0
\$\begingroup\$

I have the following:

Lead acid battery

Specs: 12V, 7.2 Ah

Model: Portalac PXL12072

Description: Kind of old, so maybe sulfated, but I won't know until I try to charge it.

Lead acid battery charger

Output specs: 10 Amps max., 14.4 VDC, 144 W max.

Model: Denryo BP-1210A

Description: New. "3-stage high tech battery charger"

The charger manual also contains a table that lists other chargers Denryo makes and their recommended batteries. My charger seems to be for batteries between 12V/30Ah (minimum capacity) ~ 12V/100Ah (maximum capacity). However, at 7.2 Ah, its capacity is below the recommended minimum capacity.

My question

Is there a safety risk to use a 7.2 Ah battery with a charger of minimum recommended capacity 30 Ah?

Similar questions

Charging lead-acid batteries?

This question states,

The charge current for small lead-acid batteries should be set between 10% and 30% of the rated capacity (30% of a 2Ah battery would be 600mA).

In my case, 30% of 7.2 Ah is 2.16 A. My charger outputs 10 A max., i.e. 0~10 A. But I don't know how smart the charger is to adjust its output.

What can happen if I try to charge lead acid battery outside charger's capacity range?

This is like my question, but it is unanswered.

\$\endgroup\$

2 Answers 2

3
\$\begingroup\$

I just stumbled in your question, I hope my comments are still useful for you (or other fellows here).

Starting from your question: Is it safe to use a 10 A charger in a 12 V 7.2 Ah Lead acid battery?

Short answer: Not as a stand alone charger. It is dangerous. You most probably will “fry” the battery, not by overvoltage, but by overcurrent. This is more prone to happen if your battery is discharged lower than 12.0~12.5 V .

Not only you will damage the battery, it is also unsafe: The incompatible charging current will be too high to be chemically converted in the electrode plates and it will generate gas, raising the internal pressure, leaking acid or worse.

Quick and dirty solution: Select a “power resistor” or even better, a brake-light lamp (12 V, 21 W), which at 12 V sinks 1.7 A (= 21W/12V) and presents a resistance of 6.9 Ω. But as the filament of the lamp is incandescent, it is non-Ohmic and will present a much lower resistance when cold (typically R.cold ~ 20~30% R.hot).

So, if your small battery is flat (close to 0 volt), the lamp limits the current to 1.7 A, which is 24 % of 7.2 Ah rating = a safe charging rate. By the way: anything higher than 20~30% might fry (or boil) most lead acid batteries, specially when voltages in the battery are close to 12~13 V .

In case the battery is partially recharged at 12 V and the charger is “smart”, it often limits its final voltage to 14.4 V. For this case, the lamp will see just 2.4 V of Delta-V. As the filament is cooler, and it might be a fraction of the hot value. As a general figure, the charging current passing through the lamp will be around:

I = ΔV/R.cold = (14.4 - 12) / (20 % of 6.9) = 1.7 A (less likely),

or

I = ΔV/R.cold = (14.4 - 12) / (30 % of 6.9) = 1.2 A (more likely).

and when fully charged at 13.5 V:

I.full = ΔV/R.cold = (14.4 - 13.5) / (20 % of 6.9) = 0.65 A (~ 9% of 7.2 Ah)

So, I’m proposing the use of the non-linear properties of a lamp to limit and to stabilize the current for your charger. The current will be maxed at 1.7 A (24 % of 7.2) for a flat battery, but will stay within 0.65~1.2 A during most of actual charging process.

Obviously this method can be adapted to other Lead-Acid Battery and Charger capacities. But unless allowed by the manufacturer, avoid charging at more than the suggested current limits. And be sure the charging voltage in the battery is always limited to 14.4 V nominal (can vary from 14.2 V @ summer to 14.6 V @ winter). And after battery is fully charged, your charger should reduce the voltage to floating values: around 13.5 V. If it does not reduce the final voltage, do not leave the charger connected longer than 24 hours for a full charge process.

I tried to summarize key figures and to propose a practical solution, but a more sophisticated circuit is always feasible (although more complex)

Final tip: Check the website “Battery University” - it provides much more battery-related information.

Supplement 17-June-2021 After a comment, I decided to double check. I got a 12V 5W car lamp and tested in somehow relaxed way. My idea was to share the data using a variable power supply and a single multimeter (DMM). Obviously if I had used 2 DMMs, and better wiring layout the results could be better. I spent more time plotting the graph than in the experiment, but it has been fun. Here are the tabulated results: enter image description here

The figure contains the tabulated data, the graph curves and the 2nd order polinomial-interpolated equation for Filament Resistance as a function of the Voltage; it shows to be generally close to the experimental data.

Last Column (“Red”) lists Resistance ratio using R.Hot as Fixed Numerator. 2nd last, column (“Indigo blue”) is for R.Cold

So in summary, we could say for this 12V 5W Car lamp:

  1. OFF resistance is about 1/10 of the Rated Resistance at 14V (R.Hot). Bruce Abbott’s comment was confirmed for the Off condition, but he did not provide more data for further comparisons.

  2. COLD resistance is about 1/7 to 1/4 of the R.Hot. Assume here as an average of 1/5 or 20%. “Cold” Filament appears “Off”, is not incandescent, but it is circulating some current.

  3. Deep Red (start of incandescence) happened at 2.5V and R = 1/3 of R.Hot.

  4. Bright Yellow is considered from 11V to 14V. For this range, Actual Resistance varied just 10% = almost constant.

  5. Major Non-Ohmic behavior happened between 5% and 50% of rated voltage.

Updating Original Equations using the R.Hot references (assumed 21W behaves as the 5W version = not really true, but that is what we have for the moment). Resistance R.cold ratios were linearly interpolated within closest tested range:

If the battery is close to “dead” (as about 6.0V):

I = ΔV/R.warm = (14.4 - 6.0) / (6.9 / 1.37) = 1.67 A (23% of rated 7.2Ah).

I = ΔV/R.cold = (14.4 - 12.0) / (6.9 / 2.7) = 0.94 A (starting to charge at 12.0V).

and when fully charged at 13.5 V:

I.full = ΔV/R.cold = (14.4 - 13.5) / (6.9 / 6.2) = 0.81 A (~ 11% of 7.2 Ah)

All the above results will most probably vary by the Wattage of the filament being used. That is why I have used the ratio of the Resistance at the rated voltage, which is expected to vary less, being a dimensionless approach. even doing so, I would expect different numerical values of reference for this 5W, when compared to a more powerful 21W of even a 50W dichroic lamp. If someone experiment with other lamps, please share or comment, ok?

\$\endgroup\$
2
  • 1
    \$\begingroup\$ +1 on the lamp as current limiter. A 12V 15W bulb I tested had only 10% of the 'hot' resistance when cold, and dropped only 0.5V at 0.3A electronics.stackexchange.com/questions/164587/… \$\endgroup\$ Jun 9, 2021 at 4:48
  • \$\begingroup\$ @Bruce, thanks for the comment. It motivated me to plot a check, which results (table and graph) I supplemented in may answer. What I found is that the OFF resistance is indeed about 10% of the RatedHot one. But with as low as 1V, the R.cold is 2x the R.off value, so about 20%. \$\endgroup\$
    – EJE
    Jun 18, 2021 at 4:21
0
\$\begingroup\$

Connect it and use a multimeter to see the voltage and then also check the current - if it is below 5A and behaves nicely then I would assume it is doing fine.

If it hammers the battery at 10A and does not do any charge control then you need another charger.

If you don’t have a meter to check what is going on, either get one or stop playing.

\$\endgroup\$
1
  • \$\begingroup\$ Ok. I will start the charge and check voltage and current. I didn't know if it was ok to connect the two for even a minute. But if a short time is safe, I will go ahead and connect them and measure what happens. \$\endgroup\$ Jun 7, 2020 at 12:29

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.