question

brook-13 avatar image
brook-13 asked

100/50 mppt wire size

  1. Hi can anyone help?I have two 245 watt 53v solar panels I'm going to wire in parallel to my 100/50 mppt. What cable size should I use from the panels to the mppt?
MPPT Controllers
2 |3000

Up to 8 attachments (including images) can be used with a maximum of 190.8 MiB each and 286.6 MiB total.

4 Answers
wkirby avatar image
wkirby answered ·

6mm² cables for each panel would be sufficient. You are planning to wire these two panels in parallel right?

2 |3000

Up to 8 attachments (including images) can be used with a maximum of 190.8 MiB each and 286.6 MiB total.

graham avatar image
graham answered ·

Brook 13 Agree with WKerby - the maximum current will be order of 18A even with 20 metres of 6mmsq cable the volt drop will be order of 1.13V which should not be a problem.


The bigger problem is getting the 12V or 24V for charging from the Controller to the batteries when a small volt drop becomes a problem. If the cable run from the controller to the batteries is about 2M I would be using 25mmsq to keep the voltdrop down


3 comments
2 |3000

Up to 8 attachments (including images) can be used with a maximum of 190.8 MiB each and 286.6 MiB total.

brook-13 avatar image brook-13 commented ·

Graham can you show me the math of how you came up with those numbers?

0 Likes 0 ·
r2d2wat avatar image r2d2wat brook-13 commented ·

There is a very simple app for IOS to do this, called wire sizer. Probably also available for other formats.

The calculation/table is also found in many books, but this is easier. Use your panels Vmp and Imp for the calculations. Since that will rarely be achieved, you have a built in safety factor. 3% voltage drop is generally accepted,

Hopefully you installed the regulator close to the battery, so for that short wire run, it is easy to up size it for minimal voltage drop.

0 Likes 0 ·
graham avatar image graham brook-13 commented ·

Brook 13 Have a read of https://www.12voltplanet.co.uk/cable-sizing-selection.html Particularly section 2 on Voltdrop

From it

Example

Using the above example of a 50W light we now know it draws 4.17A, so if we were to use a 0.5mm² cable which has a resistance of 0.037W/m and its total length from battery positive back to battery negative was 5m, then the voltage drop would be:

Vdrop = IR = 4.17A x (5m x 0.037W/m) = 0.77V or 6.4%

I would recommend that from the controller to the battery you look for a voltdrop of less than 0.5%

0 Likes 0 ·
Charlie Johnson avatar image
Charlie Johnson answered ·

@Brook 13

You have missed providing two of the key variables: allowable voltage drop (%) and the circuit length. Circuit length is the length of the path that the electrons will travel to and from the load from the source. With those variables, the size of the conductor, in circular mils, is given by this formula:

CM =(10.75 x I x L)/E

10.75 is a constant for copper

I is the current in amps

L is the circuit length in feet

E is the allowable voltage drop. For example: if allowable voltage drop is 3% with 53V solar panels, then E = 0.03 x 53V = 1.59V

Convert CM to mm^2 with: http://www.kylesconverter.com/area/circular-mils-to-square-millimeters

2 |3000

Up to 8 attachments (including images) can be used with a maximum of 190.8 MiB each and 286.6 MiB total.

sneha-rani avatar image
sneha-rani answered ·

Thanks for this info

2 |3000

Up to 8 attachments (including images) can be used with a maximum of 190.8 MiB each and 286.6 MiB total.