I am currently staying with a friend in Broome who is an expert in electrics and solar and in fact runs his entire home on power supplied from solar panels.
He has given me the following simply formula that enables people to calculate what the voltage drop of a cable will be for any given cable length and current draw. The conversion factor of 0.017 is close enough to what is used as the world standard that any difference in our calculations is not worth mentioning.
In his extensive work in this field over many years he has found that the theoretical calculations are within a per cent or two of the results of actual tests.
The formulae is as follows
D = L x I x 0.017 / C
Voltage Drop (D) = Total Length of Conductor in Metres (L) x Amps (I) x 0.017 divided by cross sectional area of cable (C) in square millimeters. The result is the actual voltage drop (in volts).
NOTE: Length of Conductor is the combined distance from battery to appliance and return to the battery via the earth.
This formula can be used to calculate the cable size for any AC or DC application, and as almost every cable is made up of copper wires, there is no need to mention other conductors (such as aluminium) that may be used for specialialised projects.
i.e. As this is a caravan
forum I have used caravan appliances to calculate cable voltage drop - this is an example.
In a caravan fitted with a 12-volt fridge, the fridge tends to be responsible for 65-75% of the total power used. Fridges are also surprisingly voltage sensitive. I have prepared the following example of voltage drops between a battery and fridge, in a hypothetical situation.
If the fridge is 10 metres from the battery (and that 10 metres is made up of six metres to the Regulator, and then four metres to fridge (which is more than in most caravans but I know of one caravan where that distance is 13 metres), because twin cable must be used, the total conductor length is 20 metres.
This example is for a fridge using 6.0 amps when running (but would use about twice for a second or so each time it cycles on.
Using Cable size 6.0 sq mm for entire length.
D = L x I x 0.017 / C
D = 20 x 6 x 0.017 / 6
D = 0.34 volts. (NOTE: This is cable voltage drop only, it does not take into account any other voltage drops in connections etc.)
If you then add in a water pump (say 6 amps), TV & DVD player (say 6 amps) and some lights (say 6 amps) you then have the following. Total current draw is now 24 amps.
Calculating using 6 sq mm cable all the way.
Battery to Regulator
D = 12 x 24 x 0.017 / 6
D = 0.82 volts.
Regulator to Fridge using 6 sq mm
D = 8 x 6 x 0.017 /6
D = 0.13 volts
Total Voltage drop at fridge = 0.95 volts.
If the fridge was running and a pump was started then the start up current draw of the pump is roughly double the running current (total 30 amps) so for a short time the voltage drop would then be 1.02 volts giving a momentary total cable voltage drop of 1.15 volts.
This is why a fridge can stop running when a pump is turned on.
If we now increase the cable size to 16 sq mm (Battery cable size) between battery and Regulator the voltage drop will be much different.
Battery to Regulator
D = 12 x 24 x 0.017 / 16
D = 0.3 volts.
Regulator to Fridge using 6 sq mm
D = 8 x 6 x 0.017 /6
D = 0.13 volts
Total drop between battery & fridge now = 0.43 volts which, whilst far from optimum, is a lot better than 0.95 volts with 6 sq mm cable all the way if several appliances are used at the same time.
My contact suggests that fridges be wired directly from the solar regulator, using at least 8 sq mm cable (auto sparkies know this as ‘8-gauge’) for distances greater than a couple of metres or so. This may seem overkill, but even 0.1 volt makes a big difference in fridge performance, especially when battery voltage is low.
My direct measurements confirm the above.
My contact says he is surprised by suggestions any of this is complex or esoteric. He says it is totally basic stuff that any electrician or auto electrician will be well aware of.
