current limiting device 6

[Sonix1]

I have a 12V dc battery, which will be fully charged (and will be charged via an alternator), and a 12V dc battery which will be at any level of charge, and when the switch is thrown, they will be connected in parallel, so the fully charged one will charge the discharged one. They will be connected via (most likely) 8AWG wire, 30' or so. I need a device to go in series to limit the current to a max of 20A or so, to prevent the wire from melting and such.

Cheap is priority

I'm not sure what the name of such a component would be, or if they're easily obtainable, or if I could build such a device with a slew of resistors/capacitors.
I am a mechanical engineer, please go easy
Thanks

 

[fsmyth]

Sure. A few headlamp bulbs. Or a lot of
tail-light bulbs. Or a 250W 0.6 ohm resistor.
But if you let the auxiliary battery become
so discharged that AWG#8 cannot handle it,
it is discharged too far. Parallel your wires,
if #8 is all you have. Fuse at 30-40A.

<als>

 

[Skogsgurra]

Cheap, simple, available and perhaps with a big indicator lamp on it?

The problem with resistors is that the current is proportional to voltage difference. You will have your 20 A when the discharged battery is discharged. But, when it charges, its voltage will rise and the voltage difference go down. Which means that the charging current also goes down a lot.

I would use an automatically variable resistor aka 12 V head lamp. It will pass 4 or 5 A when there is a 12 V voltage difference. So you will need four in parallel to get 20 A.

The fine thing with this kind of lamps is that the resistance decreases as current goes down - so the decrease in charging current will not be so big. Actually, when voltage is around one or two volts, you will measure a very low resistance. So current will flow even when the battery is almost fully charged.

You will not have 20 A all the time, but much better than a resistor would give you. And there is a very easy-to-understand indicator lamp thrown in for good measure.

There are other ways of doing this. A chopper with transistors, iron cores and diodes comes to mind. But that is not easy and not cheap. Go with the lights!

Gunnar Englund

http://www.gke.org

 

[Skogsgurra]

Two souls - one thought

Gunnar Englund

http://www.gke.org

 

[itsmoked]

Hey! Use a 12VDC lamp! They are wildly non-linear resistors. Perfect for the task.


Three souls- uno thought.


Couldn't resist.

 

[Skogsgurra]

Couldn't resist? A lamp is a resistor - albeit a bent one ;-)

Gunnar Englund

http://www.gke.org

 

[Sonix1]

Wow. Thanks Gunnar (and others)!

Lightbulbs eh? just simple 12V bulbs? wired in parallel to each other, series inbetween the batteries?, like 4 or so say?

I'm guessing the lights will see the voltage difference, say 12V max, and will light up, (and create resistance to make the light?), and if it's bright, then it means it's charging quickly (high amps), and when it dims, the battery is close to full? As the battery becomes charged, the voltage difference (what the lights see), becomes less, and their resistance drops? (really? is this normal?)

Thanks a bunch

 

[Skogsgurra]

Yupp! You got it. This is how ordinary lamps work.

Gunnar Englund

http://www.gke.org

 

[itsmoked]

Yes it is normal. But don't expect to actually see much light because if you are discharging the "too be charged" battery to less than 12V you are murdering it!! This means the actual voltage difference between your "too be charged" battery[12V] and the fully charged battery[13.6V] is only 1.6V which won't light the bulb except at midnight in a cows stomach.

You could use 6V bulbs to actually see what's happening but if you ever drop your "too be charged" battery below 6V the 6V bulbs transition to fuse mode. If the case really is that you will never flatten the "too be charged" battery below 12V then you could reasonably use 6V bulbs and then keep spares for when you really do flatten the battery.

 

[Skogsgurra]

Hear! Hear! Smoked is perfectly right. And with 55 W 6 V you need only two bulbs - makes things even easier. PLS for Smoked, too.

Gunnar Englund

http://www.gke.org

 

[Sonix1]

I was thinking the "to be charged" battery would be a marine deep cycle, or an Optima or the like. And would most likely drop it's voltage, but not sure how low, time will tell.

I won't need the bulbs for lighting purposes most likely, it probably won't be visible, but I may use tiny lights, then mount them in a visible location.

Can I use any light bulb meant for 12V? I imagine I should buy one (say a cars dome light or something), then measure it's resistance when hooked straight up to a 12V source, then design the amount of bulbs I need based on that eh? I know the basics of circuits (ohms law so I should be able to figure the rest out. I knew there was an easy solution, I was just drawing a blank.
Thanks for all the help.

 

[itsmoked]

What you don't understand is that a 12V car battery, deep cycle or otherwise is functionally dead at 12V and being damaged every hour it sits that way. A fully charged battery is 13.6V. which decays down to about 12.2 or thereabouts when dead.

skigsgurra's selection of bulbs was to meet your requirements. If you want to use "other" bulbs you may need many more to meet your stated requirements.

Note his 55Watts? at 12V? That ohms-laws the resistance.

Head lights are sized in watts, your common tail/dome/courtesy/brake lights usually are not. So yes, you would need to do some testing.

 

[Sonix1]

Yes, I wasn't clear that a head lamp was a head light, or a dome (above your head) light... Anyway, i'll pick some and test, since I could fit 4 (or 10) small courtesy lights in easier then 4 head lights.
Functionally dead at 12V? huh, learn something new every day, I thought it'd still be ok at as low as 10V and such... Ok, i'll test out some courtesy lights I have around, check the resistance and such, and plan from there.
Thanks

 

[Skogsgurra]

Excellent as Smoked might be, I wouldn't take all his words absolutely seriously. You may go below 12 V on a car battery - I think he meant to say that you shouldn't go down to zero. That would be bad. Really bad.

Gunnar Englund

http://www.gke.org

 

[fsmyth]

Take smoke's words seriously. In general,
an open terminal voltage of less than 2.1v/cell
(12.6v) means that the typical lead-acid battery
is at less than 75% capacity - what I would consider
the absolute lowest you want to go if you expect any
decent battery life. Temperature matters, also.
If this is a constant current drain application, use
a deep-discharge battery type.
<als>

 

[Sonix1]

Yes, I will be running it low often, most likely, and so a deep-cycle, or Optima style (also deep cycle), would probably be what i'll be using.

 

[itsmoked]

Thank you fsmyth!

skogs....

If you are *running* it lower than about 70% of its capacity then you need to put in *more* battery!! Or instead of 500 cycles you are going to have 200.. or 100... or 50 cycles before the battery goes dead in minutes. Now if this is a camper that you use 10 times a year then I guess 50 cycles before you have to replace the battery might be acceptable.

Here is a list of battery rules (some of us need reminding of)<not mentioning any names skogs>
********************************************************
1. Shallow discharges will result in a longer battery life.

2. 50% (or less) discharges are recommended.

3. 80% discharge is the maximum safe discharge.

4. Do not fully discharge flooded batteries (80% or more). This will damage (or kill) the battery.

5. Many experts recommend operating batteries only between the 50% to 85% of full charge range. A periodic equalization charge is a must when using this practice.

6. Do not leave batteries deeply discharged for any length of time.

7. lead acid batteries do not develop a memory and need not be fully discharged before recharging.

8. Batteries should be charged after each period of use.

9. Batteries that charge up but cannot support a load are most likely bad and should be tested. Refer to the Testing section for proper procedure.
*******************************************************

 

[Sonix1]

Exactly, a camper. Yea, I was thinking of mayber running more then one, i'll have to see what kind of drains i'm putting on it.
Your rules here (very helpful, thanks!) is that mainly for lead acid, or do the first 4 go for deep-cycle/gel/spiral grid /Optima...

 

[BrianR]

The second battery will come up to voltage almost immediately when the charge starts, limiting the current and with 30' of #8 wire as series impedance that will be enough by itself. I have done this installing a 2nd battery in a 4wd and it works fine. Of course #8 wire will not allow the 2nd battery to act as a starting battery, far too much impedance. You need #2 or thicker for starting.

 

[itsmoked]

Sonix1, that list was from the Trojan (huge name in batts)DEEP CYCLE battery page. It is true for all lead-acid chemistry batteries with minor differences in things like temperature and 1/100's of a volt cell voltages.

Brian is correct mostly.. But if that battery is damaged, a common thing, or stone dead you would smoke the wiring(bad) if not fused. Furthermore if you ever accidently start your vehicle and the engine battery is dead you will again smoke the wires if not fused. If you have the lamps in both cases they will light up brightly in these cases. Also if you have only wire and a fuse you have the problem of what do you do if the deep cell battery is what's called "flat" and you draw so much current that the fuse blows.. Then you stick another in and it blows... You are kind of trapped with no current limiter.