Novice Cable Gauge Question

[narlo]

I have a hobbyist's, caged 5v 40A (200W) output, power supply that I want to use to power a 28-hub USB2.0/3.0 data/power hub. I calculated the max current draw to be 15.8A based on:

24 x 500mA (USB 2.0 Ports) = 12,000mA (12A)
4 x 900mA (USB 3.0 Ports) = 3,600mA (3.6A)

and confirmed with the manufacturer that the board is designed to handle it.

I'm pretty good with soldering and rewiring projects such as this and generally understand how to calculate cable sizes. For this project, I calculated that I'll need 14AWG to handle up to 16A of current at a length < 2 meters (6 ft).

I'd simply like to get confirmation on the cable size. Thanks

 

[spec]

Hi narlo,

Welcome to ETO.

I see you are from the US: which state. Care to put it by 'Location' in your user page so that it displays in the window at the left of your posts.

There are three main aspects to cable selection: safety, reliability, and electrical performance.

(1) Safety
There are standard tables for ensuring safety of cables in different environments. This basically involves maximum allowable current and voltage.

(2) Reliability
There are standard tables for ensuring reliability (life) of cables in different environment: temperature, humidity, vibration, etc

(3) Electrical performance
This is the area which is often overlooked and people are included to select a cable from the standard cable tables based on voltage and current, but this is a misnomer.

To select a suitable cable to perform adequately in an application you need to establish how the cable characteristics relate to the intended function.
The main parameter is to establish an acceptable voltage drop under maximum current conditions. To do this you need to work out the total voltage drop from the voltage source to the equipment, taking into account the resistance of the cable (3mili Oms per foot for 14AWG copper wire), both supply and return (earth), and the resistance of any switches, solder joints, fuses, connector contacts etc.

When a voltage drop budget is complete you will normally find that cable of a much thicker cross section than specified in standard cable tables for a particular current is required.

An interesting exercise is to measure the actual voltage across the source terminals and compare that with the actual voltage across the sink (equipment) terminals.

To give you a clue, 2 * 6 feet = 12 feet of 14 AWG copper wire at 15.8 Amps will give a drop of 12 feet * 3 mili Ohms * 15.8 Amps= 569mV, and that does not include other voltage drops.

spec

 

[narlo]

Hi Spec,

Thanks for your reply. I've updated the profile to show the State, great suggestion.

I appreciate the homework exercise and will take measurements once I have the unit powered up.

Narlo

 

[spec]

Hi Spec,

Thanks for your reply. I've updated the profile to show the State, great suggestion.

I appreciate the homework exercise and will take measurements once I have the unit powered up.

Narlo

Hi narlo,

Nice to see where people are. It also helps with answers to requests for information because we then know what components you have access to and also your mains supply characteristics. Apart from that it is just plain interesting to know where people are.

In my experience USB power is fraught with problems caused by voltage drops between the voltage source and equipment. For example smart phones will not charge at a full rate if the input USB voltage is on the low side.

In your case, where you are supplying many USB sockets from a single power supply, you will need to minimize voltage drops to be successful. One approach is to use remote sensing and, if your power supply has that capability, that would be the way to go.

Perhaps if you could post a diagram of your proposed layout we could advise some more.

spec

 

[KeepItSimpleStupid]

I'd be interested in knowing what that board is.

I would assume that each port has the ability to enumerate and that each port has it's own DC-DC converter. It's also possible that the board needs more than 5V to operate.
If I use the first assumption, the there has to be an efficiency associated. It's probably high, but nonetheless, you should be aware of it.

 

[Tony Stewart]

For reference in future a change of AWG of 3 is a binary(2) multiple of resistance/length. either 1/2 or 2x in each direction.

For example AWG24 Phone or CAT X wire. which is 84.22 mΩ/m

Say you want 0.1V drop worst case with 2m * 2wire @ 16A then you need resistance of;
R= 0.1V/(16A*4m) = 1.56 mΩ/m

remembering AWG24=84 mΩ/m and -3AWG =1/2R
84/1.56=54 ~64= 6th power of 2. thus 6x3AWG= 18 gauges lower than 24 or AWG 8

and rising 6 gauges to AWG means voltage drop will be 6/3awg=2 powers of 2 greater voltage drop

or 0.4V max... My son in law does this in his head in a few seconds.