Yet another battery charging question!

jepomer · Sep 23, 2014

My boat has two battery banks: one 24 volt, 210 amp-hour for the electric motor; the other a Group 24 12 volt battery to run the lights, electronics, etc.

I use a single array of 3 solar panels (Ganz semi-flexible mounted on the deck) to feed two separate charge controllers - one for each battery bank.

This eliminates the problems of charging two two different types of battery setups.

Maine Sail · Sep 23, 2014

weinie said:
Assume one has a 200 AH house bank and a small solar panel of say, about 35 watts producing about 1 amp or so (with a controller, of course).
The house bank is connected to much smaller reserve or start battery via an ACR or other similar combiner.
Say the house bank is very depleted (but still above the cutoff voltage level for the ACR to close the circuit). The combiner senses the charging voltage in the system as provided by the solar panel and combines the house bank and the starting bank.
What prevents a fully charged start battery from discharging into the house bank since the internal resistance of the house bank is so low in comparison to the fully charged start battery?

The solar panel should be wired to the house bank. It will not combine until the house bank can be maintained at 13.0V by the solar arrays available current. The fully charged start battery will be at 12.7V +/- thus the house bank voltage is already above the idle start battery voltage.. Current won't flow uphill to the higher voltage source......

By the time a 1A source can bring a house bank to 13V the start battery will not dump current into the house bank. This is why it is important to lead all charge sources directly to the house bank not the start bank....

SOC & battery terminal voltage is directly proportional to the current avaible to drive it there. With a 1A current source you will not come into the combine range until very late in the SOC curve. With a 200Ah bank in good health I would not expect you will attain combine voltage until the upper 90's as an SOC. With a 1A current source absorption voltage will likely not be attained until 99% + SOC...

Maine Sail · Sep 23, 2014

weinie said:
So say I had the same 200AH house battery at a certain state of charge where it's voltage is 12.0 volts as measured on a voltmeter when disconnected it from the system entirely.

I then take the two disconnected leads from my solar panel, (which is now disconnected from the house battery) and measure the potential on my voltmeter and observe 13.4 volts.

I then connect the solar panel to the house battery and place my voltmeter at the terminals. What would my voltmeter "see"?

As the battery begins to charge, I would assume the internal resistance creeps up also causing the voltage to rise. Amirite?

#1 If your solar panel only reads 13.4V it is likely bad. Open circuit voltage should be in the 17-19V range depending upon cell temperature.

#2 It should read 12.0 volts for a period of time, same as the resting battery, and will SLOWLY creep upwards but may take days to hit 13V with a sub 1A current source....

#3 The amount of available current determines where in the SOC curve you will hit voltage points.

For example the 110Ah AGM I have on my test bench right now can take .4C or 44A of charge current and attains absorption voltage / 14.4V at approx 70% SOC when doing so.

If I fed that same battery .09C / 1A it would not attain 14.4V until approx 99.5% SOC. You are feeding a 200Ah bank approx .5% of its Ah capacity in current.

#4 If a battery comes up to voltage quickly and easily from a low SOC, with a low current source, the battery is TOAST!!!!

This is actual data from a relatively healthy flooded group 31 battery at .2C:

125Ah Group 31 Deep Cycle battery fed a 20% of "C" charge rate, or 25A, after being discharged to 50% SOC. It took 1.7 Hours in bulk or constant current charging to hit absorption voltage. If your batteries come up to limiting voltage, with 20% of "C" faster, then there is a voltage or battery issue.

2 Min = 12.59V / 25A = BULK / Constant Current
10 Min = 13.0V / 25A
20 Min = 13.12V / 25A
30 Min = 13.22V / 25A
40 Min = 13.33V / 25A
50 Min = 13.45 / 25A
60 Min = 13.6V / 25A
70 Min = 13.76 / 25A
80 Min = 13.95 / 25A
90 Min = 14.16V / 25A
100 Min = 14.39 / 25A
102 Min = 14.4V / 24.5A = ABSORPTION VOLTAGE / Attained Voltage Limiting

Remember that was at .2C or 20% of Ah capacity in charge current, and you are at .05C or 0.5% of Ah capacity in charge current, or less.....

Stu Jackson · Sep 23, 2014

weinie, I think another way to state the message is: regardless of how high the voltage is on a very small charging source like your example, the voltage will be what the house bank voltage is until the SOC gets much, much closer to full.

Stu Jackson · Sep 23, 2014

With no charging source present, the higher voltage battery would drain into the lower one. Eventually the voltages would equalize to most likely somewhere between the two, if no the lowest. IIRC, this was covered in high school physics or chemistry, too long ago for me to remember.

Maine Sail · Sep 23, 2014

weinie said:
Gotcha.... now....hypothetically, let's say you had a closed circuit of just two identical deep charge batteries in parallel however, when you initially connect the batteries one had a charge of 12.5V and the other 11.5V. What would be the potential difference as measured initially and what would it be after the system came to equilibrium?

Initially you will see a brief inrush of current from one battery to the other but the voltages level out pretty quickly. As the voltage differential becomes almost nothing current moves between the batteries quite slowly, due to the lack of a differential, and they can take hours and hours to become the same "SOC". If you hit them with a charger the current will simply flow where it is needed despite both batteries technically being at the same voltage. When I top balance LiFePO4 batteries in parallel it can take days before the current stops flowing between the cells, but we are talking mA level current movement..

shemandr · Sep 23, 2014

Stu Jackson said:
With no charging source present, the higher voltage battery would drain into the lower one. Eventually the voltages would equalize to most likely somewhere between the two, if no the lowest. IIRC, this was covered in high school physics or chemistry, too long ago for me to remember.

High school physics? Not mine. Our high school physics was F=MA and you can't push a rope.