OP
Indygunworks
Silver Member
Would I still need a controller? What is the best method to run the wiring (type?) the 40 foot that would be needed. Will that length hurt anything?
Would I still need a controller? What is the best method to run the wiring (type?) the 40 foot that would be needed. Will that length hurt anything?
Well lets find out. with some basic math, some formulas and data we can look up, and let it guide us to a decision, OK, ready to start?
In general you are not going to need a charge controller for these small 1.5 watt trickle chargers, which is probably all you are going to need, but as for the 40 foot of extension wire, we kinda need to figure out what the voltage drop is going to be over 40 feet of "some sized" wire, you did not specify what gauge of wire you were considering. Obviously the larger the wire the less loss (voltage drop). Here is an example of a simple 1.5 watt solar charger.
1.5 Watt Solar Battery Charger
We are going to make some logical assumptions that the float voltage on the battery in full sunlight will be about 13.2 volts. So now time for some math.
Ohm's law V=IR So Voltage equals Current (the symbol I is used) times Resistance (R is the symbol)
Now for the power formula since the specifications of the charger is listed in power (watts) instead of current.
P=VA meaning Power is equal to Voltage (Volts ) times Amps (current). OK, so far so good, Lets figure using this power formula by plugging in the values we know to solve for the value we don't know the Current. Here we go:
1.5=13.2 x ? Meaning we know the power in watts as 1.5 from the specification of the maximum delivery of power on the spec's from the web site of the product.. I estimated the float voltage on the battery at 13.2 as an ideal reading. This is just an educated guess, (and also what I have seen from my use of this product) OK so now all we need to do is solve for the current by dividing both sides of the equation thru by 13.2 to get the current all by itself on one side of the equals sign. Clear as mud? 9th grade algebra here.. so stay with me.
1.5/13.2 =.1136 Amps. This is our maximum current. We will just leave it in the Amps format even though it is a fractional portion of an Amp.
Now we need to make some more decisions and assumptions. But they need to be logical, and as educated guesses as we can perform. Lets say we don't want to have a voltage drop (from the resistance of the wire) any more than .2 volts, Leaving our float voltage on the battery at 13.0 volts. So lets go back to our Ohms law formula. V=IR and plug in what we have and solve for what we would like to know (in this case the resistance that will cause no more than a .2 volt drop at the given current of .1136 amp. Hang in there, this is pretty easy.
So the Volts in this case is .2 the I (or current) is .1136 Amps and we solve for the resistance (R) .2=.1136(R) now lets solve for R by dividing thru both sides of the equation by .1136 and the result is .2/.1136=1.76 so R could be up to 1.76 Ohms.
Now lets look up some values for some standard gauges of copper wire. From this table we can get the resistance in ohms per thousand foot of various wire gauges. lets look at 18 gauge wire.
American wire gauge - Wikipedia, the free encyclopedia
18 gauge wire appears to be about 6.385 ohms per thousand foot. You say you need 40 foot of 2 conductor wire, so that is 80 foot of wire for the electrons to traverse (two conductors right?) So what is the resistance of 80 foot of this wire?. So 6.385 ohms divided by 1000 = .006385 per foot x 80 foot =.5108 ohms. Remember we said we could have as much as 1.76 ohms and still only have a .2 volt drop at the rated current of .1136 amps.. So our .5108 ohms is well under our maximum resistance we could still have a useful float voltage. We might could get away even with a smaller gauge like 20 gauge, but 24 gauge would probably have too much drop. 18 Gauge would be a pretty good safe, and economical choice for your 40 foot extension run providing all connections are tight and lossless. If you wanted to use a larger gauge like 16 or 14 even, sure that would be fine too, but 18 would seem adequate.
So a long way around to making a simple decision on a wire size for a given length and a given (although very small ) current you are trying to pass through that wire. The hope is that now you have some basis for making these sort of calculations in the future..
OR you could just go here to help figure it out. Have fun, stay young, and drink Pepsi!
Wire Resistance and Voltage Drop Calculator
I would not be using that black wire that jumps across the disconnect. I have nothing I need to power on the tractor while I am not there.
Lets cover another topic... I wouldn't even need 40 feet of line if I could just mount this in the rafters under the skylights in the barn. Wouldn't receive a TON of sunlight and certainly no real PEAK hours, but would be in the daylight for most of the day. I wouldn't need any wire at all and could mount it directly above where the battery is for the tractor and just leave the (included) alligator clips hanging down to get hooked up before I leave.