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I'm just trying to get my head around the above question. It doesn't seem to follow Ohm's law. Could someone please explain what I am missing in this?

Below is a chart I found of the changing resistance of a lead acid battery compared to state of charge, however, the charge acceptance is higher when it is discharged compared to when it is charged. How does this happen with a higher resistance that gradually gets lower?

I'm also assuming a constant charging voltage from an alternator.

enter image description here

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    \$\begingroup\$ Ohms law always works, no exceptions. As @DKNguyen said, a battery is not a resistor, it is something more complex. So there are other parameters along with Ohm's law going on there. (Just like a capacitor charges faster in the beginning and slower at the end, it does not look like it follows ohm's law, but it does, along with other parameters) \$\endgroup\$ Mar 26 at 8:34
  • \$\begingroup\$ OCV after charging is not an accurate indication of SoC. Rather a small preload or a long resting period 1 day \$\endgroup\$ Mar 26 at 13:58
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Not everything has to follow Ohm's law because not everything is a resistor. A battery is not a resistor. It is a battery. A typical beginner mistake when they first learn Ohm's Law is that they think everything follows Ohm's Law. I don't know why since it is clearly taught that Ohm's Law is for resistors, but I guess when the only tool you have is a hammer, everything looks like a nail.

Consider this: when a battery is discharged the internal battery voltage is lower, meaning there is a larger voltage difference between the battery voltage and the charging voltage. More voltage difference = more current. If that voltage difference is large enough the resulting increase in current can offset the decrease in current due to the higher resistance.

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  • \$\begingroup\$ I understand the larger potential difference means more current, so when a battery is in a poor state of health but there is still a large potential difference, how come in that situation it draws little current? \$\endgroup\$
    – TMax
    Mar 26 at 9:45
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    \$\begingroup\$ Do beginners really think that? I've never seen that at all, such simple behaviour. In my experience, most beginners think they've learned one really simple equation so all the complicated stuff's yet to come. If you're alluding to the non-apparence of impedance to a newcomer, that's not what you've written. \$\endgroup\$
    – TonyM
    Mar 26 at 12:15
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    \$\begingroup\$ @TMax Well batteries in a bad state tend to have a higher internal resistance than normal. You have two parameters balancing against each other. The chemistry inside a battery has a lot more degrees of freedom to play with than a simple lumped-element circuit model. \$\endgroup\$
    – DKNguyen
    Mar 26 at 20:58
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    \$\begingroup\$ @TonyM Yes, I thought that. \$\endgroup\$ Mar 27 at 1:50
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    \$\begingroup\$ @DKNguyen I was 6. I had just learnt about equilateral triangles, and insisted that all triangles were equilateral. Face-offs with arithmetic teachers, I wouldn't be told. \$\endgroup\$
    – Neil_UK
    Mar 27 at 22:14
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How does a lead acid battery accept more current when it is discharged than when it is charged if the resistance is higher when it is discharged?

Just consider the picture in your question and put some numbers on things: -

enter image description here

If the SoC implies the battery OCV is at 13 volts AND the charge voltage is 13 volts then no matter how you work-it, the current into the battery is going to be approximately zero.

If the SoC voltage implies the battery OCV is only 12 volts and the charger is putting out 13 volts then clearly there is going to be a charge current. In this scenario the battery has 13 milli ohms and there's a voltage difference of 1 volt hence, the charge current is going to be around 77 amps.

If the SoC voltage implies the battery OCV is 12.5 volts and the charger is putting out 13 volts then the battery has about 10.5 milli ohms and the implied current will be 0.5/0.0105 = 47.6 amps.

In other words, using your graph, for a given fixed charging voltage, if the SoC is low, you get more charge current.

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  • \$\begingroup\$ Thanks Andy, however I would to know if the battery has a poor state of health, how come it won't accept charge even with a high potential difference? \$\endgroup\$
    – TMax
    Mar 26 at 9:58
  • \$\begingroup\$ @TMax that sounds unusual - maybe the battery has become damaged due to its poor state of health? Or maybe your charger won't deliver current into a battery with a SoC below a certain level i.e. it has a "feature"? \$\endgroup\$
    – Andy aka
    Mar 26 at 10:01
  • \$\begingroup\$ Its quite common when the lead acid batteries have a poor state of health they stop accepting a high amount of current when there is large potential difference, I just wanted to know what factors are influencing that? is it because they can't absorb a charge? I'm not sure \$\endgroup\$
    – TMax
    Mar 26 at 10:06
  • \$\begingroup\$ @TMax there may be some extreme limit of the battery where the internal series resistance goes a lot higher than what the graph implies. If you have a data sheet for that particular battery it would help a lot. \$\endgroup\$
    – Andy aka
    Mar 26 at 10:08
  • \$\begingroup\$ I don't have anything for that situation unfortunately, usually in that situation the plates have permanent crystallisation which reduces the surface and that's possibly why \$\endgroup\$
    – TMax
    Mar 26 at 10:17
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A lead-acid battery like all batteries has memory. (Some more than others) It is due to a double layer capacitance effect and often called something else. When you examine SoC voltages there is a difference of about 1/2V between the OCV and the voltage after some load is applied, so the resting voltage after this load for a Full range SoC is from 12.6 to 11.5. Yet the charging voltage from alternators is 14.2V. In regions where batteries rest at high temperatures, lead plates tend to get sulphated which increases the ESR of the battery and requires either pulse charging over DC or 14.5V which is often seen on some vehicles to agitate the plates enough to reduce the lead sulphate buildup. This is a tradeoff between plate corrosion and acid boiling from overcharge and plate oxidation from undercharging or over temperature.

Thus (if you) examine charge current from a CV alternator when discharged to a low state, you may experience a lower than maximum charge current until the battery has enough heat or charge level to lower the ESR then the current will rise to maximum and start reducing as the charge builds up. This rise to peak might only occur on dead batteries after say after an interior light was left on all night and day.

The CA @ 0'C & CCA @ 0'F ratings for a battery only apply when new and fully charged. Typically battery manufacturers specify ratings at freezing temp for water where the maximum current it can supply for 30 s allowing a maximum voltage drop to 7.5V. This translates to either a 5.5 V drop from OCV or a 5V drop from preloaded (e.g. <1A for 1 minute) for a battery typ with 50Ah capacity at 100 % SOC.

Thus a 500 CCA battery dropping 5V has an ESR of 10m Ohms. And a 1000 CCA battery has an ESR of 5 mOhms. This is partially due to the plate temperature rising above freezing and reducing the ESR after 30 seconds of perhaps 500A* 5V = 2500 W of self-heating and potential acid boiling. yet a dead battery barely able to turn on the headlights might only be supplying 3A* 5V drop = 15W with an ESR of 1667 mohms (hypothetical example)

Sometimes in between these examples, you may have experienced in winter as I have in Winnipeg(Winterpeg) that on rare occasions an almost dead battery will “come back to life” just enough to start the vehicle. My belief is it is the self-healing that reduces the ESR to make the battery produce more current with less internal ESR voltage drop.

All these variables are pretty dynamic and affect the SOC, expected lifespan etc. and especially reduce MTBF is from deep discharging a normal lead-acid battery for too many hours. (Some cars in Arizona have a battery life of ~1 yr from high temps under the hood)

Desulphating by high pulse charges can restore 2/3 of so-called dead batteries according to an acquaintance in the Caribbean who designed an automated reclamation tester with 200A pulse capability, but this was for batteries that did not sit for more than a year. The longer it remains undercharged, the faster it insulates the plates the lower probability of restoring it) The same is true for most rechargeable batteries except ceramic plate and deep cycle batteries are designed to withstand better.

Anecdotal

Normally acid s.g., specific gravity is the best indicator of, SOC and apparently also ESR from sulphation. I used to be Mfg Eng Mgr at C-MAC in Winnipeg where SOLARTECH units were made for truck and motive power batteries that used <50ns pulses >10A > 25 kHz only when the battery voltage was > 13.2V I I.e. when the car was running, as it drew an average of only 5W to desulphate the batteries as prevention. Generally, series connections cells are perfectly balanced in batteries when new then, only 1 cell degrades 1st as it has the weakest capacitance and thus overcharges and undercharges the fastest causing a runaway condition of rapid cell death. (Same is true for LiPo and Li-Ion) in Days or months instead of years. I tested Doug Eryou’s Solartech product on a 50k$ motive power battery for airplane tractors at the airport with a DSO and s.g. Tester and after a week of testing a battery that was performing poorly with a full charge had all “like new” s.g. readings that rose to become well-balanced and high acidic levels of a normal battery while left on float charge with the SOLARTECH box over the battery. It works! The customer was happy with the results and bought many. 1 for each battery. ( but it also interferes with trucker driver’ AM radio signals (like old copper spark plug wires did) on long hauls between cities ;) it also worked on NiCad and other batteries which were of great interest to the US military and Truck/tractor dealers.

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  • \$\begingroup\$ There seems to be a site bug where quick edits are seen as new answers. (Host concurrency latency perhaps) \$\endgroup\$ Mar 26 at 13:21
  • \$\begingroup\$ Quick sanity check - isn't CCA measured at -18C, and CA at 0C? \$\endgroup\$
    – SusanW
    Mar 27 at 0:49
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    \$\begingroup\$ You could be right Susan -17.7778 'C aka 0'F 0'C is really cold for Canadians ;) \$\endgroup\$ Mar 27 at 9:11

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