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Solar Wiring Amperage, Wire Size and Voltage Drop Info (Read 3943 times)
JB Solar
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Solar Wiring Amperage, Wire Size and Voltage Drop Info
Jun 30th, 2014 at 7:28pm
Solar Panel Primer:

OK. This is actually for all persons who want to know more about voltage drop and ampacity. For reference, most if not all of this information is from the NEC 2011 or equivalent. I design commercial solar PV professionally, so I am at least familiar with these topics.


Ampacity for solar depends on some factors including conduit, ambient temperature, wire and terminal temperature rating, and PV input. What we use in our calculations is very similar to the following:

Wire size should be able to handle Isc (short circuit current) X 1.56. Any panel will have a short circuit current. This is the current if you connect the positive lead to the negative lead and measure the current passing through around noon. Typically this is around 4-10 Amps DC. Multiply by 1.56 to get the ampacity needed for the next step.

Adjust for conditions of use.  If you are putting the panels on a roof, you need to increase the ampacity to account for the higher temperatures on a roof during operation. This is usually expressed as a temperature adder to the ambient temp. If the wires are direct on the roof for example, add 33 degrees to the operating temp for the correct adjustment. I live in an area where ambient is 87F, so I add 87 and 33 to get 120F. If we look at the NEC table 310.15(B)(2)(A), we get a factor (in this case 0.75) for conductors rated 75C in operation (this is where things fall often). So I would do the following: Ampacity needed = (Isc x 1.56) / 0.75 to get what my conductor needs to be rated for to not burn up. For fun, let's say that the Isc is 9 amps. Ampacity needed = (9 x 1.56) / 0.75 = 18.75 A. If I then look at NEC 2011 Table 310.15(B)(16), it says I can use 14 awg wire. So for one pair of wires, 14 awg would work in terms of ampacity, but what about voltage drop? What will we lose in heating up those wires?


Voltage drop is the voltage lost in the wires between two devices. Voltage drop multiplied by ampacity will give you the power lost in the wire heating up during operation. The voltage drop formula commonly used is as follows:

Vdrop (volts) = (2 x L x I x R) / 1000 = (2 x One Way Length of Wire x Nominal Ampacity (no adjustments) x Resistance of the Wire) / 1000

This comes from NEC 2011 Chapter 9, Table 8

So using our Ampacity numbers and a little more info, here is an example. We have a pair of panels that are connected in series to make our solar array, the controller is 50 ft distance from the array and we want to know what kind of Voltage drop we will get under normal operating conditions. We look on the panels and find that the Imp (Current at MPPT) is 6 A. Then we decide to review what wire size would give us a reasonable voltage drop. Assumptions: Imp = 6 A, Vmp (per panel) = 25 V, L = 50ft.

So we start with 14 Awg which we got from the ampacity. The resistance per 1000 feet of this size copper wire is 3.07 for solid. This gives us the following:

Vdrop = (2 x 50 x 6 x 3.07) / 1000 = 1.842 volts drop.

Now if we multiply this drop by the ampacity, we get the power loss in the wire, Power = Voltage x Current = 1.842 x 6 = 11 watts loss. This seems high for a pair of panels, but this is about right for a set of two 220 watt panels. The percent voltage drop would be calculated by taking the Vdrop and dividing by the Vmp. Since the panels are wired in series, the voltage is Vmp x 2 for two panels, or 50 Volts. So %Vdrop is 1.842 / 50 or 3.684%. Where I work we like to keep voltage drop under 2%. So I would pick a wire with less voltage drop.

Let's try 12 Awg next, which is the next size larger wire. The resistance per 1000 feet of this size copper wire is 1.93 for solid. This gives us the following:

Vdrop = (2 x 50 x 6 x 1.93) / 1000 = 1.158 volts drop. %Vdrop = 1.158 / 50 = 2.316%. Power loss = 1.158 x 6 = 6.948 Watts loss

The power loss is a lot less, but the voltage drop is still a little too high, let's try 10 awg:

Vdrop = (2 x 50 x 6 x 1.21) / 1000 = 0.726 volts drop. %Vdrop = 0.726 / 50 = 1.452%. Power loss = 0.726 x 6 = 4.356 Watts loss

Now we have something! We have reduced our voltage drop to an acceptable number and are confident that even if the wires are direct on the roof, we will not have ampacity issues. While we are code compliant with 14 awg wire for ampacity, it will eat up a fair share of our solar power (2.5%). Where if we use 10 awg, it reduces our power loss to just under 1%. More power to the controller means more power to the heater.

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