Charging a battery from a battery bank...

[windstuffnow]

  I have a 12v deep cycle battery that I use for portable power along with a 600w inverter.  Its used for drilling, sawing or what ever I need out in the field where there is no power.  I've been using some small solar panels with it but it takes forever for them to bring it back after use.   I started using a regular battery charger plugged into the main inverter but that's really not very efficient.

  I'm wondering if I could incorporate it into my battery bank.  I don't use it alot but its handy when I need it.   I don't think it would be very safe to simply connect it directly to the bank, it seems like the in rush of amps would surly kill it... Is there a way to control the amps going into the battery? with reasonable efficiency...

  I was told at one time to use a controller, but I don't believe you can control the incomming amps this way and would lead to blowing the controller.   I really don't know exactly how they work in that respect...

  Any helpful ideas ?  A simple circuit that I could handle maybe?

Thanks

Have Fun

Windstuff Ed

 

[TomW]

Ed;

I am sure you could rig some circuit to do the job. However, if it was me I would just connect it thru a light bulb or ballast resistor to limit current into the battery when you hook it in.

Not super efficient but the resistance will limit current flow and when the bulb quits glowing you could shunt around it with a switch to direct connect for the final topping off. Not very high tech but would be safer than just hooking it up directly.

Cheers.

TomW

[Flux]

Wire a couple of headlamp bulbs in parallel and put them in series with the battery.

They will limit the current to a safe value and as the battery charges they will not waste much power.

Flux

[richhagen]

How about a dc-dc buck converter from your bank voltage to just above the battery voltage, ideally about the float voltage (as that way you could hook it up to charge and forget about it until you need it again)  Connect the converter in series with enough resistance to limit the current to be within the capabilities of the converter when the battery is deeply discharged.  Add a fuse in series for safety of a value just above current the circuit should draw for a discharged battery.  This would greatly slow the charging, but should be more efficient than just resistance in series from the bank to the battery.  Adjustable 24-12, 36-12, and 48-12 converters are widely available and can usually be obtained through e-bay as well as many other on-line sources. Rich Hagen

[windstuffnow]

   Thanks all, I'll experiment with some nichrome wire for a resister.  I like tom's idea using a breaker switch when its pretty well done charging.  I can leave it connected to the bank as extra storage until I need it... and it will of course be fused...

More fun...

Windstuff Ed

[tecker]

 Use a 10 amp power supply ( or you could make one with a a regulator and a power mosfets ) altronics smp 10  is good and has a life time warranty. Or hang it in there parallel to a 12 volt increment or your bank ,or here's a 5 amp supply using common part I use it and you can stack on more transistor resistor sets for more power I use the smp10's all the time in security installs so I got a bunch to play with.

http://www.mitedu.freeserve.co.uk/Circuits/Power/boosti.gif

[BT Humble]

Any helpful ideas ?  A simple circuit that I could handle maybe?

I bought one of these kits:

http://www.oatleyelectronics.com/kits/k072.html

...but one of the minor transistors inside didn't like the idea of running of 24VDC and went pop!

So I went back to the old low-tech method of putting a few "festoon globes"[1] in series with the "chargee" battery.  If you use 3 or 4 in series they'll never blow out, and that dropped the current to a nice reliable 300mA.  I just connect whatever battery I want to charge, take note of the time, and let it run for however many Ah I want to put into the battery.

For higher amperage I'd use a headlight bulb or two in series, or a few strands of resistance wire if you prefer.  I prefer the lightbulbs because then I'm far less likely to accidentally leave a battery on charge all night.

BTH

[1] They're the type that they use inside those running ligts down the side of trucks.

[Opera House]

Using the existing charging system is the simplest method.  I hardly think a resistance of more than 0.2 ohms is needed if that.  The hookup cables should provide enough resistance to prevent any damage.   I also prefer a boost regulator simply because there will be times when you want to charge the battery off the battery bank when it is not also being charged.  Like overnight, when you have solar panels and you need a fresh battery the next morning.   A simple boost regulator using the TL494 switcher chip can be found in the application sheets using two sense inputs.  With this you can set the max voltage and also the max charge current.  Basically it would act like a constant current charger till the terminal voltage is reached.

This chip can be found in many old computer power supplies because it is multi sourced and costs under a quarter.  The switching transformer can be used as the inductor along the 30A switching diode and heatsink are about everything you need except for the FET.

[wooferhound]

Seems like you could pulse charge directly from the battery bank into the portable battery.

[TomW]

The hookup cables should provide enough resistance to prevent any damage.

I feel I must chime in here and say this could be very dangerous advice.

Just curious if you have ever actually worked with banks of batteries and have any idea of the potential in them?

If you connect a low battery at, say, 11 volts to a mostly charged bank at 12.8 or so the higher voltage bank will do its best to blow as much current into the second battery as it can trying to raise it to the level of the big bank. This can easily exceed the second battery's ability to absorb the huge inrush of currant and a likely result will be damage to the battery or other items like the stuff that the explosion splashes acid on.

A ballast resistance solves this dangerous situation and is quite simple and safe.

T

[windstuffnow]

  Looks like a circuit that I can handle,  I believe I have the regulator and a 710 laying around, I may give it a shot.  Thanks!

hAvE FuN

Windstuff Ed

[stop4stuff]

Hi windstuff Ed,

A quick google for '12v peak charger' revealed a couple of sites...

US... http://www.hobby-lobby.com/chargers.htm

UK... http://www.gmmodels.co.uk/acatalog/Field12v_Chargers.html

Both have delta peak chargers that can be used for 2-24v lead acid batteries, though they're not cheap

A delta peak charger tests the voltage and manages the charge current to the battery, reducing the current to a trickle charge once the battery voltage has peaked.

I use a couple of different peak chargers for my r/c models so I can charge their NiCad packs from a 12v car battery (which is usually still in the car.)

An off-the-shelf solution if you have the bucks.

paul

[tecker]

You can also use a lm 317 and get your 14 to 18 volts with no problem ( the circuit is on the back of the package at radio shack) incedently I'm really interested in about five of those disks you have on your site .Whats the average wieght of one of the 12'steel disks.

   

[Tyler883]

If you go to an RV and motorhome shop they can sell you something along the lines of a 30 Amp auto resetting curcuit breaker.

I have one on my deep cycle on my motorhome so when it is at 50% capacity, and I start the engine - I don't burn up the 16 gauge wires that charge my deep cycle.

[Norm]

  How about something really low-tech? Just take one of your alternators that will put out about 10 amps or so coupled with a controller so it won't overcharge a small dc motor to drive it....some kind of sensor or even a timer to shut off the motor.....or even an overspeed device ...when the motor starts to overspeed because the controller is cutting down on the charging rate, the motor kicks out...

                 ( :>) Norm.

             

[windstuffnow]

  tecker,  I believe their about 6 lbs each.  

Windstuff Ed

[tecker]

  Sounds good

[Tyler883]

A "30 Amp auto resetting curcuit breaker." is just my descriptive name for a pretty low tech solution too. They look like a ceramic terminal block and they sell for about $10. Just hook it up, and it will control your charge current based on its rated current. They are quite reliable, too. The ones on my RV are close to 20 years old. I had to replace one, only because the connections were corroded.

[nothing to lose]

Two things I can think of.

First someone mentioned a turn signal flasher once when we were talking about pulse chargers and such. I think that would work well here. You would be sending pulses of currant to the battery to charge it. I would think for the type of use you described pulse charging might be a good option for this battery. I would think the flasher would waist less juice than a headlight, still supply full volts and amps, and maybe the pulsing would extend the battery life a little also. Though I don't think that the real pulse chargers are quite that simple, it should work.

 Also you could use the flasher plug sort of as a switch, take out flasher and circiut is dead, connect battery and insert flasher to turn on the circuit. Use an inline fuse also.

Those resetting auto breakers are good for alot of things too, cheap at the parts stores and about free from a junk yard. But they aren't very good for some things too, since they reset instead of blowing they can almost act as a flasher and that's something you don't want if you have an arcing short.

As for

"If you connect a low battery at, say, 11 volts to a mostly charged bank at 12.8 or so the higher voltage bank will do its best to blow as much current into the second battery as it can trying to raise it to the level of the big bank. This can easily exceed the second battery's ability to absorb the huge inrush of currant and a likely result will be damage to the battery or other items like the stuff that the explosion splashes acid on"

You mean sort of like connecting jumper cables to a dead car battery and a fully charged battery with the vehicle running and a 100amp altenator??

 I do that all the time and never had a problem yet! Seems like that would be about the same thing really.

 If the vehicle is hard to start I let it charge that way for awhile to get juice into the battery. If easy to start then I just connect and crank then disconnect and let the altenator charge it.

 Either way is hammering alot of amps fast into a dead battery.

Also I have booster jumpers that kick out 500 amps for starting cars with dead batteries, though I don't know if it only kicks that 500amps out when under the load of the starter or as soon as connected to the dead/weak battery.

And I often kick about 100amps into a dead battery with my 1 ton truck. It has a selector switch for bat 1, bat 2, both, or none. Often one battery is dead, REAL DEAD, so I start on the other and switch to the dead one for charging. My 100amp altenator has buried the needle on a 60amp gauge many times, and the gauge no longer works now either. Never a problem with the battery though.

Actually, will the power supplie force alot of amps at the dead battery, or will the dead battery draw amps from the power supplie?  Kinda like a 1000 amp battery bank suppling juice to an inverter? The power is not forced into the inverter, it draws the power. So 1amp is all the inverter gets if only drawing 1amp, but kick on something heavy duty and the inverter draws a sudden surge of 500 amps from the bank.

[Tyler883]

"Actually, will the power supplie force alot of amps at the dead battery, or will the dead battery draw amps from the power supplie? "

That requires a multipart answer, but it's one that is easy enough for everyone to understand....I hope this doesn't bore anyone:

1. In some ways, the dead battery is just another load, and it will have a load resistance just like any other electrical device. Therefore, the NET Voltage ( charge voltage minus the dead battery voltage) supplied to the battery, in combination with this load resistance will determine the load current.( Ohm's law: I =V/R )
   
2. If you assume that you have a very good charger that can provide the current demands of any dead battery that you hook up to it, then it is perfectly ok to say that " the dead battery draws amps from the power supply". In fact, I like this expression because it makes it perfectly clear that a 10 amp charger doesn't always supply 10 amps..i.e. if the resistance or state of charge of the battery is sufficiently high enough, a 10 amp charger might only be supplying 1 amp.
   
3. with respect to the  " will the power supplie force alot of amps at the dead battery"......this is not incorrect in all charge conditions. Keeping in mind that the charger will never defy ohm's law, it is still possible that you have a battery that is sooooooo dead that its charge current is too excessive. In other words, in cases of (relatively) dead batteries, you might want to rephrase things and say " the power supply is providing the dead battery with too many amps.
   
4. When you connect a 12.5 Volt battery(battery to be charged) to a bank of 13.8 Volt batteries, the 12,5 volt battery is a "relatively dead battery" and will probably take a current similar to the following calculation ....(13.8-12.5) / .012 ohms = 108 Amps....not difficult for battery banks to supply, however,108 amps doesnt match many charge profiles...... c/10 or c/20( typically reccomended for a battery that has more than 80% charge) ussually works out to be about 5-20 amps for most batteries that that have a capacity of 100-200 Ah. Frankly, I would prefer to see a topping up charge that is lower than 5-20 amps.....however,I'm not sure if this is overreacting.
   

[TomW]

NTL;

Hmm, I think you may be a bit confused on how loads and sources interact.

Compared to my personal bank of 1500 AH at 12 volts your 100 amp alternator is pretty puny in regards to current output capacity.

I can quite easily draw 350 amps from it into multiple inverters and conceivably it could supply a couple thousand for a time into a heavy enough load.

For instance If I hooked up my lawn mower battery to this bank and it was at say 11 volts the bigger bank would supply as much current into it as it required to try to pull it up to an equal voltage.

Internal battery resistance is a dynamic thing so while it follows Ohms law you can't really assign it a static number that works for all levels of charge.

A series resistance will limit the current to the second battery because it has a voltage drop across it that increases as current increases and at some point it will limit that voltage to the difference between the 2 battery banks because of ohms law.

Using the .012 ohm internal battery resistance figure noted in another comment and with a difference of [13.8-11] or 2.8 volts the current into this hypothetical battery would draw 230 amps or so.

Near as I can tell not many individual consumer batteries can take this type of current for any length of time without damage. It also would represent 2.8 kilowatts of power dissipation inside the battery. Well beyond my little mower battery's ability to absorb. More than likely more than any battery by itself could, too.

I hope this makes sense because I think it is a point of safety to realize the sheer potential of any large current source like a battery bank in an RE setup. I would just hate to see someone blow up a battery because it is an ugly thing to have happen.

T

[Tyler883]

Good post, Tom.

moving on to current limiting resisters:

if we wanted to limit the current to 30 amps, we would need a .4 ohm power resister( R = V/I = 12V/30A = .4).

Unfortunately, this would have to be a 36 Watt resistor ( P= IIR = 90Amps*.4ohms) while most resistors that are common in household electronics are about .25 - 2Watts. 5 and 10 watt resistors are easy enough to buy at radio shack, it would be easy enough and cheap enough to put 3-4 10 watt resistors in parallel with each other....

...but I dont like this solution.....

...because this method of charging is going to cost us 36 watts for the entire charge time....which really sucks when you consider that most of this power will be lost in the current limiting resistor during early stages of charging.

[nothing to lose]

OK I think I get it now, at least some of it.

Kinda like water seeks it's own level. But if you have a very tall full container on the left and an empty on the right of the same size and open a flood valve between the two, there will be a fast surge of rushing water from one to the other and often the empty will actually rise momentarily to a point higher than the full one, then it evens back out. In this case it was caused by the momentum and a push pull effect of the moving water.

About the same thing with amps?

[Tyler883]

NTL,

In my opinion, thats a pretty good analogy.

Its easy to see that the flow into the empty container is limited by the size of the valve

but yet the flow will also be effected by the water level in both tanks.  

if the source tank is very tall it will apply more pressure to water flowing through the valve.

And, as the the empty tank fills, it water level will apply a counter pressure to the water flowing through the valve.

BTW. water analogies often work quite well....

Consider a garden hose;

the water pressure is similar to voltage

the water flow( quantified amount of water past a certain point in the hose per unit of time) is similar to current

the spray nozzle: if operated in an open/close fashion is the same as a switch

the spray nozzle: if partially opened is similar to a resistor. The orifice restricts the flow. ( howeever, there is no analogy that compares to the way that the water comes out at a higher pressure)

The tap at the house is similar to a regulator.

A diode is similar to a one way valve, in one direction it stops the flow( amps) unless the pressure(volts) is so great that it causes the valve to break open(break over voltage),

But a valve that that is made to compare with a diode operates a bit differently than a tyopical valve. The typical valve when opened behaves like any other orifice designed to restrict flow. But a valve that imitates a diode provides a constant pressure drop ( .7 Volt drop). By doing so, it inherently allowis differing current flows( Amps) whenever it is being subjected to a different pressure( volts)

.....er...I think this is a good point to stop.......if I continue making these analogies, I'm quickly going to start screwing up - ha ha ha.

cheers

Tyler