Battery testing
Intro
Nothing lasts forever. That includes UPS batteries, which seem to have a useful life of about 3-5 years. UPSs should be set up to self-test like every week or so. During the tests, the UPS will go on battery until some threshold is met, then return to line power. If it finds the batteries are too worn, it will signal for replacement.
Once a year, or on full charge after installation, battery calibration should also be done. The UPS will then be able to provide a somewhat reliable figure on runtime, based on current load.
If maintenance is not done in time, batteries will be charged at higher currents than is appropriate for their condition. They will start gassing hydrogen internally, and swell up. Eventually the case will fail, and a spark could ignite the hydrogen (insert picture of firemen in my machine room here). Annoyingly, the battery tray will probably be very hard to remove from the unit because of the swelling. Add possibly very high temperatures to that annoyance.
Generally, the batteries are an array of 12 V lead acid batteries, though there are some 6 V battery systems around. I'm just going to talk about 12 V batteries here, but the tester below can be used on those batteries too. The current through the resistor will be halved, but you will still get an internal resistance value.
Voltages for 12 V lead acid batteries are said to range from 11.85 V at discharge, to 12.85 V fully charged. You may see higher measured values, but that overhead should disappear pretty fast after you connect a load. But you don't want a wild swing upward when the load is disconnected. Consumers are going to not work if the battery voltage falls too low. Long story short: When connecting a sensible load to the battery, you don't want a brown-out.
Besides the obvious unloaded voltage measurement, which can give you an accurate idea of the charge level of the battery, the remainder of this writeup will deal with internal resistance, which will govern the usable amount of current that the load can draw. Battery manufactures will publish specs on their web sites; Power Sonic for example states values of 10-40 milliohms for internal resistance.
Test hardware
Battery testing hardware:
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(Battery - clamp) ---/\/\/\/\-------* *----|OOO|--- (Battery + clamp)
100W 12O/ 1.5A fastThe parts were available at Radio Shaft and Home Depot, except for the power resistor, which was ordered. The aim was to pull a current of about 1 A, which the 12 O/ resistor will get out of a 12 V battery. The 1.5 A fuse is for your protection if you mess up in connecting the system.
Test example
Tested battery: old, but unused 12V 7.2 Ah battery
Model: Portelac PX12072-HG with 2007-05-15 sticker
Unloaded potential across terminals as measured with multimeter: E_0 = 13.65 V
Resistance of the testing hardware: R = 12.65 O/
Set up the test: Make sure the switch is open (no continuity)! Attach the positive connector of the tester to the positive terminal of the battery, then do the same for the negative side. Attach multimeter leads to the terminals along with the tester. Set the multimeter to DC volts, 20 V range.
On closing the switch, battery voltage decayed exponentially, after about 2 minutes to 12.65 V. The resistor also starts getting hot. The switch was opened, and after a while, battery voltage was 12.85 V, but continued to climb by exponential decay.
Potential under load, measured across terminals when I felt that it was a good time to measure: E_L = 12.65 V
Unloaded potential, measured across terminals after opening the switch, when I felt that it was a good time to measure: E_N = 12.85 V
What is our internal resistance:
R_int = ((E_N / E_L) - 1) * R_L
= (12.85 V / 12.65 V - 1) * 12.65 O/
R_int = 0.200 O/
I don't like measuring things by feel, but I wasn't going to wait for an age waiting for the voltages to stabilize either. There's also chemistry involved here, meaning that some of the recovery in voltage is due to redeposition of ions. I think I favor close measurements, taken before and after the load is disconnected. To get a somewhat stable value, I choose to leave the power on for 10 seconds.
Turn on switch for 10 seconds, read voltage, turn off switch, read voltage Using the above formula:
E_L = 12.85 V
E_N = 12.96 V
R_int = 0.108 O/
How much current can we draw, and still have good voltage?
E_L = E_N - (I * R_int)
At 9 A, this battery won't be able to supply 12 V.
The projected load is not going to be even close to 9 A, so this battery is showing its age, but is still fine for the application.
References
EE Times Batteries in a portable world Power Sonic has a lot of specs for their batteries More to follow.