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I have a 12 V, 55 Ah lead acid deep cycle battery that I am using to power a small cooler/fridge that will hold medicine for a community project that I'm working on. The goal is to provide continuous power to the fridge despite any power outages caused by natural disasters, and there will eventually be a solar system connected to lessen the load on the battery and increase the uptime. The full system will also have a switching module that will automatically detect when the grid (wall outlet) stops working and will switch to this battery (it will also do this in the other direction), like a UPS but much cheaper due to budget constraints.

Now, I am going to do an experiment where I run the 12 V, ~48 Wh cooler directly from the battery and log the voltage vs. time across the terminals so that I can tell how long the battery can power the cooler before it turns off and compare that to the expected result.

I have found plenty of tables from google that showing the depth of discharge percentage vs. battery voltage, and it says it shouldn't drop below 12 V until the battery is about 60% depleted. This is desirable because I do NOT want to completely deplete the battery and reduce its lifespan.

Here is what I am unsure of, the fridge operating manual does not provide any information about the lowest operating voltage. Can I assume that the lowest operating voltage is close to 12 V, and that the fridge will shut off once the battery reaches, say, 11.5 V? If yes, then I am good to go. If not, then what is an average operating voltage range for a system that is designed for 12 V?

(I know that it is likely very dependent on the system, but I am looking for more of an anecdotal answer. If an answer is not obvious, then I can attempt to use a variable power supply in the lab to find out the voltage at which the fridge stops drawing current, though I am asking here because I'm in a bit of a time crunch to assemble the first prototype and would like to be able to run this experiment as soon as possible)

Edit: for reference, here is the only information I could find about the cooler https://secure.img1-fg.wfcdn.com/docresources/0/156/1564815.pdf

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  • \$\begingroup\$ No, you will have to arrange to cut off the load to prevent deep depletion. And if you DO deplete the battery, it's important to recharge ASAP to prevent sulphation and long term damage. \$\endgroup\$
    – user16324
    Mar 7, 2021 at 21:22
  • \$\begingroup\$ I see. So you're suspecting that the minimum operating voltage of the fridge is below 11.5 V? \$\endgroup\$
    – amateurhour
    Mar 8, 2021 at 1:25

2 Answers 2

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Yes deep-discharge batteries can handle many more cycles than 50 for flooded cell lead-acid batteries ( some barely survive a few )

But they still oxidize the plates with a thin coating from the calcium-lead plates over time from under-voltage.

This can be reversed with very fast rise-time pulse charging, but otherwise, they will become high ESR and unable to prove the amps you need.

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  • \$\begingroup\$ Where is calcium from in lead acid battery? It is lead and sulphuric acid. \$\endgroup\$
    – user263983
    Mar 7, 2021 at 23:21
  • \$\begingroup\$ But is the deep cycle battery likely to chemically limit itself so that it does not reach <30% capacity? As in, if I left this thing connected to the fridge to do some measurements overnight, would I wake up and find that the battery voltage is exactly 12 V as opposed to the ~14, or will it have dropped to 10.5? My thought process is since the fridge needs a voltage drop to draw current from the battery, and the battery voltage is going to drop in relation to the depth of discharge, once the battery voltage is below 12 V then the fridge should just stop operating, correct? \$\endgroup\$
    – amateurhour
    Mar 8, 2021 at 1:05
  • \$\begingroup\$ No the battery charge can be depleted fully but that may not work well with a compressor. The battery ESR rises sharply at low charge which causes more voltage drop when adequately charged. The load must have it's own proper UV protection and hysteresis for restarting to prevent cycling. So it could have dropped well below 12V trying to start weak, then cut out and then slowly restore voltage but low capacity ... or not \$\endgroup\$
    – Tony Stewart EE75
    Mar 8, 2021 at 1:31
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I see:

Not unplugging your unit when car engine is off can drain the car’s battery.

So there is probably no internal logic to cut off. As no one see what is inside, what you can is to measure the bahavior of fridge at lower voltages and try to estimate/measure the efficiency as the voltage goes down with some power-meter and temperature sensor inside. Don’t forgett to put something into fridge (beers are good) and ensure constant ambient temperature.

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  • \$\begingroup\$ Yeah unfortunately I don't suspect there's any internal logic either. The thing is though, the car wouldn't start if the battery was below 12 V, but since this is an external battery I should be able to take it closer to 11 V. So if the lowest operating voltage of the fridge was close to 11.75 V, then the battery would stop providing current at a safe capacity just because of its chemical nature and I'd be fine without adding another circuit. If the fridge operating voltage is under 11 V I would have to set something up to automatically stop the battery at 11.75 V \$\endgroup\$
    – amateurhour
    Mar 8, 2021 at 1:20

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