How to measure IR of LiPo batteries?

[Jack_K]

I use LiPo batteries in my RC airplanes. The batteries are 3, 4, 5, and 6 cells with mAh ratings of 1500 to 6000 (1.5 to 6 Ah). The batteries have labels indicating the discharge rate in C. They vary from 20C to 60C.

I want to measure the internal resistance of the batteries. I'll be using a Volt/Amp/Watt meter in series between the battery pack and the load resistor. I'll first read the no load voltage, then apply the load, and read the voltage and current. To calculate IR, I subtract loaded voltage from unloaded voltage and divide that by the amps. I can estimate the individual IRs by dividing the total IR by the number of cells. OK so far.

I used a 1 ohm 100 watt resistor on some 3 cell packs (12.6 volts fully charged). I waited about 10 seconds for the voltage to settle before taking the readings. The IR calculations are way too high. I assume I should use a much higher resistance?

My question is how much of a load to put on the battery and how long to leave it connected before taking the reading?
Jack

 

[Nigel Goodwin]

I think you've already found your answer, the internal resistance will vary considerably depending on a number of factors.

However, drawing 12A from 6Ah (or less) batteries is really hammering the batteries and will make internal resistance read higher than expected.

I would suggest that pulsing the load on and off, and reading the changes with a scope would be a better way of approximating internal resistance, rather than a continuous heavy load.

Why do you want to know anyway?.

 

[dknguyen]

I would suggest that pulsing the load on and off, and reading the changes with a scope would be a better way of approximating internal resistance, rather than a continuous heavy load.

I concur with this. Why are you waiting 10 seconds to allow anything to settle? 1 second is already within human time scales which means as far as something electrical goes, it's probably eternity. And you're using 10 seconds.

I believe internal resistance also changes with temperature and remaining battery capacity. It's probably not a fixed number like you seem to be imagining it to be. So combined with heating up due to running things at 10 seconds, and the reduction in voltage due to reduction capacity from draining it for 10 seconds under high load (especially for high cell count, low mAh packs), and due to any IR change that results from a different capacity than what you started out with, you would come up with inflated internal resistances. But if that's the conditions it's operating under in use, then those are the conditions you should be worried about.

Also, maybe measure your no load voltage AFTER you measure the loaded voltage. That way you can maybe eliminate some of the error due to draining and heating as mentioned above. Also maybe do it at half battery capacity if you're only taking a single reading since that's more representative of where the battery will be most of the time than full charge.

 

[Ylli]

Decades ago, when I worked at a major lead-acid battery manufacturer, we commonly measured internal battery resistance. We did it with an ohmmeter - but an ac based ohmmeter. Apply a know AC current source to the battery and measure the resulting ac voltage drop. Calculate Rin.

 

[dknguyen]

Decades ago, when I worked at a major lead-acid battery manufacturer, we commonly measured internal battery resistance. We did it with an ohmmeter - but an ac based ohmmeter. Apply a know AC current source to the battery and measure the resulting ac voltage drop. Calculate Rin.

How much of an AC voltage relative to battery voltage did you guys apply so you didn't blow the battery out?

 

[Tony Stewart]

All Batteries have memory, but some more than than others. So in effect there is a range of Rin depending on the pulse duration. Imagine R1C1||R2C2 where R1C1 is much lower due to both values being less than R2,C2.

 

[Ylli]

We used an actual meter designed for this purpose, don't recall what the output level was.

Try this: Take a 12.6 vac filament transformer and connect one of the secondary leads to the negative of the battery. Take the other secondary lead and connect to a 100 ohm 2W resistor. Connect the other end of the resistor to the negative lead of a 220 uF capacitor (100 uF - 470 uF OK)(voltage rating > battery voltage). Then the positive lead of the capacitor goes to the battery +. Fire up the filament transformer.

Measure the ac voltage across the 100 ohm resistor and calculate the loop current.
Measure the ac voltage across the battery and use that and the loop current to calculate Rin.

Anyway, that is what I would do.

 

[Pommie]

I don't think I'd be wanting to apply AC to Lipo batteries.

Mike.

 

[Ylli]

Well, I did say we did it to lead acid batteries, but let's runs some numbers....

We are applying about 100 ma of ac current to the lipo. If it has an internal resistance of 1 ohm (unreasonably high), then the ac will develop a voltage of 0.1 vac across the battery. This is 0.144 v peak. If the open circuit terminal voltage is sitting at 3.8 volts DC, then the ac drive might take that to 3.95 volts. Well under the 4.2 volt limit. I think it would be safe.

 

[rjenkinsgb]

Well, I did say we did it to lead acid batteries, but let's runs some numbers....

We are applying about 100 ma of ac current to the lipo. If it has an internal resistance of 1 ohm (unreasonably high), then the ac will develop a voltage of 0.1 vac across the battery. This is 0.144 v peak. If the open circuit terminal voltage is sitting at 3.8 volts DC, then the ac drive might take that to 3.95 volts. Well under the 4.2 volt limit. I think it would be safe.

I've never seen that but it is an interesting idea and I can see how it could work.

It would need a big electrolytic cap in series to block any DC path through the transformer, plus a resistance to set the test current.

As long as that current is within the safe limits for the cell or battery, it's just equally charging and discharging with rather short cycles and should have no net effect or cause any harm.

Measure the test current via the voltage across the limiting resistor then measure the AC component across the cell to calculate it's resistance.

 

[dr pepper]

60C at 6Ah is 360 amps, a value found on some high power Rc motors.
What impedance do you get?, I'd expect a fairly high drop in voltage at very high loads.

 

[audioguru]

Me and most other RC users simply use our LiPO batteries until their performance is not good enough. Poor performance is obvious and testing of their internal resistance is not needed. Not enough power and low capacity are shown during their use.
I have some batteries that are 5 years old and were worn out a few years ago but I still use them in a few of my models that do not need high power.
Most of my old batteries had intermittent power until I replaced their worn out end-of-the-cable where the connector is attached.

 

[Jack_K]

I concur with this. Why are you waiting 10 seconds to allow anything to settle? 1 second is already within human time scales which means as far as something electrical goes, it's probably eternity. And you're using 10 seconds.

I read a post on an RC forum where the guy had said that's the proper way to do. He used two 12V 50 watt halogen bulbs in parallel for testing 3 cell LiPo batteries. Maybe 1 ohm is too big a load? That's what I'm trying to find out.

 

[Jack_K]

60C at 6Ah is 360 amps, a value found on some high power Rc motors.
What impedance do you get?, I'd expect a fairly high drop in voltage at very high loads.

I measured 50 to 120 milliohms for several 3 cell LiPo's. Divided by 3 to estimate individual cell IR. That's way too high!

 

[dknguyen]

I read a post on an RC forum where the guy had said that's the proper way to do. He used two 12V 50 watt halogen bulbs in parallel for testing 3 cell LiPo batteries. Maybe 1 ohm is too big a load? That's what I'm trying to find out.

The only reason I can see for that is becuase the batteries were fully charged and he didn't want to test them at that level. But you can just test the batteries when they are half-charged if that's the case.