Measuring the Internal Resistance of a SLA 12V9AH battery - Help!

[davidabc]

Hi, I have an old and a new SLA battery 12V9AH, and the specs says that it should have an internal resistance of about 18mOhm.  But for the life of me I get results of 770 mOhm for the old battery, and about 400 m Ohm for the new battery.

I am measuring the Open Circuit Voltage (V1) and then putting a 20 Ohm (10W) resister across it and measuring the reduced Voltage (V2).  I am also measuring the current (I) when under the 20 Ohm load, just to be sure.

Old battery:  V1 = 12.73V, V2 = 12.25V, I = 0.62A
New battery: V1 = 12.52V, V2 = 12.27V, I = 0.62A

The new battery has not been charged since I purchased it yesterday.  I would have thought that 0.6A is a reasonable current to determine the internal resistance.

What am I doing wrong?  We didn't do this type of thing in Engineering.  Dave, or anyone, Help! :-)

[BeBuLamar]

I don't think you do anything wrong. I think the specs is wrong.

[bdunham7]

A battery is not a resistor and measurements of effective internal resistance are going to vary with test parameters and battery condition.  Any specification of internal resistance should specify the exact conditions used.  One common way of measuring cell impedance is using AC, another is to use a very short high-current load pulse.  For lead-acid batteries, these tests would typically be done on a fully charged battery.

https://batteryuniversity.com/article/bu-902-how-to-measure-internal-resistance

[jmelson]

If the battery has been in storage for a while, a sulfate layer has built up on the plates.  You need to cycle the battery a few times to break that down.  Possibly, an overnigbt charge will do that.  Then, make your measurement again.  I think you will find the resistance of the new battery improves quite a bit.  Maybe the old battery won't improve much as it may be at the end of life.
Jon

[Michael Craft]

The internal resistance of a battery is a function of, well, just about everything (temperature, current, % of charge, constant DC vs. pulsed DC, etc.).

A lot of online guides recommend measuring the OC voltage, measure the voltage with a known load, and then calculate the internal resistance. But a more accurate method is to not measure the OC voltage, but instead measure the voltage for two known loads (or equivalently, at two known current levels). Using a little bit of algebra, you can then calculate the average internal resistance of the battery for loads that are between those two values.

[davidabc]

Thanks everyone for your replies.  I will cycle the charge of the batteries, just to be sure. But can you suggest what resistance (or current) I should be aiming for when determining the Internal Resistance of the battery.  I'll also try multiple resistor values and not use the OC voltage as was suggested.

It just seems odd to me, all things considered, that my calculations for the internal resistance appears to be an order of magnitude larger that the specifications of the new battery just manufactured months ago. I would like to know why. 

I have also just purchased a battery tester which measures internal resistance.  I'll let you know what results I get with that.  It should be in the same ball-park as with my measurements, I would hope.

[IanB]

Old battery:  V1 = 12.73V, V2 = 12.25V, I = 0.62A
New battery: V1 = 12.52V, V2 = 12.27V, I = 0.62A

The new battery has not been charged since I purchased it yesterday.

According to the internet, the open circuit voltage seems a bit low for an SLA battery. It seems it should be closer to 12.9 V. As others have said, maybe give it a good thorough charge before measuring?

[bdunham7]

It just seems odd to me, all things considered, that my calculations for the internal resistance appears to be an order of magnitude larger that the specifications of the new battery just manufactured months ago. I would like to know why. 

I have also just purchased a battery tester which measures internal resistance.  I'll let you know what results I get with that.  It should be in the same ball-park as with my measurements, I would hope.

Your results will probably be (very) different because they likely used an AC measurement method.  An OCV of 12.6V and an effective internal resistance of 18mR would imply a short circuit current of 700 amperes.  I doubt it can produce that, at least not long enough for you to measure it.  It is an interesting experiment, but don't be shocked if the manufacturers spec, your battery tester and your load-derived results are all quite different.

[radiolistener]

lead acid battery internal resistance depends on it's charge state.

Also note that lead acid battery parameters also depends on the battery temperature. Specification parameters for lead acid batteries are listed for 25 °C. If your battery has other temperature, you're needs to heat or cool it to 25 degrees before testing.

Since your batteries have different open circuit Voltage, they probably have different charge state. So you can't compare them. First you're needs to charge it for 100%, withstand them for two days after charge, bring the temperature to 25 degrees and only after that you can test it.

Also, I think 20 Ω load for 9 Ah battery is too small for internal resistance measurement. I think it's better to use something like 1 Ω load.

Also note that your current meter has some resistance and it affects your current measurement.

Also note that resistor has some error, so you needs to measure it's exact resistance.

[Vovk_Z]

What am I doing wrong?  We didn't do this type of thing in Engineering.  Dave, or anyone, Help! :-)

I guess you are doing it right, but there are several different methods how to define and how to measure internal resistance. You may get closer to specs if you pre-load battery. For example, measure V1 with some 100 R load, and then V2 with 20 R load. So you'll have delta V and delta I.

[MrAl]

Hi, I have an old and a new SLA battery 12V9AH, and the specs says that it should have an internal resistance of about 18mOhm.  But for the life of me I get results of 770 mOhm for the old battery, and about 400 m Ohm for the new battery.

I am measuring the Open Circuit Voltage (V1) and then putting a 20 Ohm (10W) resister across it and measuring the reduced Voltage (V2).  I am also measuring the current (I) when under the 20 Ohm load, just to be sure.

Old battery:  V1 = 12.73V, V2 = 12.25V, I = 0.62A
New battery: V1 = 12.52V, V2 = 12.27V, I = 0.62A

The new battery has not been charged since I purchased it yesterday.  I would have thought that 0.6A is a reasonable current to determine the internal resistance.

What am I doing wrong?  We didn't do this type of thing in Engineering.  Dave, or anyone, Help! :-)

Just curious, what do you need to know the internal resistance for?

The best test i think is not an absolute measurement but a comparative method.
If you get a new fully charged battery you can measure the voltage and apply a load for a number of seconds, let's say 10 seconds, and immediately note the voltage and write the two down.
Now sometime in the future you do the test again, and write the voltages down.
Then again some time in the future from that test do the test again.
You would repeat this test maybe once a month or something like that and you would accumulate a set of measurements over time and these would be all for the same battery not a different one even if the same make and model.

After a while the voltage at the end of the 10 seconds will start to decrease.  This tells you the battery is getting old.  Once it gets down so low that it no longer works in the application you wanted it to work in you know you need a new battery.

This tests bypasses the need to know the internal resistance as well as the need to know the current fully charged charge capacity because it measures the most practical aspects of the battery as it relates to the application it is used for.

For a car battery it would be noted that as the final voltage gets too low the engine starts to crank over very slowly.  This is especially true in winter so that is a good time to test it.

[radiolistener]

Note that lead acid battery Voltage depends on electrolyte density, which depends on battery temperature. So, if you doing test at different environment temperature you will get different Voltage for the same state of the battery. Values listed in the battery specification is relevant for 25 °C.

This is a time to prepare for a nuclear Armageddon and power blackout, so there is a note: lead acid battery can be damaged if you discharge it at low environment temperature. Electrolyte in discharged battery can freeze and damage battery case. So, be careful, it will be hard to find battery replacement after nuclear Armageddon, so take care to keep the battery in charged state, especially at low environment temperature.

[2N2222A]

What would happen if you had two identical equally charged batteries and hooked them back to back? Could you measure the resistance with a regular multimeter?

[Vovk_Z]

No, you can't measure the resistance of any battery with a conventional ohmmeter.

[bdunham7]

What would happen if you had two identical equally charged batteries and hooked them back to back? Could you measure the resistance with a regular multimeter?

Theoretically perhaps, practically I really doubt it with a conventional multimeter ohm function.  You might pull it off with an AC ohmmeter.

[radiolistener]

What would happen if you had two identical equally charged batteries and hooked them back to back?

One of them with higher Voltage will be discharged to the level of second battery Voltage. The second battery will be charged from the first one. After that they will stay at the same Voltage level.

Could you measure the resistance with a regular multimeter?

Do you mean internal battery resistance? Yes, but you're needs a load with known resistance.
You can measure load with multimeter in the Ω-meter mode.

There are two ways to measure internal resistance:
1) Measure open terminals Voltage and Voltage under load with known resistance (for example 1.00 Ω).
2) Measure Voltage under load for a two known load with different known resistance (for example 1.00 Ω and 2.00 Ω).

The second way is more precise.

[2N2222A]

1) Measure open terminals Voltage and Voltage under load with known resistance (for example 1.00 Ω).

The person who started this topic showed that batteries are not linear when measured this way. For instance, a 1A load may bring the Voltage from 12.8 to 12.5. A 2A load may bring the Voltage from 12.8 to 12.4

[radiolistener]

The person who started this topic showed that batteries are not linear when measured this way. For instance, a 1A load may bring the Voltage from 12.8 to 12.5. A 2A load may bring the Voltage from 12.8 to 12.4

This is because 1 A is too small current for 9 Ah battery. Internal resistance measurement needs about 1.5C...2C current.

For 9 Ah battery you're needs to use 15 A current. With 1 A you will not be able to see sulfation effect. And Voltage drop will be too small for 1 A, so your DMM needs to be more precise. It needs to use at least 10 A for 9 Ah.

This is why I suggested to use 1 Ω load for measurement.

And also note, that topic starter tried to compare fully charged battery with partial charged battery. He needs to use both fully charged for proper comparison, because internal resistance depends on the battery charge.

[Vovk_Z]

This is because 1 A is too small current for 9 Ah battery. Internal resistance measurement needs about 1.5C...2C current.

Possibly, you are talking about a kind of Lithium battery? Because for a lead acid battery 1.5-2C can possibly be too much.

[radiolistener]

Possibly, you are talking about a kind of Lithium battery? Because for a lead acid battery 1.5-2C can possibly be too much.

No, I'm talking about lead-acid batteries. This is a standard current for testing lead-acid batteries.

For example the load fork for testing a car battery use 100 Amps current for 55-60 Ah battery. It's about 1.8C current. You're needs to apply that current for at least 5 second and after that write down Voltage under the load.