Why can't I measure the internal resistance of a car battery? - Page 1
Forum section: electronics / beginners

author: j.a.mcguire
author_rank: Contributor
timestamp: May 29, 2015, 09:18:00 am
content: I had a flat battery on the car this week and am working away. The only thing I had was a 240V AC->12V DC 90W transformer so my plan was to bring the battery indoors and jury-rig the power-pack to the battery in the hope that the internal resistance would be enough to slow down the current draw, not blow the pack and charge the battery for an hour or so to get me going.I tried to hook up a multimeter across the termainls of the battery to check the internal resistance, but it wouldn't read? Reading online it seems that the correct method of checking this is to measure the voltage with no load, then measure the voltage with a constant load and use Ohms law to figure out the internal resistance of the battery.Why isn't the multimeter capable of measuring this? Is it as simple as, the multimeter only outputs a few volts and so the difference in potential prevents the meter from taking a reading? DMM was a Fluke 177.

author: j.a.mcguire
author_rank: Contributor
timestamp: May 29, 2015, 09:39:20 am
content: The differential voltage is used to calculate the internal impedence?So if my differential voltage is 1V it would be 1V/1A=1 Ohm?Also can anyone advise why the DMM can't measure this?

author: Andy Watson
author_rank: Super Contributor
timestamp: May 29, 2015, 09:54:14 am
content: Quote from: j.a.mcguire on May 29, 2015, 09:39:20 amHow do you think the DMM measures resistance ?

author: j.a.mcguire
author_rank: Contributor
timestamp: May 29, 2015, 10:14:01 am
content: Apparently it uses a constant current to measure. Which would be the reverse of the above method because it's generating its own power source?So because the battery is a power source it cannot be measured by the DMM?

author: Dr. Frank
author_rank: Super Contributor
timestamp: May 29, 2015, 10:16:40 am
content: Quote from: j.a.mcguire on May 29, 2015, 09:18:00 amThe Ohm mode can only measure "naked" resistors.Instead, a battery is a voltage source in series with its internal resistance!This 12..14.4V battery voltage will definitely disturb the Ohm measurement of every DMM.In the case of more simple instruments, such an abuse might also destroy the DMM.Fortunately, a Fluke DMM is mostly safe against such misuse.Therefore, as you describe yourself, the only way is to measure the voltage drop on the battery poles when connecting a known current load. The internal resistance then is the derivative dU/dI = R (differential resistance).You can do that easily in the car. Measure the pole voltage, and switch on the headlights, when the engine is NOT running.The light bulbs have usually 110W in total, that are about 9A, and then simply divide the voltage drop by these 9A, and you're done. A healthy battery should have a few milli-Ohm only. So the voltage drop should be < 100mV @ 9A.The other, more important aspect is the battery capacity, which can not be measured so easily.Btw.: To charge the battery, you have to apply 14.4V constant voltage.12V will not charge your battery at all!The charging current has to be limited, but the internal resistance is so low, that doesn't do the job!Therefore, you have to limit it otherwise, either by a power supply with current limit, or by an external power resistor.Example: Limit to 5A, with 14.4V / 80W PSU => (14.4 -12)/5 => external resistor should have about 0.47Ohm, 12WFrank

author: Andy Watson
author_rank: Super Contributor
timestamp: May 29, 2015, 10:28:22 am
content: Quote from: j.a.mcguire on May 29, 2015, 10:14:01 amYes, but, what it actually measures is the voltage. Resistance being terminal voltage divided by the constant current. When you measure the battery you are adding the battery voltage into the equation - the meter doesn't know about this extra voltage.

author: Seekonk
author_rank: Super Contributor
timestamp: May 29, 2015, 01:19:21 pm
content: Been a while since I looked this up, but the resistance measurement would require a voltage taken at nominal current to eliminate surface charge and then a higher current.

author: IanB
author_rank: Super Contributor
timestamp: May 29, 2015, 02:28:39 pm
content: Quote from: j.a.mcguire on May 29, 2015, 09:18:00 amJust FYI, the internal resistance of a car battery is really low. Really, really low. In the order of 10 milliohms or so. If you consider that it can supply hundreds of amps to crank your starter motor, it can also sink hundreds of amps from an unregulated charger.

author: TimFox
author_rank: Super Contributor
timestamp: May 29, 2015, 02:39:37 pm
content: If you do the change in voltage due to 1 Amp load test:Assuming 10 milliohm resistance (above post), loading the battery with a 12 ohm resistor for 1 Amp current should change the voltage at the battery terminals by 10 mV.--To measure < 10 mV change at approximately 12 V requires at least 4-1/2 digits in the DVM.--The 12 ohm resistor must dissipate at least 12 watts: power wirewound required.--Keep the voltmeter leads connected directly to the battery terminals when connecting the resistor remotely (4-terminal Kelvin connection).

author: tron9000
author_rank: Frequent Contributor
timestamp: May 29, 2015, 02:50:44 pm
content: http://batteryuniversity.com/learn/article/how_to_measure_internal_resistanceThink of your battery as a big capacitor (or another battery) with a series resistor. The more current you pull, the more voltage that is dropped across the internal resistor. Measuring OHMS direct with a DMM will say O.L. as you are just passing current through from + to - terminals.Really the only thing you can determine withjusta DMM and a battery is a rough state of capacity: 13.4V is well charged, anything below 11.8V is in need of a charge, but it cannot determine if that battery has good internal resistance and weather it will charge at a decent rate or discharge without dropping voltage.

author: Mechatrommer
author_rank: Super Contributor
timestamp: May 29, 2015, 03:13:49 pm
content: Quote from: j.a.mcguire on May 29, 2015, 09:18:00 ambecause internal resistance is not a real physical resistance, but only an equivalent imaginary resistance.

author: kripton2035
author_rank: Super Contributor
timestamp: May 29, 2015, 03:43:45 pm
content: you need an AC (milli) ohm meter to do thatyour multimeter measures DC resistancehere some projects that can measure AC resistance :http://kripton2035.free.fr/digital%20esr/esr-kaspars.htmlhttp://kripton2035.free.fr/Continuity%20Meters/continuity-an220.htmlthis digital meter can do it also :http://www.ebay.com/itm/20R-Internal-Battery-Resistance-Impedance-Meter-Tester-/331466623559?

author: IanB
author_rank: Super Contributor
timestamp: May 29, 2015, 04:14:59 pm
content: Quote from: Mechatrommer on May 29, 2015, 03:13:49 pmOf course it's a real physical resistance! A battery is a real physical thing made of solid matter.

author: sdg
author_rank: Regular Contributor
timestamp: May 29, 2015, 05:38:07 pm
content: Quote from: blueskull on May 29, 2015, 09:34:43 amOn an empty battery ?The alternator is not designed for that...

author: jsquaredz
author_rank: Contributor
timestamp: May 29, 2015, 06:03:40 pm
content: Quote from: IanB on May 29, 2015, 04:14:59 pmIsn't the "resistance" really ESR or Equivalent series resistance? Similar to a capacitor. Current isn't actually flowing through the battery so you aren't measuring resistance through the electrolyte, but rather you are measuring the ability of the battery to source current at a given voltage and its being expressed with ohms law in ohms. So 12V / 800A = 15mOhms, but if that same battery were failing and could only provide 100A then it would have an ESR of 120mOhms

author: jsquaredz
author_rank: Contributor
timestamp: May 29, 2015, 06:09:49 pm
content: Quote from: jsquaredz on May 29, 2015, 06:03:40 pmIn practice I suspect you would really want to test at some fixed current and measure the voltage drop to calculate the resistance.

author: IanB
author_rank: Super Contributor
timestamp: May 29, 2015, 06:25:02 pm
content: Quote from: sdg on May 29, 2015, 05:38:07 pmWhy do you say that? Push starting or jump starting a car with a flat battery is what people have always done. Once the engine is running you just drive the car for a few miles and the battery is charged. (Of course the battery may have lost some of its useful life from being run flat, but that's a different story.)

author: SeanB
author_rank: Super Contributor
timestamp: May 29, 2015, 06:42:28 pm
content: Once started let it idle for a few minutes, and the alternator will charge the battery at a low current to some point. So long as the idiot light is off it is going to be putting in charge, and with low revs the output will be limited somewhat.Monday I went to the car, turned the key and nothing. Jump start and went to work a little late. Get there, check the paperwork and the battery is 14 months old. Check with the supplier ( bought as a breakdown spare on a Saturday afternoon as I __REALLY__ needed the car, and Midas was the only place open) and there is a 24 month warranty. Took the deceased battery in, and they say they will have to charge for 24 hours and call me back. I suggested ( best description short of flipping my lid in store) that they charge for a short time and I would be back in an hour. One hour later back and they checked, and still dead and not charging ( I told them that...........) so out I went with a new battery. Will be back with it in under 24 months if the quality is like the other, so keeping the slip and the swap info, I will have a 24 month warranty on the replacement as well.

author: TimFox
author_rank: Super Contributor
timestamp: May 29, 2015, 06:46:21 pm
content: Quote from: jsquaredz on May 29, 2015, 06:03:40 pmInternal resistance of a battery is different from ESR of a capacitor. The "current" through the dielectric of a capacitor is displacement current, caused by charge piling up on the electrodes. The current through a battery is a flow of physical charged ions inside the electrolyte.