1 Last Battery Question.... I promise

[EvanDoss]

Now that I have a 12v line running from my car to the battery on my pu I can run the Fridge while driving.  But what happends when I stop to eat or something for an hour or more?  If the car is off, the alternator is no longer recharging my car battery and I am afraid the fridge will drain both batteries to the point where I won't be able to start my car.  Is this a problem anybody else experiences?  Or do you simply disconnect the electrical hookup from the car if you plan on stopping for more than and hour?

Thanks again for everyone's help.  I wish I had found this forum last year when I bought my pu.

[wavery]

Now that I have a 12v line running from my car to the battery on my pu I can run the Fridge while driving.  But what happends when I stop to eat or something for an hour or more?  If the car is off, the alternator is no longer recharging my car battery and I am afraid the fridge will drain both batteries to the point where I won't be able to start my car.  Is this a problem anybody else experiences?  Or do you simply disconnect the electrical hookup from the car if you plan on stopping for more than and hour?

Thanks again for everyone's help.  I wish I had found this forum last year when I bought my pu.

I would unplug the trailer and turn off the 12v switch on the fridge (so that it doesn't run down the camper battery) myself. If you are concerned about the fridge, you could light the propane.

[chasd60]

I used to just unplug the camper from my truck. I had a brake controller that had an LED which would not be on if I forgot to reconnect.
 
You can always get a battery isolator and that will allow both to charge and not allow the starting battery to discharge.
http://www.bellrpg.net/soderbloom/RV/grouppart.asp?MAJ=011&PRI=010&SEC=055&GRP=%202336

[wavery]

I used to just unplug the camper from my truck. I had a brake controller that had an LED which would not be on if I forgot to reconnect.
 
You can always get a battery isolator and that will allow both to charge and not allow the starting battery to discharge.
http://www.bellrpg.net/soderbloom/RV/grouppart.asp?MAJ=011&PRI=010&SEC=055&GRP=%202336

Actually, this is all that you would need:
http://cgi.ebay.com/30-40-AMP-12V-RELAY-SOCKET-WITH-DIODE-FREE-KIT_W0QQitemZ270122689089QQihZ017QQcategoryZ75389QQssPageNameZWDVWQQrdZ1QQcmdZViewItem

It's only $10 (Incl S&H) :sombraro:

[mike4947]

But even with a solinoid/isolator the trailer battery will still be draining from the fridge. If you dry camp that means arriving with less than a full battery.

We went as far as running a second line throught the center pin for a dedicated fridge line that waasn't connected at all to the trailer wiring to power the fridge. That and a key controled solinoid made sure we arrived with a full battery, long stops didn't drain down either battery. And if over several hours we lit the propane.
If you don't want to go to extremes simply switch to propane when stopping.

[wavery]

But even with a solinoid/isolator the trailer battery will still be draining from the fridge. If you dry camp that means arriving with less than a full battery.

We went as far as running a second line throught the center pin for a dedicated fridge line that waasn't connected at all to the trailer wiring to power the fridge. That and a key controled solinoid made sure we arrived with a full battery, long stops didn't drain down either battery. And if over several hours we lit the propane.
If you don't want to go to extremes simply switch to propane when stopping.

This is true Mike but if you are stopping for an hour to eat or something, that would normally mean that you were going to be driving for a couple more hours to get to your destination. That should be time enough to replace any charge that was lost in the camper's battery.

As for us, we just prefer to tow with the propane running and haven't had a problem.

[mike4947]

The reason I did it that way was because the TV wasn't replacing the drain from sitting.
If you don't actually check what you're getting back at the trailer, both voltage and amperage you can easily end up with a flat trailer battery. Fridges draw between 8 and 12 amps and they don't care where they get the power from. If the TV supplies 6 amps the frdige will draw the rest it needs from the trailer battery.
Not many TV/trailer connections supply enough amps at a voltage needed to recharge what is lost from a trailer battery.

[mach8274]

Now that I have a 12v line running from my car to the battery on my pu I can run the Fridge while driving.  But what happends when I stop to eat or something for an hour or more?  If the car is off, the alternator is no longer recharging my car battery and I am afraid the fridge will drain both batteries to the point where I won't be able to start my car.  Is this a problem anybody else experiences?  Or do you simply disconnect the electrical hookup from the car if you plan on stopping for more than and hour?

Thanks again for everyone's help.  I wish I had found this forum last year when I bought my pu.

On your Expedition, the charge circuit is only sending power to the trailer if the ignition key is in the on position. The only way your vehicle battery will run down while sitting is if you leave the key on. The fridge will only run on the camper battery even if you leave it hooked up to the tv. I have basically the same electrical system in my F150 and I leave the fridge running on 12v hooked up to the truck all the time. I have left it on overnight before and the truck fired right up. The camper battery will run down though. Ford built the trailer charge circuit that way for that very reason.

Bottom line, leave your trailer hooked up, just keep in mind that the trailer battery may run down, but will start charging once you start the tow vehicle.

Dave

[AustinBoston]

Note: Austin's wheels are turning here, so expect anything to come out...

If you run your pop-up battery dead (using the fridge) while parked, don't expect the tow vehicle to charge the battery at all once it's started.  Here's why.

The amount of charge current that actually goes to the battery depends (in part) on the battery voltage.  A "dead" battery will have a voltage down around 11.5 volts, IIRC.  At that point, It would gladly accept 30-50 amps.  Even at the low end, we then need to add the fridge draw, 8-12 amps more.  So the minimum charge current will approach 40 amps.  Most charge circuits have a self-resetting breaker set for 25 amps.  That breaker is going to trip in short order (probably 5-10 seconds or less), stopping the charge current for 20-30 seconds.  During that time, all of the charge that went on the battery during the 5-10 second on time will be drawn off again by the still-connected fridge.

Anybody able to refine (or refute) this thinking?

Austin

[wavery]

Note: Austin's wheels are turning here, so expect anything to come out...

If you run your pop-up battery dead (using the fridge) while parked, don't expect the tow vehicle to charge the battery at all once it's started.  Here's why.

The amount of charge current that actually goes to the battery depends (in part) on the battery voltage.  A "dead" battery will have a voltage down around 11.5 volts, IIRC.  At that point, It would gladly accept 30-50 amps.  Even at the low end, we then need to add the fridge draw, 8-12 amps more.  So the minimum charge current will approach 40 amps.  Most charge circuits have a self-resetting breaker set for 25 amps.  That breaker is going to trip in short order (probably 5-10 seconds or less), stopping the charge current for 20-30 seconds.  During that time, all of the charge that went on the battery during the 5-10 second on time will be drawn off again by the still-connected fridge.

Anybody able to refine (or refute) this thinking?

Austin

The vehicle's regulator will receive a signal of the total voltage package, vehicles battery, equipment being used (draw) and the TV battery (in this case, another draw). The regulator will sense a total voltage of 12+ volts (no matter how dead the camper's battery is) and most of the charge will be directed to the vehicle's battery (path of least resistance). Although, the alternator will be putting out a slightly higher amperage but seldom more than 10A over the total draw after about 10 minutes.

The alternator won't force more amps down a wire than it can handle, as long as it has an easier place to dump that those amps (path of least resistance).

[AustinBoston]

The vehicle's regulator will receive a signal of the total voltage package, vehicles battery, equipment being used (draw) and the TV battery (in this case, another draw). The regulator will sense a total voltage of 12+ volts (no matter how dead the camper's battery is) and most of the charge will be directed to the vehicle's battery (path of least resistance).

Stop right there!  You are making a huge assuption - that all of the current would come from the charger.  Given a heavy enough connector, you could get several hundred amps to flow from a fully charged battery to a dead one, even without an alternator.

Although, the alternator will be putting out a slightly higher amperage but seldom more than 10A over the total draw after about 10 minutes.

This is also dependent on the setup.  My Astro is equipped with an oversized alternator that is supposed to provide "extra power" for towed vehicles.  The manual doesn't specify what "extra power" means.  Could be three extra amps, could be thirty.

The alternator won't force more amps down a wire than it can handle, as long as it has an easier place to dump that those amps (path of least resistance).

And it won't find an easier path than a dead battery at 11.5 volts.

However, I do think there's a hole in my thinking.  In a standard setup, the fridge is connected outside the 25 amp breaker.  (The breaker is the closest thing connected to the camper battery.)  This would mean that when the breaker trips, the fridge is disconnected from the camper battery, but not from the charge line.  In addition, the fridge current would not be part of the current that tripped the breaker as I originally stated.  The fridge would not drain the camper battery while the breaker is tripped.

With a badly drained battery, it could still get "pulses" of charge until the voltage rises enough so that the breaker stops tripping.

Austin

[wavery]

Stop right there!  You are making a huge assuption - that all of the current would come from the charger.  Given a heavy enough connector, you could get several hundred amps to flow from a fully charged battery to a dead one, even without an alternator.



This is also dependent on the setup.  My Astro is equipped with an oversized alternator that is supposed to provide "extra power" for towed vehicles.  The manual doesn't specify what "extra power" means.  Could be three extra amps, could be thirty.



And it won't find an easier path than a dead battery at 11.5 volts.

However, I do think there's a hole in my thinking.  In a standard setup, the fridge is connected outside the 25 amp breaker.  (The breaker is the closest thing connected to the camper battery.)  This would mean that when the breaker trips, the fridge is disconnected from the camper battery, but not from the charge line.  In addition, the fridge current would not be part of the current that tripped the breaker as I originally stated.  The fridge would not drain the camper battery while the breaker is tripped.

With a badly drained battery, it could still get "pulses" of charge until the voltage rises enough so that the breaker stops tripping.

Austin

AB.......

The "Path" is not the battery. The "Path" is whatever size wire that you have going TO that battery. The wire going to the vehicles battery will probably be a "0"gage or "00"gage battery cable that is connected directly from the alternator to the battery and less than 3' long (in most cases).. The wire going to the camper battery will probably be a 10 or 12 gage wire over 20' long, which will offer a lot of resistance (in comparison). Therefore, most of the amperage will go to the TV battery (path of least resistance) until the resistance evens out which would mean a pretty full charge on the TV's battery.

BTW.....I never make "assuptions" :p

[AustinBoston]

AB.......

The "Path" is not the battery. The "Path" is whatever size wire that you have going TO that battery.

The path is the entire circuit, and includes both batteries.  They form a parallel circuit, and where the current goes in a parallel ciruit is always proportional.  Read more below.

The wire going to the vehicles battery will probably be a "0"gage or "00"gage battery cable that is connected directly from the alternator to the battery and less than 3' long (in most cases).. The wire going to the camper battery will probably be a 10 or 12 gage wire over 20' long, which will offer a lot of resistance (in comparison). Therefore, most of the amperage will go to the TV battery (path of least resistance) until the resistance evens out which would mean a pretty full charge on the TV's battery.

BTW.....I never make "assuptions" :p

I wrote an extensive reply to this last night, then the site crashed.  

Your biggest assumtion is that the current just chooses a single path.  In a parallel circuit, current will always flow proportionally through all parts of that circuit.  If this were not true, then you would never be able to run more than one electrical device (A/C, headlights, radio, etc.) at a time.

The "path" is the entire circuit, and where the current flows is primarily dependent on voltage drop.  Part of the story on voltage drop is the resistance of the wires, but that is only part of the story.

Start with a 10 gauge wire 25 feet long.  10 gauge wire is 1 ohm per 1,000 feet, so 25 feet of it has a resistance of 0.025 ohms.

Now take an effectively dead battery at 11.5 volts, and a fully charged battery at 12.7 volts.  The potential difference between the two is 1.2 volts (12.7 - 11.5 = 1.2V).

Connect them through the 25 feet of 10 gauge wire?  How much current will flow?

Using Ohms law (E=IxR where E is voltage, I is current, and R is resistance), we have:
1.2 = I x 0.025

Solving for I gives us 48 amps.  That's even before the truck is started.  That truck battery isn't going to loose it's ability to run current to the camper because the alternator is running.  Instead, it will provide any current the alternator is unable to supply.  If the alternator is able to maintain a system voltage of 12.7 volts, then the truck battery will not drain or charge.  If it goes above 12.7, the truck battery will see a charge current.  But as the voltage rises, the camper battery will see more current.  Read on.

If the alternator is able to maintain the typical charging voltage of 14.4 volts, that means a voltage differential of 2.9 volts (between alternator and camper battery).  Again, solving for I, we get a charge current to the camper of 116 amps.  In order for it to maintain that 14.4 volts, it would have to be able to supply all of the current needed to charge the truck battery, all of the current to run the truck, and 116 amps for the camper battery.  If it can't do all three, the alternator's output voltage will drop, but the current will still be split proportionally between running the truck, charging the truck battery, and charging the camper battery (unless, as stated above, the alternator output drops to or below the truck battery voltage, when 100% of the current not used to run the truck will go to the camper battery, and the camper battery will see 48 amps).

Invalid assumptions I have made:

* That is that the batteries have no internal resistance.  They do have some internal resistance, and that becomes a part of the equation (essentially causing the battery voltage to rise while charging or drop while discharging), so the actual camper current would be lower than calculated, but I doubt it would drop by more than 30% of what is stated here.
* That the ground side of the circuit has no resistance at all.  If a heavier gauge wire is used for the ground, and/or good electrical contact is made through the hitch, then the numbers will not change significantly because of the ground, but the current values would be at least slightly lower.
* A third assumtion is that everything on the load side of the circuit acts as a resistor.  While I believe there is some capacitance in a battery, I don't think this changes the numbers significantly because we are talking a DC circuit.

Some reference material:
Ohm's Law on Wikipedia
Series and parallel circuits on Wikipedia

Austin

[wavery]

The path is the entire circuit, and includes both batteries.  They form a parallel circuit, and where the current goes in a parallel ciruit is always proportional.  Read more below.



I wrote an extensive reply to this last night, then the site crashed.  

Your biggest assumtion is that the current just chooses a single path.  In a parallel circuit, current will always flow proportionally through all parts of that circuit.  If this were not true, then you would never be able to run more than one electrical device (A/C, headlights, radio, etc.) at a time.

The "path" is the entire circuit, and where the current flows is primarily dependent on voltage drop.  Part of the story on voltage drop is the resistance of the wires, but that is only part of the story.

Start with a 10 gauge wire 25 feet long.  10 gauge wire is 1 ohm per 1,000 feet, so 25 feet of it has a resistance of 0.025 ohms.

Now take an effectively dead battery at 11.5 volts, and a fully charged battery at 12.7 volts.  The potential difference between the two is 1.2 volts (12.7 - 11.5 = 1.2V).

Connect them through the 25 feet of 10 gauge wire?  How much current will flow?

Using Ohms law (E=IxR where E is voltage, I is current, and R is resistance), we have:
1.2 = I x 0.025

Solving for I gives us 48 amps.  That's even before the truck is started.  That truck battery isn't going to loose it's ability to run current to the camper because the alternator is running.  Instead, it will provide any current the alternator is unable to supply.  If the alternator is able to maintain a system voltage of 12.7 volts, then the truck battery will not drain or charge.  If it goes above 12.7, the truck battery will see a charge current.  But as the voltage rises, the camper battery will see more current.  Read on.

If the alternator is able to maintain the typical charging voltage of 14.4 volts, that means a voltage differential of 2.9 volts (between alternator and camper battery).  Again, solving for I, we get a charge current to the camper of 116 amps.  In order for it to maintain that 14.4 volts, it would have to be able to supply all of the current needed to charge the truck battery, all of the current to run the truck, and 116 amps for the camper battery.  If it can't do all three, the alternator's output voltage will drop, but the current will still be split proportionally between running the truck, charging the truck battery, and charging the camper battery (unless, as stated above, the alternator output drops to or below the truck battery voltage, when 100% of the current not used to run the truck will go to the camper battery, and the camper battery will see 48 amps).

Invalid assumptions I have made:

* That is that the batteries have no internal resistance.  They do have some internal resistance, and that becomes a part of the equation (essentially causing the battery voltage to rise while charging or drop while discharging), so the actual camper current would be lower than calculated, but I doubt it would drop by more than 30% of what is stated here.
* That the ground side of the circuit has no resistance at all.  If a heavier gauge wire is used for the ground, and/or good electrical contact is made through the hitch, then the numbers will not change significantly because of the ground, but the current values would be at least slightly lower.
* A third assumtion is that everything on the load side of the circuit acts as a resistor.  While I believe there is some capacitance in a battery, I don't think this changes the numbers significantly because we are talking a DC circuit.

Some reference material:
Ohm's Law on Wikipedia
Series and parallel circuits on Wikipedia

Austin

Electricity is a lot like water and batteries are a lot like non-vented (but expandable) water reservoirs.

Let's make the batteries 2 equal size water beds made with unbreakable but flexible material. One water bed is connected to the main water supply system with a 2" diameter PVC pipe, 3' long (simulating the TV battery cable). The other water bed (#2) is connected to the exact same water supply with a 1/2" PVC pipe, 20' long (simulating the supply line to the camper battery).

Now water bed #1 is already 3/4 full and water bed #2 actually is in a vacuum state (Drawing the water to it).

Water bed #1 will obviously fill rather quickly because there will be very little resistance in the 2" supply pipe (path of least resistance) until the bed starts to reach capacity. As #1 starts to reach capacity, it's resistance will build. This will automatically turn down the pressure on the water supply (via the "Water regulator" :p ) However, #2 will ALWAYS be limited to the amount of water than can get to it by the 1/2" supply pipe that is the "Path" that has the most resistance. As the regulator turns down the pressure, #2 will also receive a proportionate less supply.

Your facts about the "pressure" (voltage) is correct. The pressure will be constant throughout the entire system. However, the volume (amperage) will be considerably less to #2 than it will be to #1. Amperage is what actually fills the battery not voltage.

If you try to force too large a volume of water down a water pipe it will burst. If you try to force to many amps down a wire, it will melt. There is no way that a 10gage wire will handle 100amps. It would simply melt. Max amperage for a 10g wire is 55A. I haven't heard of any melted battery charging wires on camper battery supply lines.

Now.....the big question is......do you want a water bed in your PU :sombraro:

[AustinBoston]

Electricity is a lot like water and batteries are a lot like non-vented (but expandable) water reservoirs.

But far from identical.

Let's make the batteries 2 equal size water beds made with unbreakable but flexible material. One water bed is connected to the main water supply system with a 2" diameter PVC pipe, 3' long (simulating the TV battery cable). The other water bed (#2) is connected to the exact same water supply with a 1/2" PVC pipe, 20' long (simulating the supply line to the camper battery).

OK, but you are missing something.

Now water bed #1 is already 3/4 full and water bed #2 actually is in a vacuum state (Drawing the water to it).

Water bed #1 will obviously fill rather quickly because there will be very little resistance in the 2" supply pipe (path of least resistance) until the bed starts to reach capacity.

This is only true if the water supply is relatively unlimited.  It is not.

As #1 starts to reach capacity, it's resistance will build. This will automatically turn down the pressure on the water supply (via the "Water regulator" :p ) However, #2 will ALWAYS be limited to the amount of water than can get to it by the 1/2" supply pipe that is the "Path" that has the most resistance.

You are still assuming the wire is the limiting factor.  I am so sure it is not that I would bet my life on it; and I am not a gambling man.

The fact is, with a 12 volt supply, 25 feet of 10 gauge wire is limited to 480 amps (I=E/R, or 480=12/0.025).  Of course, at that current, it would very quickly do other things...

As the regulator turns down the pressure, #2 will also receive a proportionate less supply.

Your analogy breaks down in a number of ways.  First, for the analogy to be the same, water bed #1 would need to be about 250 feet above the water supply.  Water bed #2 would have to be 30 feet below it.  These altitudes represent the differing voltages of the batteries.  You can not ignore this or everything else goes wrong.  

Then, the original supply would have to be a 2" pipe limited to 50 P.S.I. (equivalent to voltage), split into a 2" pipe and a 1/2" pipe at the supply point.  Water is going to flow from WB #1 into WB #2 under very high pressure even if the supply is turned off.  If the supply is then turned on, water may flow to bed #1, but the amount flowing to bed #2 will ONLY INCREASE.  Even if the pressure is turned back on the supply, water will still flow under great pressure to WB #2.  In fact, as the pressure is turned back enough, the difference in the flow rate to WB #1 will be propotionately huge and eventually stop and reverse (so that water is flowing out from bed #1 to bed #2) even with the supply on.  At the same time, the difference in the flow rate to WB #2 will hardly change at all.

If you miss the HUGE difference caused by the difference in the voltages of the two batteries, you miss everything.

Your facts about the "pressure" (voltage) is correct. The pressure will be constant throughout the entire system.

No, that's not what I've said.  If the voltage was the same throughout the system, no current would flow at all.  All three - both batteries and the alternator - are potential current sources, and so no current flows unless there is a voltage difference between them.  (this is known in electronics as bucking.)  The greater the votage difference, the more current will flow.

However, the volume (amperage) will be considerably less to #2 than it will be to #1. Amperage is what actually fills the battery not voltage.

Not in the corrected scenario.  Even so, we're talking about electricity, not water.  It follows very well known laws (well, all except microwave electricitry, which many engineers consider to be black magic  :eyecrazy: ).

If I have a 12 ohm resistor on the end of a 6" 00 cable with a 12 volt supply, that circuit will have 12 ohms of resistance and carry 1 amp.  This is Ohms Law, and it is indeed a physical law that no analogy can successfully ignore.

If I then have a 1 ohm resistor on the end of 1,000 feet total (500 feet each way) of 10 gauge wire with the same 12 volt supply, that circuit will have 2 ohms of resistance (one ohm for the resistor and one ohm for the wire) and will carry 6 amps.  Again, this is Ohms Law.

Notice that the higher resistance wire is carrying six times as much current.

If I connect both of those circuits together in parallel, the total resistance (at the supply) will be 1.72 ohms:

R(total) = 1 / (1/R1 + 1/R2)
R(total) = 1 / (1/12 + 1/2)
R(total) = 1.72 ohms

Plugging this in to Ohms Law gives a current of 7 amps (what we would expect if we were running a 1 amp circut and a 6 amp circut in parallel):

I = E / R
I = 12 / 1.72
I = 6.98 (~ 7) amps (rounding error)

Despite the size of the larger cables, most of the current will go down the smaller cable because it is the lower resistance path.  You can't pretend that the wire is the only relevant resistance here.  If you do, you again will miss everything.

If there is a limit on the supply side, so that the source can only produce 5 amps, the voltage would drop, but the majority of the current will continue to flow on the side with the lowest toal resistance.  At a current limit of 5 amps, the supply voltage would drop to 8.6 volts.  (E=IxR, 8.6 = 5 x 1.72)  The 12 ohm resistor would carry 0.72 amps (8.6/12), and the 1-ohm resistor (remeber the total with the wire is 2 ohms) would carry 4.3 amps (8.6/2), for a total of 5.02 amps (rounding error again).

But this is an oversiplification because the effective resistance of the batteries changes with the applied voltage.  If the alternator voltage dropped to the standing voltage of the good battery, it's resistance would become effectively infinte, and all current into it would stop flowing to that battery, no matter how big the wires.  Current would continue to flow to the dead battery, and lots of it.

What's important in this is not the wire sizes or lengths.  The reason the truck's internal cabling is as heavy as it is isn't because of wire resistance, it is to prevent excess heating at the design currents.  (Excess heating is also the reason the charge line has a fuse and a circuit breaker.)  What's important is the voltage of the batteries (which essentially acts like a resistance).  Everything else is almost irrelevant.

Austin