So I thought that lead-acid batteries are preferred, because you would only need to apply the maximum specified voltage (mostly 14.4 V for pure lead-acid and 14.7 V for the AGM/EFB, correct?) and everything would settle when the power supply voltage equals the battery voltage.

But as it seems, this would result in nearly unlimited current when \$V_{bat}<< V_{psu}\$. Okay, but why can you just jump-start the car batteries in winter by directly connecting two batteries without any current resistors?

Theoretically. the batteries would be damaged because of this or would catch fire when being connected for a long time, wouldn't they?

Besides this, I understood that you firstly apply CC until roughly 80% of charge and then change to max charging voltage after that, until this is reached and V_float will be applied. Isn't this basically how you also charge lithium-ion batteries? Until the max. cell voltage CC is applied and then you wait for reaching the max. cell voltage while in CV mode.

So what are the real differences in charging lead-acid vs. lithium-ion when it's basically the same procedure with CC charging until max. charging/cell voltage and then CV to reach/equal battery voltage with charger voltage in depth? Why is lithium charging way faster than lead-acid?


1 Answer 1


That's a really good question that can be answered in many ways or levels of complexity.

Let me try to explain at the equivalent circuit level, but first we need to define an approximate equivalent circuit. There will be significant resistance in jumper cables perhaps even more than the ESR of the good battery, and this is often the challenge to start a car with a dead battery in -30'C weather when the engine load is high from cold viscosity demanding the full CCA rating of xxx Amps at 7.5V. But when the engine starts, the battery is immediately charged by the alternator at 14.2V. ( I know all the tricks growing up in "WinterPeg")

How can the battery voltage jump so fast?

  • Consider an alternator rating of <= 100A, yet the starter demand say up to 800A from discharging from a vehicle running with an alternator. What this means is the ESR for a charged battery and discharged battery are quite different and is very high when dead. The battery effect series resistance is ESR and its Ah capacity from 12.5 to 11.5 can be converted to Capacitance as xx KiloFarads. Thus what we have is two pairs of series RC in parallel for a simple model.

When we short a battery with 0 Ohms for an instant the battery voltage is 0. How can it jump back so much then slowly rise a bit more to almost where it was before?

  • Consider again the twin RC model. This means a larger C restores the smaller C value, but the larger C also has a larger ESR so it does not deplete instantly while the lower C has lower ESR and draws massive currents. This simulates the behaviours fairly well.


simulate this circuit – Schematic created using CircuitLab

Lithium Ion chemistry has a similar model for oxidation when undercharged but for charging, the profiles are different due to temperature rise also increasing oxidation and failures in test (FIT) rates from the Arrhenius Effects. Yet above 100% SoC in Lithium cells, erosion of the electrode-electrolyte interface increases but the last 10 to 20% of the Capacitance must soak in with the CV limit and limit the time spent between 3.9 and 4.2V per cell to extend lifetime.

Therefore the ideal capacity charge profile for a Lithium cell is different from the ideal lifetime charge profile less commonly known. (Search Battery University for this data)

Ideal Capacity uses CV=4.2V+ (x mV) and CC cutoff = 2.5% of CC but may only get >=250 recharge cycles
Ideal Lifetime cycle CV=4.0V approx and CC cutoff = 10% of CC and only depleted to 50%. Or can be 30% to 80% SoC limits for 50% use. But then you may get > 3000 charge cycles.

There are research papers on the chemistry reasons and some design variations are better than others with Fe added. But it is clear that boiling the batteries and CV extending the limit time is bad for Lithium batteries.

Even worse is using the batteries while charging at CV then the charger never cuts off and the batteries erode usable life > xx faster. The result is high ESR in at the weakest cell in series and degraded C values.


  • With hand waving arguments, in effect I have illustrated how a double-electric layer effect is modeled to match your experience with batteries.
  • if you wish to pursue the chemical research papers now you can be aware there must be 2 dominant zones and effects charge capacity and for erosion of capacity for CV effects on residual charge and erosion.

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