current = 0.5AVoltage drop on wire = 1ohm * 0.5A = 0.5VInput power = 20V*0.5A = 10WPower at load = 19.5V*0.5A = 9.75WPower loss in wire = 0.5V * 0.5A = 0.25WEfficiency = 9.75W/10W = 97.5%So by doubling the voltage, you dropped the wiring loss to one quarter.

author: Datman
author_rank: Regular Contributor
timestamp: July 17, 2020, 06:08:07 pm
content: Here, in Italy, I would use a classic 3x2,5mmq (13AWG), with 2 of the three wires tied together:14' / 2,5mmq + 14' / 5mmq = >> 35+17=52mOhm; 5A x 52mOhm = 260mV.

author: cliffyk
author_rank: Frequent Contributor
timestamp: July 17, 2020, 06:28:40 pm
content: Quote from: DW1961 on July 17, 2020, 05:18:15 pmYup, 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 the30 W UV LEDI recentlly used in a UV curing oven I built from a salvaged microwave oven. It uses aconstant current 18-39 Vpower 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 workPutting 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...

author: DW1961
author_rank: Frequent Contributor
timestamp: July 17, 2020, 09:02:41 pm
content: Quote from: blueskull on July 17, 2020, 05:37:48 pmI did! 5V for both.

author: cliffyk
author_rank: Frequent Contributor
timestamp: July 17, 2020, 09:10:53 pm
content: Quote from: DW1961 on July 17, 2020, 09:02:41 pmI 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...

author: DW1961
author_rank: Frequent Contributor
timestamp: July 17, 2020, 09:16:33 pm
content: Quote from: cliffyk on July 17, 2020, 06:28:40 pmThat'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.

author: DW1961
author_rank: Frequent Contributor
timestamp: July 17, 2020, 09:20:55 pm
content: Quote from: cliffyk on July 17, 2020, 09:10:53 pm5 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.

author: cliffyk
author_rank: Frequent Contributor
timestamp: July 17, 2020, 10:35:58 pm
content: Quote from: DW1961 on July 17, 2020, 09:20:55 pmThat is exactly what it is. Amperes are not a measure of quantity like volts or ohms, but rather measurement of rate--therate of flow of electronsin a circuit, that's why it's calledcurrent

author: DW1961
author_rank: Frequent Contributor
timestamp: July 17, 2020, 11:02:58 pm
content: Quote from: cliffyk on July 17, 2020, 10:35:58 pmOK, I think I see my malfunction. The calculator can't change the application itself in order to benefit from higher voltage, and it could only change % voltage drop? I remember that higher voltages can dramatically drop Amps and thus wire size to get the exact same power to the application. It always seems like it was getting something for nothing. You can't do that with water, for instance. You have either higher pressure or a larger pipe. Higher pressure means--at some point--thicker pipe, better fittings, etc.

author: cliffyk
author_rank: Frequent Contributor
timestamp: July 17, 2020, 11:41:54 pm
content: Quote from: DW1961 on July 17, 2020, 11:02:58 pmSame thing; pressure is voltage, pipe is resistance, flow rate is current. Increase the pressure (same as increasing voltage) or pipe size (same as reducing resistance) and you get more flow.

author: DW1961
author_rank: Frequent Contributor
timestamp: July 18, 2020, 06:19:54 pm
content: Quote from: cliffyk on July 17, 2020, 11:41:54 pmWhere the analogy breaks down for me is that you are getting more water pressure while using a thinner pipe, which doesn't work in hydraulics. The more pressure, the thicker the pipe. e.g., and I'm probably still not getting it, you can light a 120V 1000 watt bulb at 8.3 Amps. Using the same 10000 watt bulb with 240V, you only need to calculate for 4.1 Amps to supply the same power to the bulb.. So, you can use more pressure and a smaller pipe (wire) with electricity?

author: cliffyk
author_rank: Frequent Contributor
timestamp: July 18, 2020, 09:28:33 pm
content: Quote from: DW1961 on July 18, 2020, 06:19:54 pmI believe you are confusing the "bursting" pressure of a physically thinner walled pipe ("thinner walled", not smaller I.D.) with pipe I.D.--if we define "smaller" pipe as just a smaller I.D. the analogy is spot on...

author: DW1961
author_rank: Frequent Contributor
timestamp: July 19, 2020, 01:22:43 am
content: Quote from: cliffyk on July 18, 2020, 09:28:33 pmUsing the 240V example above, you actually lower the current from 8.3 Amps to 4.1 Amps, but maintain the same power at the application. My mistake using the 240V example is that a 240 circuit is really two 120V circuits, which means the available current now has two feeds, not one. So you actually don't have smaller wire, just more of it, as in having a larger ID pipe with lower pressure but the same amount of water flowing through it. And that makes perfect sense to me.Is that it?!