- 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?
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
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
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.
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:
V_{drop} = 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%
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