Understanding a model for battery charge acceptance

[Rick]

Understanding a model for lead-acid battery charge acceptance and just what it means to you.

A fully charged battery at “standing voltage” is most easily modeled with an ideal dc voltage source in series with a resistance value equal to the internal resistance of the battery. The internal resistance consists mostly of the electrolyte resistance and the parasitic resistance of the plates and cell interconnects. The “standing voltage” is the voltage which the battery stabilizes to after disconnecting from a charging source and has allowed the “surface charge” (more on this later) to leak off. An ideal voltage source is one which has zero internal resistance and a constant voltage regardless of load. Obviously the internal resistance, Ri of our model, limits the real current to a short-circuit value equal to Vth divided by Ri.

The standing voltage for a 12 Volt battery may be most simply expressed by the following formula:

Vth = 6(0.85 + SG)

Where Vth is the ideal battery voltage for a given specific gravity of electrolyte and is also what is called the Thevinin equivalent voltage equal to open circuit voltage.

The coefficient, 6, of the expression is equal to the number of cells of the battery.

SG is the specific gravity of the electrolyte and generally varies between 1.250 and 1.32 depending upon the manufacturer’s choice for a given application specific design for the fully charged battery.

So, for a 12.7 Volt battery one ought to measure the electrolyte specific gravity to be 1.260 and one can surmise that the sealed AGM or Gel batteries having voltages of 13.0 Volts have an internal specific gravity of 1.317, not that you can actually measure it. What is interesting about this concept is that if, after overly gassing a sealed battery you notice that the standing voltage is 13.2V you might think, “Great, my battery has more voltage than yours”! BAD! What that means is that the specific gravity has increased due to irreversible loss of water from gassing out of the valves or vents and internal corrosion will be accelerated thereby degrading life.

Ri for a good quality is on the order of 2.4 milli-Ohms per 100 Amp-hours of battery capacity rating, therefore, a good 400 Amp-hour rated battery may exhibit an Ri of 0.6 milli-Ohms. Higher values are also normally encountered for thick plates.

Ri rises in a non-linear manner as the battery is discharged increasing to perhaps 5 times or more the original fully charged value, assuming the same temperature.

Initial charge acceptance is the current Ic = (Vcharge – Vth)/Ri

So, for example, if you have a charge source capable of delivering the initial charge acceptance current at a desired acceptance voltage of 14.4 V to a battery only slightly discharged having an Ri of 3 milli-Ohms you will measure close to (14.4 – 12.65)/ 0.003 = 583 A, a rather large number which is why you normally observe a bulk voltage due to the source limited in output current capability.

Now suppose that that same battery is deeply discharged to say 12.2 Volts and that the Ri has risen to 10 milli-Ohms, then you have: Ic = (14.4 – 12.2)/ 0.010 = 220 A which is not an unreasonable value for a high-output large frame cool alternator. Knowing what happens to Ri is obviously of value here.

How to capitalize on this knowledge: If you have a charge source capable of delivering an initial Ic without a bulk limit you can merely note the value of Ic and compare that with subsequent recharge values to determine battery degradation over time and multiple charge cycles. You NEED a good battery monitor to do this accurately so that you can compare Ic values for the SAME depth of discharge for comparative cycles.

You can use the TIME of charging in the bulk mode UNTIL the acceptance voltage is reached to make the quantitative comparison if your charge source will not deliver Ic immediately at the acceptance voltage.

Once you notice a significant difference in either Ic (with a large source) or TIME to acceptance (with a small source) you can CALCULATE Ri at noble point of observation. Re-arranging the equation for charge acceptance current and solving for Ri; Ri = (Vcharge – Vth)/ Ic

In the case of small charge sources you use the noble point JUST when the acceptance voltage is reached to note the current Ic. In the case of large sources you make the measurement immediately.

Note that Vth MUST be measured with no charge source, no load, and with the battery having no surface charge (see below).

What causes your Ri to degrade to higher values?
1. Normal time degradation of the battery (cure: live with it or replace battery)
2. Sulphation caused by spending and average time in a state of discharge due to normal periodic cycling (cure: spend sufficient time to fully recharge battery)
3. Failure to recover battery capacity by applying a sufficiently high acceptance voltage at a sufficiently long time and/or a sufficiently long float time (cure: increase acceptance voltage and/or acceptance charge time....increasing float time alone is probably not a cure to the problem)
4. Variation in the cell-to-cell potentials which is directly related to specific gravity (cure: equalize the battery cell-to-cell voltages by applying a proper equalization cycle)
5. Increase of parasitic battery cable and connection resistances (cure: clean and tighten all connections)

What about the supposed, “surface charge”? Bare with me, this is where it gets REALLY interesting. Our initial battery model does not accurately depict what you might observe as the Vth immediately after charging. You will notice that after charging that the terminal voltage fairly slowly decreases over time until it stabilizes to a constant value (no load, no charging). The discharge curve plotted would form a decreasing curve of Epsilon, a specific exponential found almost everywhere in nature. The energy “lost” during this standing discharge to a steady standing terminal voltage is often referred to as the surface charge energy. It is not equivalent to many Amp-hours as to compared to the potential number rated by the battery manufacturer. The phenomenon is electrochemically complex and beyond the scope of this article yet there is a relatively simple addition to our model which will depict the phenomenon observed.

The model of the recently charged battery:

Using our original simple model place imagine a parallel combination of a capacitor and a resistor. Place this combination in series with our model. A charge source now “sees” the Vth through the capacitor before “seeing” Ri and Vth. Now any charge accumulated on the capacitor will ADD to Vth. If, in fact, the charge on the capacitor makes a voltage equal to the difference between the applied charge voltage and Vth then NO current will flow into the battery except for that “leaking” around the capacitor through its parallel leakage resistance of our model.

How to use this information: The fact that you may or may not understand just what a capacitor is or what is its value in micro-Farads is not important. What is important is that a charge can accumulate sufficient to prevent much curent flow into a fully charged battery. IN ADDITION, the resulting leakage current can be used to infer the fact of a fully recovered battery capacity. What first is required is an ability to properly apply a 3-step or Amp-hour Law charging regimen AND the ability to set an abnormally high acceptance or charge voltage.

Here is how you make the test: Fully charge the battery. Allow the battery to stand at float potential until the battery temperature effectively stabilizes to ambient temperatures preferably below 80 deg. F. Remove the float voltage and apply something near 15V and measure the resulting current with your battery monitor. A good quality battery with its capacity recovered to its potential (initial rated value minus normal time degradation) capacity will charge accept only about 50 milli-Amps per 100 Amp-hour rating. Values above 200 milli-Amps indicate possible failure to recover capacity. With some AGM and Gel-cell batteries a voltage of 15.5 to 16 Volts can be used if charge acceptance currents are in the vicinity of 25 milli-Amps per 100 Amp-hour rating.

Because the charge acceptance current is expectedly low a relatively inexpensive variable power supply can be used in place of a charger to make this test. Many of them have ammeters built in with sufficient resolution. As long as they are short-circuit damage-proof or full-current limited there is no risk to applying this voltage to a poor quality battery which will attempt to charge accept much more current.

Note that the components added to make our surface-charge model valid effectively “disappear” when discharging the battery below the value of the surface charge (or if left standing sufficiently long so that the charge leaks off, usually in a few hours to as much as 24 hours).

One addition that should be made to our model making it modestly complete is a leakage resistance in parallel with Vth. This effectively models the phenomenon of lost battery capacity over time when standing for long periods without charge. The value of this resistance is highly variable with temperature, dirt and moisture.

Is this all too much?
Regards,
Rick Young, technical evangelist who still doesn’t understand how women work yet keeps working at it like a dog with a bone.

[BC Mike]

Great stuff Rick. I hope I live long enough to even grasp the basics. It is more complex than sailing. The battery in my 1990 Isuzu lasted 10 years and I don't think I even looked at it and it included one complete discharge 'cause the guys in the service station inadverdently disconnected the alternator. The battery in my new 003 Saturn lasted one day. The two batteries in the boat settle at different volts, about 12.4 and 12.7. Once I make up my mind about a bigger boat I will get the fancy regulator and monitor, for the moment I will charge them the prehistoric way.
Michael

[delmarrey]

Do you plan on doing the Seattle Boat Show?

[Rick]

I plan to be there probably Monday or Tuesday...don't know yet.
Send me e-mail if you want to meet up or whatever. There might just be a beer break.

Rick

[mvas]

I am having trouble following your logic.

You wrote ...

With a Battery Charger set to 14.4 Volts ...
a) Battery Voltage = 12.65 V allows 583 Amps Charging
b) Battery Voltage = 12.20 V allows 220 Amps Charging

You claim, the Amps will increase from 220 amps to 583 Amps
as the battery charges from 12.20 Volts to 12.65 Volts.

This is not what I see when I charge a Lead Acid Battery.

When I charge a battery with a constant 14.4 Volts,
I see the Amps decrease as the battery charges.

Please explain.

[senormechanico]

Wow. Old thread !

Hi to Rick if you're still around !

Steve from Whidbey

[mvas]

I found this thread using Google search ... "Lead Acid Battery Charge Acceptance"

Rick's states,

So, for example, if you have a charge source capable of delivering the initial charge acceptance current at a desired acceptance voltage of 14.4 V to a battery only slightly discharged having an Ri of 3 milli-Ohms you will measure close to (14.4 – 12.65)/ 0.003 = 583 Amps, a rather large number which is why you normally observe a bulk voltage due to the source limited in output current capability.

But when I connect a 14.4 Volt Charger to a 100% SOC, 12.65 Volt, Fully Charged Battery, I measure only 8 Amps.

To me, Rick's example describes the conditions near the end of Absorb Mode while recharging a Lead Acid Battery...
1) An almost fully charged battery at 12.65 Volts
2) A constant Voltage source at 14.4 volts

I did some more thinking about Rick's statement regarding 583 Amps,
and this is my interpretation of Rick's statement ...

IF we had a 14.4 Volt charger AND
IF it was capable of providing 600 Amps AND
IF we had cables big enough to carry 600 Amps AND
just when we initially connect the wires
then for a very, very, very brief time 583 Amps would flow.

Now back to the real world ...
The brief 583 Amps does not really matter to us because
the charging amps is almost immediately 8 Amps, longer term.

A Charge Acceptance of 8 Amps during Absorb Mode is much more realistic than 583 Amps.

[senormechanico]

You have not posted the internal resistance, nor the amp hour rating of your battery.
Both variables in the equation.

You are correct that you would need a supply capable of delivering such current AND theoretically zero ohms wiring to the battery.

[mvas]

I have a fully charged 12.7 Volt 100AH battery.

And Per Rick ...
The Internal Resistance would be approx 3 milliohms,
and then ( 14.4 – 12.65 ) / 0.003 = 583 Amps

Clearly Rick is discussing only the possibility of a huge, but very brief, initial instantaneous amperage.
If this does occur, it is for such a very brief moment that it does not matter to us.

The Acceptance Rate that concerns us is the +/- 8 amps the we can measure on our ammeter.

How many amps do your batteries draw near the end of Absorb Mode?
At that point the Acceptance Rate is typically, 1% to 3% of the C20 Hour rating.
Certainly not ... 583 Amps.