Measure the Internal Resistance of a Battery?

Discussion in 'Power Electronics' started by TBennettcc, Nov 7, 2011.

1. TBennettcc

292
2
Dec 4, 2010
I'm doing some troubleshooting on 12-volt sealed lead-acid (SLA) batteries (out of APC battery backup units).

I've got a battery that gives me 13.2 volts.

It is an Exide PowerSafe EP1229W. On the side, it reads:

This battery is rejected by the battery backup as no good. I'm trying to figure out why. I figured I would start with trying to determine the battery's internal resistance. (If you think there's a better way to test these batteries, please let me know!)

I have one DMM and four 10-watt resistors (two 100-ohm and two 50-ohm, all 10% tolerance).

First, I test the voltage of the battery with no load. Open-circuit voltage is 13.2 volts.

I only have one DMM at the moment, so I am making one measurement at a time. When I hook up one 100-ohm resistor across the battery, the terminal voltage drops slowly. It seems to hold steady at 12.79V Current appears to be 140 mA at that point. The resistor I am using has a measured value of 99.3 ohms.

So, wouldn't it be a simple case to apply Ohm's Law? V= I*(R+Rbat)?

12.79 = .14 * (99.3 + X)

X = -7.94 ohms?

Am I doing something wrong?

Thank you for your time.

69
20
Dec 2, 2010
The battery can be thought of as an ideal voltage source with the internal resistance in series. An open-circuit voltage measurement would have negligible current flowing through this internal resistance, and thus negligible voltage drop, so your battery's ideal voltage source is 13.2V. Under load, you get 12.79V across 99.3 ohms, giving you 129mA. This means the internal resistance dropped the voltage from 13.2V to 12.79 when it had 129mA flowing through. Thus, the value of the internal resistance is (13.2 - 12.79) / .129 = 3.18 ohms.

3. TBennettcc

292
2
Dec 4, 2010
Thanks so much. I KNOW I know Ohm's law, but I must've been half asleep yesterday. The way I did it works IF I actually put in the right numbers. Not sure why I measured 140mA instead of 129mA. I probably should have calculated the current instead of trying to measure it. Looks good. Thanks again. 69
20
Dec 2, 2010  