Solar Batter Tenders

   / Solar Batter Tenders
  • Thread Starter
#21  
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?
 
   / Solar Batter Tenders #22  
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!:D

Wire Resistance and Voltage Drop Calculator
 
   / Solar Batter Tenders #23  
Normally if you have less than 15 watt solar panel you do not need a controller.

These are widely used in boats, since the bilge pump is wired to run directly from the battery even if the main switch is off: better to have a run down battery in a floating boat. A friend of mine is an airline pilot who is gone for days at a time and uses one.

I know any number of people who use them on cars and trucks that sit out on cold weather. Cabelas has several from around $20. Ditto Radio Shack, Coleman and Schumaker make them as well.

Cabela

BUT, voltage drop is a serious problem and 40 feet of adequate cable would be far more than a charger! Easier to buy a jump starter....
 
   / Solar Batter Tenders #24  
Wont get much easier of an install than this either


If there is a fuse in that holder, then the parasitic current draw is just going to flatten your battery again, even with the Green stud Off, or removed.

That fused Keep Alive lead is supplied with those battery disconnect switches so that if you have a specific load (say, an alarm system) that you want to always have power, then you can run a separate dedicated power line to that one load through that fused lead.

Just pull the fuse out, or remove the Keep Alive lead at least until you get the problem traced.

Rgds, D.
 
   / Solar Batter Tenders
  • Thread Starter
#25  
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.
 
   / Solar Batter Tenders
  • Thread Starter
#26  
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!:D

Wire Resistance and Voltage Drop Calculator


So if I understand this right I can buy this

1.5 Watt Solar Battery Charger

and add 40 something feet of this

Shop 18-AWG 2-Conductor Thermostat Wire (By-the-Foot) at Lowes.com

and prevent any batter loss with this

http://www.harborfreight.com/battery-disconnect-switch-97853.html

and for less than 50 (would still have to buy the alligator clamps to hook it to the battery) bucks I will have a battery that is always topped off and doesn't bleed energy while I am gone.

Is the a valid understanding? If so I will pick up that solar charger TOMORROW, as they will be digging the footers on Thursday and I can get all this hooked up then. I have to switch to the grader blade anyways for incoming snow and I might as well do this stuff so I don't have to jump start the tractor the next time I am out there.
 
   / Solar Batter Tenders
  • Thread Starter
#27  
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.
 
   / Solar Batter Tenders #28  
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.

Good to hear - wasn't sure if that was your battery, or a random pic.

Rgds, D.
 
   / Solar Batter Tenders #29  
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.

That might work.. I would want to know how much if any leakage current your tractor is drawing from the battery. You are assuming that is the case, but are you really sure that it is? Have you tried to measure the leakage current with a VOM? (multimeter) set to measure amps then down to the milliamp range. You may not really need the battery disconnect device. I don't know if you do or not.

While your 18 gauge thermostat wire will work, keep in mind that it is solid wire and is designed to be installed once in a house. Automotive type wire is stranded and is designed to be flexible. Thermostat wire is more rigid and if it is bent or moved a lot of times it can fail due to stress. Maybe not a big deal, if not stressed too bad, but just keep it in mind. 18 gauge stranded speaker wire or red/black automotive wire like you buy at an auto parts store might be a better choice. It is up to you. If you cut and splice the provided wire on each end of your 40 foot run, perhaps the solid wire would not need to move and could be fastened to the building? Just a thought. What ever you do, be sure to observe polarity, and I would always check the polarity of the voltage coming out of the final thing I attach to the battery before I attach it. There again that VOM multimeter is going to come in handy. Your gonna need one sooner or later, and might as well get one if you don't have one.:)
 
   / Solar Batter Tenders
  • Thread Starter
#30  
I have one, just not real versed in how to use it.

Once I add the disconnect would it even matter if I have leakage? It wouldn't be leaking while the tractor is parked.
 

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