Author Topic: Need help to understand battery internal resistance pls (Read 4477 times)

Chriss · « **on:** October 27, 2020, 12:46:50 pm »

Hi!
I'm "playing" with measuring of lead acid battery internal resistance.
On som datasheet it is writing lets say for a 12v 7Ah UPS battery the internal resistance should be ~18m $\$\Omega\$$ .

On the net I found writing's like "high internal resistance is leading in bad battery..." etc...

This is not really clear for me.

If a battery have a low internal resistance that means it is more close to short circuit.
If the internal resistance is let's say in M $\$\Omega\$$ range it means for me far away from short circuit.

Short circuit means for me a state where the battery internally discharge itself.
High internal resistance means for me the battery will last longer cos it will not discharge itself so fast.

So, I'm a bit confusing with this.

Can somebody clarifie it a bit for me please?

Thank you.

RoGeorge · « **Reply #1 on:** October 27, 2020, 01:24:19 pm »

Quote from: Chriss on October 27, 2020, 12:46:50 pm

If a battery have a low internal resistance that means it is more close to short circuit.

No, because the internal resistance of a voltage source is considered to be in series (not in parallel) with the battery.

ledtester · « **Reply #2 on:** October 27, 2020, 01:36:07 pm »

Quote

If a battery have a low internal resistance that means it is more close to short circuit.
If the internal resistance is let's say in M $\$\Omega\$$ range it means for me far away from short circuit.

Short circuit means for me a state where the battery internally discharge itself.

A short circuit usually means a low resistance path from (+) to (-) created by the circuit attached to the battery.

If there was a low resistance path from (+) to (-) within the battery itself wouldn't last very long - it would get hot and soon become a dead battery.

Here's some articles on internal resistance and self-discharge:

https://batteryuniversity.com/learn/article/rising_internal_resistance

https://batteryuniversity.com/learn/article/elevating_self_discharge

Chriss · « **Reply #3 on:** October 27, 2020, 02:07:43 pm »

Thank you so much!
I understand it completly how you interpreted it.

How high resistance would be in a bad battery?
I mean, if I measure and calc the internal resistance, how can I know if it is ok or not?

Thank you very much.
I got it now.

TimFox · « **Reply #4 on:** October 27, 2020, 04:15:42 pm »

You need to consider how much current the battery supplies to your application. As the internal (series) resistance increases, the voltage at the load deceases until it is too low for your purpose.

Chriss · « **Reply #5 on:** October 27, 2020, 04:47:56 pm »

Thanks guys for the help.
It is completely clear for me now how the internal-series resistor is acting in the battery.

At the moment when I realized based on your writing does that internal resistance is in series actually then I
understand why is it bad if this resistance is high...

tunk · « **Reply #6 on:** October 27, 2020, 04:49:27 pm »

It will also heat up the battery.

tooki · « **Reply #7 on:** October 27, 2020, 08:59:02 pm »

Quote from: TimFox on October 27, 2020, 04:15:42 pm

You need to consider how much current the battery supplies to your application. As the internal (series) resistance increases, the voltage at the load deceases until it is too low for your purpose.

Well said!

There’s even the case where a high internal resistance is actually desirable: built-in current limiting. The perfect example of this is putting a single LED on a 2032 button cell. The 3V nominal voltage would theoretically fry many an LED. But the internal resistance limits the current, meaning that in practice, you can put any LED* on it without a current-limiting resistor, making a perfect LED tester or mini-flashlight.

*Even traditional green LEDs with a Vf of just 1.7V!

Chriss · « **Reply #8 on:** October 30, 2020, 08:24:03 pm »

Ok guys here I am again with some confusing think.

I ordered one new Tp12-7 type battery for my UPS and played a bit with...
I can not find a datasheet for this battery, but similar batteries datasheets is writing in the datasheet
an internal resistance of the battery around 18mOhm.

I didn't get anything close when I was measured.

Here is how I made the calc and what result I got, which is for me a bit high, but probably the error is laying by me.
I used a voltmeter and a resistor R_L = 330 Ohm/16w.

I measured the battery voltage without the R_L = 330 Ohm and it was U_b1=12.92V
After that I attached the R_L and measured the U_b2 = 12.88V
Then I calculated the current I = U_b1/R_L = 0.0392A
Then I calculated the U_d = U_b1 - U_b2 = 0.04V
Then I calculated the internal resistance of the battery R_s = U_d / I = 1.02Ohm

So, I come to a result does my new battery on ambient temperature of ~22C has an internal resistance of 1.02 Ohm.
That is for me somehow really high, or maybe I'm wrong? Or maybe I did something wrong in my calc?

TimFox · « **Reply #9 on:** October 30, 2020, 09:15:04 pm »

Did you measure the battery voltage directly at the battery terminals, not downstream at the load resistor?

Chriss · « **Reply #10 on:** October 30, 2020, 09:16:38 pm »

Yes, on the battery terminals.

TimFox · « **Reply #11 on:** October 30, 2020, 09:39:23 pm »

Technically, the current I = Ub2/R, but that makes no practical difference to your calculation.
Your voltmeter resolution is not very good for this measurement, but if the resistance really were 0.018 ohm, the difference in voltage would be only 0.0007 V, and the meter couldn't measure it at all.
Did you verify the resistor value? Was the battery fully charged when you did the measurement?

Chriss · « **Reply #12 on:** October 30, 2020, 09:46:27 pm »

What you wrote about my DMM makes sense. The battery is fully charged. I asked the company who shipped the battery if they will full charge the battery or not before shipping. And they told every battery are full charged and tested before shipping.

I also measured the load resistor and the result was 332 ohm what makes not a big difference.

What you prefer what resistor should I maybe use maybe to make the measurement more accurate?

TimFox · « **Reply #13 on:** October 30, 2020, 10:41:07 pm »

To measure the resistance, you should pull as much current as is safe, for as long as it takes for the meter to read the loaded voltage. Do you have access to a voltmeter with more than 3-1/2 digits? How much current is expected from the battery in normal use? Of course, the load resistor will need to be a high power part.

Chriss · « **Reply #14 on:** October 30, 2020, 11:08:36 pm »

I will see if I can manage one dmm as you preferred.
In normal use the battery will be loaded around 6-7A.

bsfeechannel · « **Reply #15 on:** October 31, 2020, 01:14:30 am »

According to this article the internal resistance of a lead-acid battery is a function of the load current. For low currents the internal resistance is high and decreases exponentially until approx. 0.69C. Since C is 7Ah for your battery, according to the article, you should test it at around 5A. Internal resistance also changes with charge, time and temperature.

BravoV · « **Reply #16 on:** October 31, 2020, 05:31:25 am »

Quote from: Chriss on October 27, 2020, 02:07:43 pm

How high resistance would be in a bad battery?

See the internal resistance as the bottleneck size, the smaller one (high resistance) will slow down the flow, and it will affect the flow no matter how big the bottle's capacity.

Of course the bottleneck size only matter when you pouring out (drain) or pumping in (charge) the bottle (battery).

If bottle sits idling, not in pouring nor at filling state, then it does NOT matter what the bottleneck size is, even the neck as small as human hair size.

Now to add more complexity as bsfeechannel pointed above, the bottleneck size diameter will change depends on how fast you pour/drain the bottle, and that is the nature of that lead-acid battery.

Hope this analogy helps.

Siwastaja · « **Reply #17 on:** October 31, 2020, 08:24:21 am »

M&M analogy is great, but let's also put that back in electrical numbers avoiding ohms if that's causing more confusion than helping:

Case 1: Assume a good imaginary battery reading 12.0V with no load. You connect a 10A load, and the voltage drops to, say, 11.9V. It's quite good.

Case 2: Assume a worse battery reading 12.0V with no load. You connect a 10A load, and the voltage drops to, say, 10.0V. It's quite bad, isn't it?

Further, we can calculate heating in the battery:

Case 1: 10A * (12.0 - 11.9V ) = 1W
Case 2: 10A * (12.0 - 10.0V) = 20W

And finally, to put that in ohms:
R = dV/dI = 0.1V / 10A = 10mOhms;
R = dV/dI = 2.0V / 10A = 200mOhms.

There is no "actual" series resistor inside the battery; this is an imaginary model. The simple resistor we draw includes all effects inside battery causing voltage drop and heating relative to current drawn.

The simplest somewhat usable model of battery is a perfect voltage source + a series resistor. We have just modeled this imaginary battery using this simple model of 12V voltage source plus a series resistor of 10mOhms (or 200 mOhms).

Real batteries behave in much more complex ways, and more complex models are needed, but for many beginner purposes, such simple model is good enough.

Do note that the battery equivalent series resistance is highly dependent on temperature, state-of-charge and battery aging.

Chriss · « **Reply #18 on:** October 31, 2020, 08:50:03 am »

I really appreciate your time what all of you spend to write such nice explanations.

I understand every part you are explained, and it is fine for me now.

Here is my new thinking relation which is not clear for me.

For first I was just playing to learn, but now I'm trying to use in practical meaning what I learned
to lets say manage the problem "how to know if a lead acid battery is good or bad based on the internal resistance."

Why I come to this idea is because I know and I own a car battery tester with huge load resistor on, something like this tester:
Schumacher BT-100 100 Amp Battery Load Tester, but I also saw something like this:
MOTOPOWER MP0514A 12V Digital Battery Tester etc.
Some of them can also calculate the internal resistance and will show on a display or print out.

The Schumacher BT-100 is a robust tester with a heavy-duty resistor and will really pull the s..t out of a battery.
But the mentioned Motopower MP0514A is a toy compared to the BT-100 tester.
On the Motopower tester the wires from the clamps to the tester are so thin does they can't even hold 10A or less.
So, they can not pull a huge amount current even for 1sec. The whole unit would be burn down to hel.

But the Motopower tester is still a tester and the manufacturer guarantee does the unit will meet the purpose to test batteries.

Then I was thinking how the hack does this lightway testers are designed and how they can do the test.
What comes to my mind was:
1. they can somehow calculate the internal resistance. ( but how they do it? )
2. they measure the actual voltage of the bat and if it is under 12.2v or so then they will say "bad", "recharge" or similar...
...
But now, as I can see, it is not simple to measure and say what is the real internal resistance of the battery, how this
small like a toy battery testers can do the measurement and how precise they are in real?

on the net

Siwastaja · « **Reply #19 on:** October 31, 2020, 11:05:46 am »

Well, it gets really complex and ugly, and as EEVblog forums have shown a few times, almost religious sometimes. People who have bought, or even just seen a Youtube video of a nice 1kHz AC ESR meter have hard time admitting the numbers it show might not be particularly useful, or the context needs to be understood.

The truth is, the simple model of battery being an ideal voltage source + a series resistor does not catch the actual battery behavior very well.

Hence, there is no single well-defined value for "battery internal resistance" that is good for all uses.

If you want to test if the battery is able to supply some 200A cranking current to the engine starter motor at -30 degC, obviously the best way to measure is to pull 200A at -30degC and see if the voltage is enough to turn the starter properly.

It's also equivalently obvious such test would be tedious to arrange.

Using very short pulses to measure the voltage drop allows quick measurement and lower power dissipation. Going short enough, we can easily alternatively pull current from the battery, then push it back, while storing the energy in the capacitors of the tester. Say we do quick millisecond long discharge and charge cycles, measuring the voltage drop both ways. Done with sine waves, this approaches a so-called 1kHz AC ESR measurement. Which produces a nice value that can be compared against, but it doesn't capture the reality how much the voltage actually drops under a load that lasts for longer than 1ms; it captures the capacitor-like behavior of the battery.

So there is no simple answer. Many ways of measuring battery ESR exist. My favorite is running three pulses with different discharge currents, say at 5A, 10A, and back to 5A, say 10 seconds each, then measure the relative voltage drop between the end of the 5A steps and the 10A step, and calculate R = dV/dI. This could be approximately called a 0.1Hz ESR, and for many purposes, it's "DC enough".

And of course, sometimes you want to know how well the battery can supply 1ms pulses. A GSM or Wifi device might do power-hungry transmission of pulses at that length, and the designer doesn't want to spend in capacitors handling this 1kHz power need; they would need to be quite big! In this case, the 1kHz ESR value could be an usable parameter, and the fact that 1kHz ESR is less affected by temperature and state-of-charge than DC ESR plays in the advantage of the designer here.

OTOH, if the designer doesn't know that, and tries to use 1kHz ESR value to determine undervoltage lockout value or maximum power dissipation for DC application, they can be shocked to find real DC ESR is easily 2 times higher under specific conditions (near the end of discharge, exactly where the voltage drop is very important to understand properly for triggering undervoltage lockout not too early and not too late).

SpecialK · « **Reply #20 on:** October 31, 2020, 02:42:43 pm »

Quote from: Siwastaja on October 31, 2020, 08:24:21 am

M&M analogy is great, but let's also put that back in electrical numbers avoiding ohms if that's causing more confusion than helping:

Case 1: Assume a good imaginary battery reading 12.0V with no load. You connect a 10A load, and the voltage drops to, say, 11.9V. It's quite good.

Case 2: Assume a worse battery reading 12.0V with no load. You connect a 10A load, and the voltage drops to, say, 10.0V. It's quite bad, isn't it?

Further, we can calculate heating in the battery:

Case 1: 10A * (12.0 - 11.9V ) = 1W
Case 2: 10A * (12.0 - 10.0V) = 20W

And finally, to put that in ohms:
R = dV/dI = 0.1V / 10A = 10mOhms;
R = dV/dI = 2.0V / 10A = 200mOhms.

There is no "actual" series resistor inside the battery; this is an imaginary model. The simple resistor we draw includes all effects inside battery causing voltage drop and heating relative to current drawn.

The simplest somewhat usable model of battery is a perfect voltage source + a series resistor. We have just modeled this imaginary battery using this simple model of 12V voltage source plus a series resistor of 10mOhms (or 200 mOhms).

Real batteries behave in much more complex ways, and more complex models are needed, but for many beginner purposes, such simple model is good enough.

Do note that the battery equivalent series resistance is highly dependent on temperature, state-of-charge and battery aging.

This is very helpful. Do note that you would need a 1.2 ohm resistor that can handle 120W (at least) to pull 10A for any significant amount of time. This is very different from the OP's 330R resistor. Also, the 200mR battery ESR becomes a significant voltage divider and would throttle that 10A to 8.6A. I suspect you could just compare the current produced between known good battery and the test subject battery.

TimFox · « **Reply #21 on:** October 31, 2020, 03:11:36 pm »

The voltage directly across that 1.2 ohm resistor will give the current, but a separate measurement of the voltage at the battery terminals is needed for the change in voltage. A simple voltmeter suffices for the voltage at the resistor, but a higher-resolution meter should be used (simultaneously) for the battery. At high current, the resistance in the connecting wires can give errors that must be avoided.

Chriss · « **Reply #22 on:** November 01, 2020, 08:19:59 am »

What about to use a scope vs a dmm?

TimFox · « **Reply #23 on:** November 01, 2020, 02:43:46 pm »

You can get 4.5 digit resolution from a reasonable DMM, that allows a much better measurement of a small decrement in battery voltage than with a DSO (usually 8 or maybe 12 bits). 8 bits is only 2.5 digits (roughly).

Chriss · « **Reply #24 on:** November 02, 2020, 04:15:53 pm »

Hmmm....
Conclusion, there is no simple way to quick test an ordinary car or UPS battery for it's healthy.
The best way is to check how old is the battery and if it is several years old, and giving some issue to any of the systems
just order a new one and be sure does the battery is no more the reason for any possible issues in the system where it is installed.

So, a car battery should be changed in every 3-5 years and in a UPS for PC every 1 - 2 years.
As I can stich out from your writings, reading on the net and testing by me, this is the only secure option to be sure the battery is ok.


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Author Topic: Need help to understand battery internal resistance pls (Read 4477 times)

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