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I am using an AGM deep cycle battery 130Ah 12V connected to 200W solar panels and a 500W inverter to power electrical devices such as fans, laptops and lamps. The inverter has an automatic cutoff feature at a nominated voltage. I understand that deep cycle batteries should not be discharged below 12V (approximately 50% SOC). I have noticed that when the inverter is in a state of providing charge that the measured battery voltage is considerably lower than that measured if I switch the inverter off. Sometimes the measured battery voltage can jump as high as 0.5 to 1V when the inverter is powered off while it is powering devices. On this basis, I have set the inverter's auto cut off setting to 11.5V. Sometimes the voltage drops to 11.5V or less and I lose power even though the battery's actual voltage is much higher.

If the battery should not be discharged to less than 12V, does this refer to the battery's resting voltage or the lower voltage that I am seeing while it is powering devices?

What is a safe level to set for the inverter's auto cutoff voltage so that I don't destroy the battery and also to prevent devices from cutting out while the battery still has a sufficient SOC?

UPDATE:

I don't have the battery's datasheet handy. This information is available on the battery itself:

Valve Regulated AGM Rechargeable

BATT130LED

Charging Details:

Float 13.6-13.8V

Equalisation 14.4-14.7V

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    \$\begingroup\$ Resting voltage. In general, the higher the current, the lower you can set the cutoff. \$\endgroup\$ – mkeith Dec 27 '17 at 7:38
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    \$\begingroup\$ Where is datasheet? What do you prefer more short-term usable life or long-term life? \$\endgroup\$ – Sunnyskyguy EE75 Dec 27 '17 at 8:15
  • \$\begingroup\$ @TonyStewart.EEsince'75 Datasheet for the inverter? I want the right balance between battery longevity and being able to use it for the purposes for which I purchased the solution. \$\endgroup\$ – Guru Josh Dec 27 '17 at 8:22
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    \$\begingroup\$ He means the battery datasheet. I would guess that if the battery is rebounding to over 12V, you are doing right by the battery. If you are OK with the battery life, then leave it as is. If you just need a tiny bit more life, it is probably easiest to just lower the voltage slightly. If you are far short of the battery life you need, then you probably need to add another battery. \$\endgroup\$ – mkeith Dec 27 '17 at 8:27
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    \$\begingroup\$ Battery damage is not a simple yes/no thing. Discharging the battery to 11.3V instead of 11.5V could e.g. half the life of the battery - but you won't see this for at least weeks. Plus, a completely stuffed battery can typically still be 'fully charged' (14V+) - it just has no useful capacity. \$\endgroup\$ – Someone Somewhere Dec 27 '17 at 9:45
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Battery aging is inevitable. It occurs during charge transfers and life cycles reduce 50% for every 10'C rise above room temp and to some extent by the minimum % DOD or cutoff voltage you choose.

You must define a spec to choose desired %DoD and thus capacity and thus lower cutoff voltage and thus lower cumulative Ah capacity* charge cycles.

As always, refer to your AGM datasheet.

This datasheet from Victron shows the the number of Charge cycle always increases by using a lower %Depth of Discharge (DoD). This implies a higher "resting" cutoff voltage gives longer life. But how much? When I multiply the # of cycles from the barchart by the amount of charge used (DoD) the total life cycle charge goes up from 300 effective total charges to 435.

enter image description here

( measurements taken from bargraph by counting pixels with I_view)

Conclusion: Reducing your cutoff will extend inverter runtime but also reduce battery number of days of charge life, so you lose time later.

Aging is directly related to total charge transferred during life. For longest lifespan, keep at room temp! (20~23'C) and choose Mfg's advice wisely.

If you need more capacity, use more batteries rather than lower the cutoff, but if desperate, you can reduce cutoff to 10.8V for the best short term capacity.

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  • \$\begingroup\$ My main imperatives are portability and longest possible short-term use. I don't want to burn my battery out in less than 6 months but from the statistics that you have provided, I would be happy to get 37.5% less recharge cycles out of the battery if it meant that I could get an extra 30% DoD out of each cycle. This might be enough extra juice to save from having to add an extra battery and solar panels. If I decided I was prepared to run with an absolute maximum DoD of 80%, roughly what would this equate to in terms of the cutoff voltage that I should set? \$\endgroup\$ – Guru Josh Dec 27 '17 at 10:33
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    \$\begingroup\$ From what others have said, 12.0V correlates with %85~80%DOD and 10.6V to 100%DoD \$\endgroup\$ – Sunnyskyguy EE75 Dec 27 '17 at 17:39
  • \$\begingroup\$ The chart above seems to contradict the fact that deeper DoD results in lower life when you compare the 50% and 80% DoD figures which have the same 300 "full cycles" per your calculation. yes, it does show a modest difference versus the 30% case, but it's the deeper discharges that supposedly show the biggest difference so I'd expect 80% to show the largest gap and 50% and 30% to be relatively close. Can we be sure those graphs are in "partial charge cycles" (i.e., from 100% SoC to 70% then then back to 100% counts as 1) or are they in "full charge cycles" ... \$\endgroup\$ – BeeOnRope Apr 5 '18 at 4:07
  • \$\begingroup\$ ... (i.e., where 100 -> 70 -> 100 would count as 0.3)? The numbers make a more sense if they are full cycles, since here the various DoD levels all have the expected relationship with life. I though battery life was always generally reported in "full cycles", but I can't be sure. \$\endgroup\$ – BeeOnRope Apr 5 '18 at 4:09
  • \$\begingroup\$ yes 1 cycle at 30% DoD counts as ~ 30% mAh used but actual mAh was computed. then accumulated for total life of mAh delivered so if you had 3 batteries and rotated them using only 33% DoD, you gain ~45% total lifespan energy delivered. But 50% and 80% DoD made no difference \$\endgroup\$ – Sunnyskyguy EE75 Apr 5 '18 at 4:20
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Essentially, what you are trying to do is cutoff before the state of charge (SOC) gets too low. The problem is that the correspondence between battery voltage and SOC depends on several variables, including details of the battery chemistry, temperature, load, etc.

If I were in your situation, I would just lower the cutoff a bit to get a bit deeper discharge. I think most likely you are not damaging your battery excessively. But for the record, here are some more systematic ways to approach this.

What you want to do, is somehow figure out what is the SOC when you cutoff at 11.5V with your current load. One way to do this would be to isolate the battery after the inverter cuts off. Let it sit at room temperature for 24 hours and measure the open circuit voltage at the end of that period. Then look at a chart for your battery or a very similar battery to determine SOC from voltage.

Because it is an AGM battery, you do not have access to the electrolyte and cannot measure density of the electrolyte as you could with a flooded lead acid battery. But for a flooded battery, that would be an option to determine true SOC. They sell tools for this (Battery Hydrometer).

Another idea is to actually measure and integrate net discharge while you are using the battery, from the time it is full to the time the inverter cuts off. This will tell you how many amp-hours have been removed from the battery, and you can use the battery capacity to compute state of charge. For example, 5A for 20 hours = 100 Amp-hours. Since the capacity is 130 Amp-hours, the SOC would be (130Ah - 100Ah) / 130Ah = 23%. Note that capacity depends on discharge rate. The listed capacity of 130Ah is based on a 20 hour discharge time (6.5A for 20 hours). So to be accurate, you would want to use the capacity based on actual discharge rate, if that information is available. So that could complicate matters.

The other idea is to estimate capacity during recharge. I have done this in the past with batteries that were recharged with no load on them. If the charger does constant current of 20A for 5 hours, then transitions to constant voltage, I would estimate that it was about 120 Ah of restored capacity (5 hours @ 20A + 20% to allow for the CV top-off phase). This is probably the least accurate way to do this. But sometimes it is the easiest way, because you don't need to get at the battery terminals. You just watch LED's on the charger, and assume that the charger's CC rating is relatively accurate.

Hope this helps.

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    \$\begingroup\$ I have just tested the battery’s voltage some 9 hours after the inverter cutoff at 11.5V (and was switched off) and also without receiving any further charge and the voltage has rebounded to 12.2V. So this is some indication that 11.5V is a conservative LVC value to implement. With the charger method of determining SOC, are there chargers available that display the battery’s voltage on an LED screen as it is recharging? \$\endgroup\$ – Guru Josh Dec 27 '17 at 19:01
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    \$\begingroup\$ The charger I was talking about had simple LED lights to indicate what stage the charger was in. Typically they charge at constant current up to some high voltage (over 14V), then they charge at that high constant voltage for a limited time, then they fall back down to a lower, safe float voltage (a bit over 13V). The time spent in constant current is a rough indication of how much charge was added to the battery. \$\endgroup\$ – mkeith Dec 27 '17 at 19:32

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