I've trying to find information on amperage and wire sizes but every chart I find is slightly different.
For example, can I use 18 AWG (stranded) for 6V / 10A at 1 foot distance?
What size would I need for 20V / 5A at 3 inches?
Voltage determines insulation type and min. thickness. Current determines how much copper cross section area you need to carry it given certain constraints like maximum temperature rise and permissible voltage drop.
Depending on the max. ambient temperature and the insulation type, its highly probable that 18AWG wire in free air can carry 10A. If its in a bundle, clipped to a surface or tightly enclosed by extra insulation, YMMV and you'll need to do the math, not just look up a table!
18AWG is already about as thin as is ever convenient to work with so although you could possibly go down to 20AWG for your 20V 5A case, you might as well use the same 18AWG wire.
A wire carrying a “high” current in a “low” voltage circuit: The wire tables give you the resistance per unit length. Be sure to calculate the voltage drop and see if is acceptable for the application.
I guess it also depends on how big of a voltage drop you're willing to tolerate.
For example, a 18 AWG wire is supposed to have around 6.385 mOhm per ft. or 20.95 mOhm per meter according to Wikipedia :
https://en.wikipedia.org/wiki/American_wire_gauge#Tables_of_AWG_wire_sizesSo if you output 6v at 10A and you have 1ft between source and consumer, then you have 2 ft. in total so a total resistance of 12.77 mOhm or 0.01277 ohm
If you apply Ohm's law : voltage = current x resistance, then it's quite easy to figure out you have a voltage drop of v = 10A x 0.01277 = 0.1277 volts on the wires and therefore you may get approx. 5.85v at the other end of the cable.
You can also calculate the amount of power dissipated in these 2 ft of wire : power = Current
2 x resistance = 100 x 0.01277 = 1.277 watts
so you have to think how warm the wires will be above ambient (temperature inside a product), or how will this heat affect connectors, will heat be heatsinked into circuit boards through connectors?
If you use a pair of awg 18 wires, the current will be spread across the pair of wires, so each wire will do 5A and will warm up less and you'll have a lower voltage drop.
It should be as thick as possible. More thick wire means less resistance, less voltage drop, less power loss and less heating. But very thick wire is not flexible and inconvenient to use. So you're needs optimal balance which depends on your needs
For example if you're expect 0.5 V voltage drop with 6 Amp current (about 3 W loss), then you're needs R = Udrop / I = 0.5 / 6 = 0.084 Ω
If the required total length of wires (sum of both) is 10 meters, then your wire needs about 0.084/10 = 0.0084 Ω per meter.
Now you can open some table with AWG / resistance specification, for example
this one.
And select proper wire which resistance should not exceed your needs.
For 0.0084 Ω/meter you can select AWG14, it has 0.00828 Ω/meter.
The wire tables give you the resistance per unit length.
And if you cant find it in the tables R = ρ L/a were ρ is the resistivity of the conductor material (copper is around 1.7 x 10
-8 ohms per meter).
I've trying to find information on amperage and wire sizes but every chart I find is slightly different.
For example, can I use 18 AWG (stranded) for 6V / 10A at 1 foot distance?
What size would I need for 20V / 5A at 3 inches?
That’s because they’re either general guidelines, or the specifics of a particular standard.
Ultimately, it comes down to two things:
1. How much voltage drop can you tolerate?
2. How hot can the wire get?
Radiolistener did a great job explaining how to calculate the first.
The second is a bit harder to calculate, insofar as the temperature of the wire will depend on the heating (easy to calculate how many watts are being dissipated), ambient temperature (easy to measure), airflow (a lot harder to quantify), the properties of the wire insulation material, etc. Hence the existence of ampacity guidelines.
30 AWG in liquid nitrogen would work fine.
Ultimately, it comes down to two things:
1. How much voltage drop can you tolerate?
2. How hot can the wire get?
Actually three things, add:
3. How thick a wire do you need for adequate mechanical strength?
Thank you so much for the help everybody!
One last question... how does this apply when using a buck converter?
For example, I'm going to use 20V / 5A PSU on a different project with a buck converter. How would I calculate the wire size for such a load? If I understand typical buck converters, this would allow 5V @ 20A (assuming 100% efficiency which is unrealistic, I know). I'm still confused on how buck converters actually work but 20A is much higher than the PSU's rated 5A?
You determine how much voltage drop you're willing to tolerate between the power supply and your voltage regulators or whatever you have in the device.
For example, let's say you have a laptop adapter that outputs 20v at max 5A but comes with a 2 meters length of cable and a barrel jack connector at the end... at 5A you may have already 19.9v at the barrel connector, because there's 0.1v drop on the cable between laptop adapter and barrel jack, unless the laptop adapter has voltage sense or intentionally outputs more than 20v to counteract that.
From the barrel jack connector, size the wires so that you'd have as low of voltage drop as you need.
As for 20v 5A to 5v 20A ...
You will get maybe 96-97% efficiency. So 20v at 5A = 100w , assume max. 95w output ... at 5v that would be 19A.
The inductor, diodes and capacitors that are part of the buck regulator which converts the high voltage to low voltage, making it possible to get higher current on the output.
From there, it's like me and others have explained ... Voltage = Current x Resistance (ohm's law) so depending on thickness of the wire, you will get less voltage drop in the cables. \
You can also use multiple wires in parallel, like computer power supplies do with pci-e 6 pin cables, where there's 2-3 pairs of AWG16 or AWG18 wires giving 12v to the video card.
I hear ya on lack of consistency across various sources.
Being in the Aerospace and Defense sector here in the US, I always fall back to the tables in
FAA AC 43.13-1B Chapter 11. The tables are approved for selection of conductors used in aircraft based on current flow (intermittent/continuous) and whether they are bundled or in free air. Given these are the basic tables that all avionics/mechanic technicians can employ on all aircraft from small single engine piston poppers to multi-engine jet airliners, they are conservative. PDF Page 511 is the start of the electrical wire rating section.
One thing to be cognitive of is that those tables assume Tefzel insulation (MS22759/blah PTFE/ETFE or some such shite), which isn't inexpensive when contrasted with other wiring that has different insulation. Tefzel insulated wire is the defacto standard here in the US Aerospace due to it's mechanical and thermal properties.
Anyways, I just wanted to go out of my way to muck things up for you further, by providing the source that I make use of professionally and personally.
Good luck.
-MHz
30 AWG in liquid nitrogen would work fine.
And broadens the sound stage whilst simultaneously clarifying the tonality of the high end without losing any depth in the lows to give the ear a sumptuous display of the aural feast?
Thank you so much for the help everybody!
One last question... how does this apply when using a buck converter?
For example, I'm going to use 20V / 5A PSU on a different project with a buck converter. How would I calculate the wire size for such a load? If I understand typical buck converters, this would allow 5V @ 20A (assuming 100% efficiency which is unrealistic, I know). I'm still confused on how buck converters actually work but 20A is much higher than the PSU's rated 5A?
Buck converters work like a transformer for DC. That doesn't describe the actual internal workings, but for simplicity's sake it can be thought of as a black box that exchanges voltage for current. If we assume 100% efficiency again for simplicity you could take 10V at 2A and drop it down to 5V at 4A. In the real world 80% is a reasonable estimate for a general purpose buck converter, in which case 10V 5A input would give you 5V at 3.2A. The input and output voltages can be anything you want within the constraints of the converter but the output voltage will by definition always be lower than the input.
If your interested here are some of the temperature ratings of various wire insulation's
https://adiwire.com/technical-data/typical-temperature-range/Also to play around here's a calculator that will tell you the approximate maximum current rating of copper wire
Wire fusing in free air You'll see that copper can handle a fair bit of current before fusing .
If you use between 5 to 15% the maximum fusing current you generally fall within an acceptable working current of the wire your using without excessive temperature increase bearing in mind the temperature rating of the wire insulation .
Many of the current rating charts you'll find are for NEC (National Electrical Code ) standards for industrial and residential wiring with ambient temperature of 30C . These current ratings are low for good reason but not practical for other applications . For example 14 gauge wire will fuse at around 200 amps so if you use 10% of that you could put 20 Amps across it with moderate temperature increase. But NEC will say 15 Amps maximum for 60 degree wire . That's less than 8% the fusible rating .
The biggest issue is voltage drop . You want to keep that as low as practically possible .
How would I calculate the wire size for such a load?
Just google "current to AWG".