I have a 0.25A@14V solar panel that I want to use to float charge two 12V flooded lead-acid deep-cycle batteries (boat). I would like to use a current-splitter to feed the two batteries so that if either battery were to develop a shorted cell, the other battery still gets half the available solar current.

I have used a "battery isolator" consisting of two diodes connected common anode, with the two cathodes connected to the positive pole of the two batteries, respectively. However, that just means that one battery can hog all the available current. In the event the battery goes bad, it hogs all the current and the other (good) battery gets none.

The current splitter circuit should have less than 1V drop between the solar panel input and the positive pole of the battery it is connected to.

 

You're going to need to monitor the current in and out to both batteries to be able to split it effectively. A duel switcher which limits Iout for each battery to be Iin/2.

I've been pondering how hard it would be to setup a charge system that charges each battery on a bank individually, and this is the basis of it. Would like to hear other ideas here too.

 

If you have a variable voltage input source the simplest way I can think of is the standard LM317 voltage regulator set up in constant current mode.
Once its set right each regulator will always give its load the same charging current regardless of if the input voltage goes high or the battery voltage drops low.

the attached spec sheet shows how you can set it up.

Attached Files

LM317t Adjustable voltage regulator..pdf (732.6 KB, 32 views)

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The problem with a LM317 limiter is that is requires a 1.2V drop before it starts limiting. You may want one that limits at a lower voltage.

You also could use a fold-back type limiter that would actually reduce the current limit if the battery voltage drops significantly below the charging voltage. A Google of "current limit circuit" will generate numerous hits.

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Carl
Curmudgeon Elektroniker

 

And that's just the voltage across the sense resistor. The total headroom required is closer to 3V.

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Ron

 

How much current will your solar panel provide?

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Ron

 

Open Circuit, 19V
0.25A@14V.
0.5A into a short.

 

How about this: Multiplex the solar cell with a 50% duty cycle, low frequency (e.g., 1Hz) multivibrator, driving two P-channel (or N-channel, if it is convenient to switch the low sides of the batteries) to switch between batteries. I don't think you would need any diodes for isolation, because the body diodes do that for you, and the intrinsic diode in the solar cell would prevent it from draining the batteries during darkness (I think ).
With a voltage monitor and a timer, you could probably isolate a bad battery and stop the multi, diverting all the current to the good battery until you replace the bad one. This sounds like a job best done using a microcontroller.

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Ron

 

The solar cell diode polarity, when generating current, is positive at the anode (that's why a single solar cell has such a low voltage). Thus, when dark, the battery will be seeing a series of forward biased diodes. This may cause a small current drain from the battery, depending upon the number of solar cells in series. I believe some solar cells have a forward biased diode in series with the output to prevent this drain.

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Carl
Curmudgeon Elektroniker

 

You're right. My bad. A Schottky diode in series with the solar cell is probably required.

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Ron

 

Clever! Much easier and no loss of power due to proper current sensing. And the current pulses should have some anti-sulphating effects which seems useful given the OP's fear of a battery fail problem.

Only needs a couple of pfets and a logic chip which can do both 50:50 oscillator and inverter/driver?

 

MikeMI, did you ever get this problem solved?

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Ron