I'll soon have an analog battery load tester which draws 100A fixed. Given I have knowledge of the CCA (or other such rating) of the lead acid battery, open voltage and the voltage under load, is it possible to estimate the internal battery resistance, even an approximation? I suppose I look for an equation that takes into account all these parameters (or maybe some more variables I can't think of right now) and gives a mΩ output.
Measuring Lead Acid battery resistance indirectly?
- Thread starter apagogeas
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Power Me Up
Enlightened
Yes: The resistance can be worked out by dividing the voltage drop by the current.
E.g. if the Voltage drops from 12.7V to 12V under load, the difference is 0.7V
0.7 V / 100 Amps = 0.007 ohms or about 7 milliohms.
E.g. if the Voltage drops from 12.7V to 12V under load, the difference is 0.7V
0.7 V / 100 Amps = 0.007 ohms or about 7 milliohms.
Thanks for the reply although I'm not sure if it is that simple V=I*R rule but it is indeed the first that came to my mind. Shouldn't CCA (or another parameter that indicates current ability) be in there? I actually try to make an excel with various input parameters like CCA, mΩ etc and estimate the condition of the battery based on voltage drop observed given I know the current drawn is 100A. My confusion comes down to the fact that all car batteries, regardless of capacity/CCA ability should have a similar Ω rating when new (obviously affected by the quality of the materials used), lets say 10mΩ (I have read a range of 8mΩ to 20mΩ when new). However if I test an 100CCA battery, my voltage drop would be quite bigger than testing an 500CCA one. Therefore this test would suggest that the 100CCA battery has already 5 times higher resistance which contradicts to what I just said above. So either this 8mΩ to 20mΩ when new is wrong or the CCA (or similar) has to be taken into account.
Power Me Up is correct. The general formula is R = (V1 - V2) / (I2 - I1). In your case I2 = 100, I1 = 0, so it's simply the drop in voltage divided by 100.
EDIT: Cold cranking amperes (CCA) is the amount of current a battery can provide at 0 °F (−18 °C) for 30 seconds and maintain at least 1.2 volts per cell (7.2 volts for a 12-volt battery). Theoretically, you could measure V1 as the open circuit voltage at 0 °F , and calculate R = (V1 - 7.2) / CCA.
EDIT: Cold cranking amperes (CCA) is the amount of current a battery can provide at 0 °F (−18 °C) for 30 seconds and maintain at least 1.2 volts per cell (7.2 volts for a 12-volt battery). Theoretically, you could measure V1 as the open circuit voltage at 0 °F , and calculate R = (V1 - 7.2) / CCA.
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Ohm's Law is immutable; It's a law of physics and always applies. Figuring out how to apply is where the variables come in.
For example, the R in the equation really needs to be replaced by a complex impedance and not a simple resistive value. As I recall from long ago, lead acid batteries have an inductive component as well as resistive. The whole thing is just a model to try to explain how a battery behaves.
But what this really boils down to is, what you're trying to accomplish. What does knowing the equivalent series resistance figure do for you? If it's just for fun, that's awesome. But if you're trying to measure or accomplish something else, then more analysis might be necessary.
Finally, you mentioned battery "state". If you simply mean State Of Charge (how fully charged the battery is), it's pretty straight forward for lead acid. The entire charged range of a 12V LA battery is over 1.5 Volts. 12.9 Volts represents a 100% charge. Subtracting 1.5 Volts, 11.4 V represents 0% charge. It varies linearly with Voltage between these numbers; so 12.15 Volts is 50% charge. I'm parroting information I read from Deltran. You can read this same thing and more in their FAQ section. Their other technical articles are quite good about lead acid batteries. They taught me a lot.
Note that the Voltages above represent unloaded readings on batteries that have settled long enough to be correct. 24 hours of rest is the standard, but I think 15 to 20 minutes makes a huge difference.
Brian.
For example, the R in the equation really needs to be replaced by a complex impedance and not a simple resistive value. As I recall from long ago, lead acid batteries have an inductive component as well as resistive. The whole thing is just a model to try to explain how a battery behaves.
But what this really boils down to is, what you're trying to accomplish. What does knowing the equivalent series resistance figure do for you? If it's just for fun, that's awesome. But if you're trying to measure or accomplish something else, then more analysis might be necessary.
Finally, you mentioned battery "state". If you simply mean State Of Charge (how fully charged the battery is), it's pretty straight forward for lead acid. The entire charged range of a 12V LA battery is over 1.5 Volts. 12.9 Volts represents a 100% charge. Subtracting 1.5 Volts, 11.4 V represents 0% charge. It varies linearly with Voltage between these numbers; so 12.15 Volts is 50% charge. I'm parroting information I read from Deltran. You can read this same thing and more in their FAQ section. Their other technical articles are quite good about lead acid batteries. They taught me a lot.
Note that the Voltages above represent unloaded readings on batteries that have settled long enough to be correct. 24 hours of rest is the standard, but I think 15 to 20 minutes makes a huge difference.
Brian.
StorminMatt
Flashlight Enthusiast
Ohm's law is really more of a mathematical model than it is a physical law. After all, it defines something that doesn't really exist in the real world: a perfectly linear resistor. Although the approximation of a perfectly linear resistor applies VERY well to many devices under many conditions, a battery is not really one of them. Internal resistance of a battery can vary quite widely, depending on such factors as the current draw, state of charge, and temperature. In order to get a general ballpark figure, it would probably be best to calculate internal resistance under a variety of loads on a freshly charged battery.
Ok, I'll try to fully explain what I want to measure. When we get a brand new fully charged battery, the overall resistance is R = R1 + R2 where R1 is the metal resistance (includes voltmeter contact, posts, internal grid and probably something more) and R2 is the resistance due to electrolyte contact to the plates (amount of available contact e.g. due to sulfation, smaller plates etc).
R1, based on what I have read should be very small let's say 10mΩ and should be almost the same in all batteries (the range 8-20mΩ I said above and primarily affected by the quality of materials and the connections used). R2 would vary greatly based on the capacity of the battery and CCA ability (smaller/bigger plates exposure and electrolyte density).
So, what I plan to do, and thus my question, is to establish a way to measure R1 which is a direct measurement to estimate the battery condition. So, when I get a new brand battery, which I will assume it is "perfect" without corrosion/sulfation etc, I'll measure the resistance using the load and estimate the overall R using V=I*R.
In this first R result, I'll consider R1 to be minimal to R2, perhaps equal to 0 because all the actual measured resistance will be due to the available current that can be drawn by the electrolyte towards the plates. I'll keep a note of this R2. Actually this R2 would possibly be derived by CCA (another equation) since CCA in practice is a measurement of available plates exposure to the electrolyte and transferable energy ability (approximately).
Now, after some time when I'll measure the battery with a load again, the battery will possibly has suffered corrosion, sulfation etc. Assuming I have fully charged the battery, I'll get a new combined measurement of R = R1 + R2 where R2 is the initial current deliverable ability of the electrolyte to the fully exposed plates (when no sulfation or other damage has occurred) and given the electrolyte condition does not change (assumption again - I mean its ability to deliver current is not affected), any increase to resistance measurement should be attributed to R1, which includes internal corrosion, sulfation buildup etc. Therefore I have a way to directly measure the internal condition of the battery (not the actual charge in there).
Now, R1 in brand new batteries shouldn't be affected by the capacity/CCA ability since it is only metal and any variation should be attributed to the quality of metals but it should be small i.e. in the range 8-20mΩ. However, if I use directly the V=I*R, a low CCA battery will really show me the combined resistance which includes R2. However, the real measurement of the condition of a battery is R1 only. I hope I made more clear what I'm after. R1 in practice will tell me (approximately) how much corroded/sulfated etc the battery is. The reason I expect some form of CCA (or some other such measurement) to be in an equation is to eliminate the influence of R2 in the combined resistance.
And yet another question: those battery testers that measure battery resistance, what do they actually measure? R1, R2 or R1+R2?
R1, based on what I have read should be very small let's say 10mΩ and should be almost the same in all batteries (the range 8-20mΩ I said above and primarily affected by the quality of materials and the connections used). R2 would vary greatly based on the capacity of the battery and CCA ability (smaller/bigger plates exposure and electrolyte density).
So, what I plan to do, and thus my question, is to establish a way to measure R1 which is a direct measurement to estimate the battery condition. So, when I get a new brand battery, which I will assume it is "perfect" without corrosion/sulfation etc, I'll measure the resistance using the load and estimate the overall R using V=I*R.
In this first R result, I'll consider R1 to be minimal to R2, perhaps equal to 0 because all the actual measured resistance will be due to the available current that can be drawn by the electrolyte towards the plates. I'll keep a note of this R2. Actually this R2 would possibly be derived by CCA (another equation) since CCA in practice is a measurement of available plates exposure to the electrolyte and transferable energy ability (approximately).
Now, after some time when I'll measure the battery with a load again, the battery will possibly has suffered corrosion, sulfation etc. Assuming I have fully charged the battery, I'll get a new combined measurement of R = R1 + R2 where R2 is the initial current deliverable ability of the electrolyte to the fully exposed plates (when no sulfation or other damage has occurred) and given the electrolyte condition does not change (assumption again - I mean its ability to deliver current is not affected), any increase to resistance measurement should be attributed to R1, which includes internal corrosion, sulfation buildup etc. Therefore I have a way to directly measure the internal condition of the battery (not the actual charge in there).
Now, R1 in brand new batteries shouldn't be affected by the capacity/CCA ability since it is only metal and any variation should be attributed to the quality of metals but it should be small i.e. in the range 8-20mΩ. However, if I use directly the V=I*R, a low CCA battery will really show me the combined resistance which includes R2. However, the real measurement of the condition of a battery is R1 only. I hope I made more clear what I'm after. R1 in practice will tell me (approximately) how much corroded/sulfated etc the battery is. The reason I expect some form of CCA (or some other such measurement) to be in an equation is to eliminate the influence of R2 in the combined resistance.
And yet another question: those battery testers that measure battery resistance, what do they actually measure? R1, R2 or R1+R2?
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CCA is a function of R. For a 12 V battery, CCA = (V1 - 7.2) / R. A battery has high CCA for the same reason it has low R ... lots of surface area on the plates; the same as any conductor with larger cross-sectional area has lower resistance than a smaller conductor of the same material.
R1 + R2 The tester can't tell the proportions of the two factors.
Old thread, but did not notice it before.
A simple resistance measurement can be a very poor method of measuring either the capacity of a battery and generally poor for state of charge as well.
- In a battery that is well maintained over it's life, i.e. it is properly charged, never left long in a discharged state, etc., then the resistance of the battery will not decrease very much as it wears out until its capacity has been significantly reduced
-In a battery of good condition as well, resistance does change over the state of charge, but the change is not as predictable and will vary from battery to battery.
- What resistance is a very good indication of is sulfation as it removes plate area and directly increases resistance
Battery testers measure the sum of all resistances in a battery. More sophisticated test methods have some ability to isolate.
Good resistance testing should be done with two resistances. A single resistance does not give as accurate of measurement as you will be reading from the open circuit voltage. A two resistance step measurement is more accurate as it measures from two load positions.
Keep in mind there is a fairly large capacitance in a battery (farads) and hence measurements need to be taken over seconds to be accurate.
Semiman
A simple resistance measurement can be a very poor method of measuring either the capacity of a battery and generally poor for state of charge as well.
- In a battery that is well maintained over it's life, i.e. it is properly charged, never left long in a discharged state, etc., then the resistance of the battery will not decrease very much as it wears out until its capacity has been significantly reduced
-In a battery of good condition as well, resistance does change over the state of charge, but the change is not as predictable and will vary from battery to battery.
- What resistance is a very good indication of is sulfation as it removes plate area and directly increases resistance
Battery testers measure the sum of all resistances in a battery. More sophisticated test methods have some ability to isolate.
Good resistance testing should be done with two resistances. A single resistance does not give as accurate of measurement as you will be reading from the open circuit voltage. A two resistance step measurement is more accurate as it measures from two load positions.
Keep in mind there is a fairly large capacitance in a battery (farads) and hence measurements need to be taken over seconds to be accurate.
Semiman
Russel
Enlightened
True, measuring the difference in voltage drop between two loads, such as 5 amps and 100 amp for a standard lead acid car battery will result in a more accurate internal resistance calculation. But, if you use the voltage drop difference between a 100 amp load and open circuit, taken after the 100 amp load, you should still be able to calculate internal resistance with reasonable accuracy for the pupose of future IR comparison.
I will agree with your statement if you let the battery sit for quite some time before making the OC measurement as that voltage can change a lot. I would say for a large battery, a 5A load may not even provide enough initial loading if the battery was initially charged. I have looked at far too many discharge curves for lead acid batteries and seen how much voltages can change based on when charging/discharging occurred, etc. ..... actually right now I have an AGM battery in my home lab doing charge/discharge tests.
Look at some of the cool equipment from Cadex using new techniques for battery measurement.
Look at some of the cool equipment from Cadex using new techniques for battery measurement.
Russel
Enlightened
Agreed. I would assume a 24 hour rest period after charging, before testing IR. The Cadex equipement is very interesting, as well as "Batteries in a portable world."
