I wanted to choose a load that would discharge the battery in a reasonable period of time, yet give me enough data points for my test. Looking at the curves above, It can be seen that a load of about 7.2A should discharge the SLA in about one hour. My definition of “discharged” was where the voltage started to drop off suddenly and headed south of 11V.
A 1 ohm power resistor would draw 12 amps (I=E/R) but it would also have to dissipate 144 watts. I didn’t have one, and I’m sure it would get extremely hot anyway. Then I got the bright idea to use 12 volt automotive lamps for my load. They would dissipate the heat well, are rated between 35 and 60 watts or more, and could be combined to add up to most any load I wanted. One other nice feature of using a filament as a load (at least in theory), is as the voltage drops, the resistance of the filament decreases and keeps the current drain more or less constant.
Since they were readily available, I chose my car’s fog lights for the load. Two 55W fog lamps in parallel = 110 watts. I (amps) = P (watts) / E (volts) so 110 watts / 12 V = 9.2A. I charged one of my new SLAs and took it off the charger for an hour to allow it to come to a “resting” voltage. Then, using an inexpensive digital voltmeter I recorded the resting voltage, hooked the battery to my fog lamps and started my stopwatch. I took one reading at 1 minute, then 5 minutes, and at five minute intervals after that.
I repeated the tests for the other battery as well as another new, but known good 12Ah SLA. My suspicions were confirmed when the suspect pair failed after 2530 minutes, compared with 55 minutes for the good one. I entered the data in Microsoft Excel and generated a graph with all three curves on it.
