Unless I've missed something with your first measurement (it is after 1am) I think you need to measure/calculate the current with the load connected. Your calculation for current was based on 4.193v being across the 5ohm resistor, which it isn't. That's across the internal and the 5ohm. Only 4.083v is across the 5ohm. Although that actually makes the result further from the charger's measurement at 134mOhm.

@senceryazici

5 жыл бұрын

Steve Evans yeah, I was wondering that too!

Adam, both circuits are poor methods of measuring the internal resistance of the battery. The AC method is preferred, since it does not discharge the battery with a DC load (thus changing the internal resistance during the test due to partial discharge). Simply swap R1 and the blocking capacitor positions (making the +ve of the battery terminal connected directly to R1 and the capacitor to the AC source) and connect the common terminal of your measuring instrument to the battery +ve terminal also. The internal resistance of the battery is simply :- (The AC voltage across the battery terminals divided by the AC voltage across the reference resistor R1) multiplied by the Reference Resistor Value. This is so because both resistances pass the same current and the capacitor reactances do not come into the equation. Even the calibration accuracy of the measuring AC voltmeter is not important providing it is linear !

@mntngeek

7 ай бұрын

Another way is to just use a signal generator with a known 50 ohm output impedance. Compute RMS output. Measure signal voltage across battery. 50 * VBAT / VRMS. It's not perfect, but very close if the battery's impedance is in the 100mOhm range (well within 1%). 2V P/P = 1.4142 VRMS, VBAT will be ~3mV. One can first measure the true RMS output value with a DMM. One can also use a known resistor value to calculate the true output impedance if needed.

@davidpacholok8935

4 ай бұрын

Yes in other words use the 5 ohm resistor as a current shunt to measure current. At 100 hz his 220pf!!! Caps are about 6 orders of magnitude to small.

The current going thru the resistor isn’t the open circuit voltage / 5Ohm but the measured voltage under load. So: 4.080 / 5V = 0.816A So the internal resistance equals: (4.193-4.080)/0.816=0.138mOhm

Very helpful video 👍

Great to see you comparing all the options. I fully agree with your advice to stick to the same method for all your cells. And use IR tests as yet another opportunity to reject the worst outliers.

@AdamWelchUK

5 жыл бұрын

Cheers Paul. Glad you liked the video.

Your own sig. gen. (you've done your test with) was set for 100Hz only but the commercial internal battery resistance meter clearly shown the test done @ 1KHz (on the LCD screen) and I think that plays a crucial role in such a difference

@anthonyvolkman2338

Жыл бұрын

You are absolutely correct! The DC test and the AC Capacitively coupled tests are vastly different. As you said his was home built version was 100 Hz whereas the commercial one was 1khz (1,000 Hz). The AC test is what is called an impedance test which is extremely important especially for loads that draw high frequency current (PWMs, VFDs, ESCs and other circuits that draw massive dynamic current). It is exceptionally important for applications such as Regen Braking. Where you can be pulling 10 of amps or in my case thousands of amps where in an instant you can be trying to recharge the battery through the Regen Braking protocol. This can be an issue if the "impedance" is high. Think of it as an ac current (not voltage!!). That is due to pulling current out of the cell and then putting current back into it. Yes current because of how a Lithium battery chemistry is charged.

Thanks Adam!

Plenty of thanks to you! Really nice video! Btw, you've got such a nice accent!

Excellent!

Thank you for the video

Thanks Adam for your video appreciated your performing the different methods, on the differences on results its like I tell my wife when we go on a diet and she's says what is my weight at the doctors office not on my scale at home. My comment was allway doesn't matter as long as you know what your weight was when you started you will be able to tell how much you lost, the only thing is our scale at home might not be calibrated as the doctors office scale. My point is which ever your method you will be able to monitor your batteries degradation from time to time, you can buy a new battery and see what its internal resistance is and you can compare one batteries internal resistance to the other batteries in your collection of batteries.

From what I've seen in multiple videos that demonstrate the first method (although all the others I've watched used a much lower value resistor), if you simply assume the "real" internal resistance is 80% of your calculated value, you'll likely be much closer to the value displayed by most of the tools that people seem to use to measure and display internal resistance (e.g. battery charger/dischargers) and in this case, virtually dead-on the value provided by the AC method commercial tester you demonstrated. Example: multiply your 131mohm value x 0.8 ~105mohm.

I use the LiitoKala Lii-500 . Anything with in 100mR works for me. Never had Balance issues doing so. Usually if a Cell has higher Resistance it will show up as Excessive Heating during Charging.

*A spiral bound notepad, big Clive style, makes an appearance. The internet waits for Adam to start blowing things up.*

One thing I've learned from my own testing and some research online is that it's better to DC load test when bat voltage is near to nominal voltage (~3.7). Also if you measure the load voltage first then the open voltage a few seconds after then you take the extra voltage out of the equation. I can't remember what that extra voltage is called but it's basically like a static build up that adds error.

Every DMM calculates the AC voltage a little differently, it does depend on the algorithm in the meter. That said, your first two were within 20%. Only the second commercial one had a discrepancy that was significant. I'm sure if you did a sample of ten or 20 batteries you would find the error to be linear. Also the commercial unit said the batteries were better than the home brew meter, I call that a win win. Nice tutorial !

Interesting! My method is to use two different resistors as load and calculate the internal resistance from the difference in voltage over the difference in load resistance. I imagine this wll take away the weak top voltage and bring you down at the flat spot on the battery voltage before testing. Would like to see the result of this compared to the methods you showed us. Rudy

As someone suggested, the simplest way to deal with the capacitors influence is to measure the ac voltage across R1. Rint = R2×V2/VR1

On the first method you need to actually measure the current. It's flowing through the IntR plus the test R and hence will be less. Ex: 4.2V / 5.0R = 0.840A 4.2V / 5.1R = 0.824A If you build a constant current source with an LM358 and an Nch MOSFET you can easily measure test resistors for more accurate results. Better still, use the constant current source to draw a known current from the battery and then determine the internal resistance that way. Lots of very practical and easy ways to do this.

@Mansare94

3 жыл бұрын

But you don't know the internal resistance until you know the current. You can get the true current by using ohms law across the load resistor: V=IR V/R = I I = 4.083/5 = 0.8166 amps.

.003a difference when adjusted for the machines draw of .8 vs calculated draw of .8386. No need to redo its spot on. Approx 4.5 % difference. Doesn’t look like it off hand but easy to ignore with small numbers. Great video

I am sure your capacitors (220pf) are way too small. The Xc of the caps at 100Hz is 7.2M ohms and means that their value becomes very significant when compared with the the DVM input impedance.

@davidpacholok8935

4 ай бұрын

Thank you for pointing that out! You too must have stayed awake in high-school AC circuit analysis. That he got a reading at all, let alone one that within a Decade of reality is pure luck. Journal of Irreproducable Results stuff!

Use resonance formula and you will get it thanks for the video

Greeting Adam, I found out that while your're measuring the battery, you haven't connect your function generator to the battery. So, i think that's the reason your calculation and measuring value are quite different a lot.

Thanks. For new lifepo4 cells in 4s do you think it is better to test IR for each cell at high SOC? I have a situation where one of the sales in my pack goes higher than the rest if I exceed pack voltage of about 13.8. Otherwise the voltages are the same at lower pack voltage.

Thank you for your videos! I really enjoy them and learn alot. I've always been interested in what it means to have a high or low resistance. Can you tell if a cell is bad from the numbers? How can that number relate to the application it is used in? For example if you are building a power wall.

@borayurt66

5 жыл бұрын

The higher the internal resistance the more power is lost to it. Batteries with high internal resistance are gone bad, or made bad. best to discard them.

Your first calculated current assumes internal resistance is zero. You should have used a measured current calculated from the drop across the 5 ohm resistor. Thanks for the info though.

Calculate the time constant of the RC circuit to determine the correct measuring frequency. That is more than likely how it is done in the commercial one. Also the initial experiment is only accurate within micro second after connecting the load as the voltage instantly discharges once connected especially at full charge. Then best thing would be to have a switch connected and pulse it long enough to make a measurement which is next to impossible on a low cost multimeter.

I read in some IC datasheet, can't remember which, that your method is somewhat correct, but to make measurement more accurate, it should be done with two different load currents and evaluating the difference: Ri = (Vl1 - Vl2) / (Il2 - Il1)

@MrSummitville

4 жыл бұрын

This is *HIS* Internal Resistance measurement method, *HE* will use *HIS* values to compare *HIS* batteries ...

You have neglected the impedance of the capacitor. Use a bigger one at higher frequency to reduce its effect or just add it in your calculations.

@AdamWelchUK

5 жыл бұрын

Good idea. I’ll look into that. Thanks.

@NiHaoMike64

5 жыл бұрын

Or measure the voltage across the resistor.

@andybarnard4575

5 жыл бұрын

Impedance of 220pF capacitor at 100Hz is about 7ohm, assuming internal resistance of meter can be ignored use 7+5 = 12 for R1 and you get 125m ohm using that method. So they all agree....

@AdamWelchUK

5 жыл бұрын

Andy Barnard Crumbs - is it that much? That’s really helpful, thank you.

@andybarnard4575

5 жыл бұрын

It would have been more helpful if I had added the phase of the reactance as well! So calculation is not right, but the principle is there....

All you need to do is create an AC current source using a 1uF capacitor between signal generator and battery. If you set signal source to 1.59V rms at 100Hz it will generate exactly a 1mA current source. If you then measure the AC voltage across the battery, the meter voltage will directly read the internal resistance in kohms or 1mV/ohm. To get higher sensitivity use 10uF and 10kHz to give a sensitivity of 1V/ohm.

What would be the correlation between ac impedance and dc resitance?

Perhaps 0.838A vs 0.800A could account for the difference in resistance. Also, doing one test after the other (potentially) effects the Vo figure, but still the resistance should be fairly static.

There is resistance in the wires and obviously the connections in the battery holder as well.

@1959Berre

4 жыл бұрын

Indeed, do not underestimate the resistance of the wires and weak contact points. It all adds up.

Several errors here. Some already pointed out (eg 1kHz NOT 100 Hz - it makes a big difference!) AC IR is great for low power electronics batteries. For higher power such as EV or Hybrid or RC, DC IR is best. There's a standard UL test. It involves measuring difference between two loads (0.2C and 1.0C), for specific time frames etc... at 85% and 20% SOC (that too makes a difference). And then there's the uncertainties and measurement errors...

Why we dont measure directly the current (for the first exp)? I did this experiment with my batteries and two of them are really close but I wonder which one is the best? My vote for measuring directly the current btw.

Internal AC resistance measuring frequency is 1kHZ

Try using 1KHz to match the AC meter.

Oh, so many variables in the second method. The capacitors have their own reactance, which is considerably higher than the battery internal resistance at 100Hz and should be taken into account. The Vici meter isn't the most accurate and without zero-ing out the lead resistance that can add 10% error by itself (when measuring the 5 ohm resistor). Even the wave form of the generator can skew results (which is only theoretically a perfect sine wave), as you're not using a RMS meter. Furthermore, the internal resistance changes (increases) very quickly in the first few seconds of applying a continuous load (you've seen how fast it drifts when using the first method), but the AC method only applies very short alternating pulses that doesn't allow the internal resistance to drift too much. I think the method is still valid for comparing different cells, but not for actual measurements.

adam make an arduino internal resistance measuring device version 2 , i think many wait for this!

@borayurt66

5 жыл бұрын

+1

@illchmann

5 жыл бұрын

8:40 220pF and 100Hz and you are afraid of the precision you have measured the 5Ohm power resistor?? Xc is at 100Hz around 7Mega Ohm !!!! i think you have to tune the values and frequency a little to come into a useful range, e.g. 220nF and 100kHz Xc is then a million times lower at 7Ohm. Actually i would go even futher ro let Xc as low as possible to not take it into account compared with the DC resistance 5Ohm so a big 1000uF and 10kHz so the capacitors are in milloohm reactance and the 5Ohm is dominate. The do the math again you should come close to what the devrices is measuring in about 100mOhm what anyway your first methode results in. Keep it up

The overall battery resistance consists of ohmic resistance, as well as inductive and capacitive reactance. batteryuniversity.com/learn/article/how_to_measure_internal_resistance True impedance of a cell can only be determined by injecting a 1khz signal into the battery. Pure resistance is usually fine for determining really bad cells though.

Hi Adam i know 9 months has past since this video catching up on all your great vid's. Can i ask if only building say a 3s 4p all 18650's charged 4.2v and rested for a month , if all cell voltages come in between 4.10 to 4.13 volts would that prove the internal resistance are pretty close ?. Internal resistance would show the drop in voltage same time period and ambien temperature ?. Thanks in advance.

@MrSummitville

4 жыл бұрын

No ...

I think in the first test it is difficult to calculate the current I accurately because the internal resistance itself has not been allowed for. When the current is flowing the 4.193V will actually correspond to the cell side of the internal resistance. That presmably means the internal resistance has been under calculated slightly.

@unlost117

5 жыл бұрын

Yes on his 3rd line, I=V/(Rinternal+Rload)

@jerrys9426

5 жыл бұрын

@@unlost117 Yes that's right

@unlost117

5 жыл бұрын

@@jerrys9426 Or he could simply have used Vload in same calc instead of Voc on the same line to make it correct ie I=Vload/5 instead of I=Voc/5

Hi Alan great video.. Coukd you tell me what you would regard as high internal resistance and therefore a doggy cell. Thanks for any reply. Regards David

@AdamWelchUK

5 жыл бұрын

Thanks David. As I said in the video - it’s a matter of bench marking your own cells. I’d like to build packs where all the cells have an IR less than say 80 milliohms. But when I planned to use 140 cells in my pack, and only 120 of them were within this limit, I had to revise my limit! I charge and discharge my pack at very low currents, so IR isn’t a huge factor for me - perhaps except after a poor winter period when self discharge can be more of an issue.

Those 220 pF caps have a hopelessly high impedance at 100 Hz, rendering the AC measurement totally inaccurate. The first DC measurement also included the holder resistance, which is not insignificant, better to probe across the cell buttons directly.

I'd like to ask a question since you are a very precise kind of presenter / based on math and reasoned thought - as illustrated again (thank you) on this video. I'm trying to measure capacity (loss) over time/cycles of an 18650 battery bank while its live - charged by PV array / discharged to power the inverter - each day. I'm hoping to get 3,000 - 7,000 cycles out of my 780ah @ 48v 18650 battery bank instead of 100(s) with a design where of 40% DOD in the middle of the battery voltage range. As part of this, it would be nice to have a measure of overall battery bank capacity loss over time - and again, a measure of the LIVE system rather than take it offline and spend several days doing static pack charge/discharge. To take a measure - I'm recording ah / volt to a fixed inverter cut-off of 48v(3.42v/cell) after the PV array power goes to 0 (no charging - discharge only). So far, I'm averaging 86ah per volt of drop from the low 52v(3.71v/cell) to 48v(3.42v/cell) range - but the top voltage varies day by day somewhat depending on the amount PV charging for the day. I'm predicting I'll see this "86ah/volt" go down to 80, 70, 60, 50... over time indicating battery capacity degredation. Do you have any thoughts on this and if so, would you be willing to share? Thank you sir.

@AdamWelchUK

5 жыл бұрын

I ran a set of lithium cells for a year - charging and using them every day. I tested the cells before and after that year and documented my results here... kzread.info/dash/bejne/hpmKl9SGgLOxkbA.html

the frequency was different, you used 100Hz, while the tester had 1000Hz

I am wondering why I got results way higher than what I have with my Miboxer C8 charger ????

@MrSummitville

4 жыл бұрын

What TEST Method did you do vs what TEST Method did the Miboxer C8 charger do?

Why author and some of you mix symbols and units? U=V, I=A, R=ohm, so in formula should be Uoc=something V instead Voc

👍

FYI, you will find that by attaching THOSE LEADS your reading will be +/- 2 Ohms Meausure the resistance of the leads, you'll see what i mean

@NicholasSouris

5 жыл бұрын

Agreed, your clips are providing some resistance

@nfsmith51

4 жыл бұрын

Yeah, but the resistance of the leads is negligible. I tried to measure mine, and even hooking both leads in series, the meter shows 0 ohms resistance, which is far from 2 ohms and infinitely far from -2 ohms. How they would ever indicate -2 ohms is beyond me, even if they were cooled by liquid nitrogen I don't see how you could imagine a -2 ohm resistance. In other words, using +/- is just flat wrong, and +2 ohms is unrealistically skeptical. Any kind of quality of the leads would be effectively 0 ohms.

@martinkuliza

4 жыл бұрын

@@nfsmith51 LOL...... i suppose i now have 2 options 1. i can explain this to you (But.. at the risk of you being argumentative) 2. i can let you figure it out given your response and the tone in your response, I'LL CHOOSE TO TAKE OPTION 2. let's leave it at that

@nfsmith51

4 жыл бұрын

@@martinkuliza What's to figure out? In theory, and to a negligible degree, in fact there will be some resistance, but it's not measurable with most meters and is conventionally ignored. Furthermore, unless you live in an alternate universe, it can never be negative. +2 ohms is entirely too much for any kind reasonable quality lead. If you have anything rational to say, go ahead, but stop attacking me personally.

@martinkuliza

4 жыл бұрын

@@nfsmith51 mate, Listen up YOU'RE AN IDIOT, ok now, that being the case, why should i continue this conversation let's reverse engineer this conversation, ok, Just so you can see why you're an idiot in your last comment you basically ended up by saying - that i shouldn't attack you personally FOR THIS YOU ARE AN IDIOT allow me to explain... 1. i have not attacked you at all, to say i have is stupid 2. if anything you have done it t me, but i didn't say anything , because it doesn't matter and you're just basically being a little princess now 3. look at how the conversation started..... fact is.... you responded to my comment the the op first and in that comment your tone was on the side of BULLSHIT and again.... i didn't say too much about it because it doesn't matter after that and given your tone i had 2 options i told you you'd be argumentative i also said LEAVE IT AT THAT, didn't i ? but you didn't so.. WHAT FUCKING DRUGS ARE ON when you say t me that i'm attacking you mate, you're an idiot, that's what it is. SIMPLE QUESTION Why couldn't you just leave it as it was ? WHY WAS THAT HARD ? now... you have yourself a shitfight. again.... since you're an idiot, i now choose to no longer continue this because there is only one road this can go down and i'm not interested in going down that road, OK you have taken a simple topic like resistance and turned it into a shit fight go away and think about it and rethink life, ok have a nice day

Well, I knew I wasn't going to be the only one to spot the flaw with your DC method i.e. you weren't allowing for the current to be affected by the internal resistor. However, here is the solution: Put an ammeter in series with your load resistance and that will give you the actual current flowing. Then measure the voltage across the *ammeter and resistor combined*. You can then calculate the real load resistor value as it will include the internal resistance of the ammeter. Then you can just plug all this into your formula as before, although you won't need to calculate the current as you'll already know it. :) But of course now you have a couple of commercial metering options so no need to pfaff around with formulas! Which one is more correct? Who cares? As long as you use the same instrument when grading a bunch of cells as it is the relative value that is truly important. Four days out from the Winter Solstice here...last few days have been amazingly almost wall-to-wall sunshine! I hope you have an outstanding Summer Solstice! :)

@MrSummitville

4 жыл бұрын

He does no't need a ammeter, because he measured the Voltage across the resistor while under load, therefore he can, and should have, calculated the actual amps flowing through the resistor ...

Hi Adam, you have to measure at a 1000Hz frequency not 100Hz, all cells data sheets that I've seen use a 1KHz sine wave frequency please re-run this video using your frequency generator at 1000Hz, that is where the discrepancies come into play. I would like to see the result of that compared to the store bought version it should then be MUCH closer..... Also if you look at the store bought resistance meter you'll actually see that it shows it's test frequency to be 1KHz, everyone harping on about how connections were made and assumptions upon the internal resistance being zero are completely missing this one VERY important factor and have evidently never completely read an 18650 data sheet to understand this, implore you sir to please redo this video to see the difference, thank you for your valuable input .

Don't not open link of battery internal resistor meter

Hi, in the second metode you used different frequencies, in the diy it was a 100Hz sine wave and in the comercial one it was a 1KHz lets suppose a sine wave This is probably the difference in the tests

@AdamWelchUK

5 жыл бұрын

That occurred to me in the edit. I’ll be sure to test out that theory. Cheers

@jonander6811

5 жыл бұрын

@@AdamWelchUK 👍

@TheEmbeddedHobbyist

5 жыл бұрын

I second that, if your using an AC component in the measurement then the frequency is going to play a part in the result.

E o mica gresala in calcul. Curentul prin rezistenta externa de 5 ohm, este 4.083 v / 5 ohm = 0.8166 A => Rezistenta interna este 0.11 V / 0.8166 A = 134.70 mohm

Great post, but way over my head lol.

@ponkuna

5 жыл бұрын

Arliss Young I second it. lol.

ADAM WELCH, //// Eric Johnson is saying that overdrive & fuzz guitar pedals that take batteries will sound different when using different types of 9volt batteries NON-Rechargeable 9 volt batteries like Alkaline, Carbon zinc, Lithium, Mercury because of the DC resistance of the 9 volt battery. The sound will be different if you use an Alkaline, Carbon zinc, Lithium, Mercury battery each will sound differently because of the batteries DC resistance /// 1.) Why does each 9 volt battery type Alkaline, Carbon zinc, Lithium, Mercury have a different battery DC resistance? Why different battery types will have a different DC resistance compared to another battery type, any reasons why? Yes I know that the batteries DC resistance will drop the 9volts when the guitar pedal is drawing more current but what Eric Johnson is saying is that 9 volt Alkaline, Carbon zinc, Lithium, Mercury will have a different DC resistance which will drop the 9 volt down differently when the overdrive or fuzz pedal is drawing more or less current.

It seems all the Lithium ion cell fanatics get 4.2 volts or close to it out of charged cells. I've been following a lot of them and experimenting with dozens of recycled notebook cells myself for the past year and have never gotten exactly 4.2 volts out of any of them. 4.10 4.15 4.12 4,25 are typical measurements. This is the real world

It's a little mistake in mind. The current through external resistance of 5 ohm is 4.083 v / 5 ohm = 0.8166 A => Internal resistance is 0.11 V / 0.8166 A = 134.70 mohm

@AdamWelchUK

5 жыл бұрын

There are a few little mistakes in the video but hope the main point is still valid. Thank you for taking the time to post - I appreciate it.

@nfsmith51

4 жыл бұрын

You should acknowledge this mistake in your original post. It's a significant gaff in theory, and it really does alter your main point. I agree with Emil's calculation.

@1959Berre

4 жыл бұрын

Do not simply calculate the current by dividing the open voltage by the resistance, you have to measure the current under load.

@MrSummitville

4 жыл бұрын

@@1959Berre No, we do not need to measure the actual amps under load, because we can calculate the actual amps under load by measuring the actual voltage under load...

I is not =4.193/5 , but 4.088/5

Incomprehensable due to the equations being much to small on screen. Go to a white board to get your message across.

Internal resistance of a battery is far from stable, it is related to STOC (state of charge) and temperature, and it can be very different for the same battery under different conditions. Besides, internal resistance is of no real importance for assessing the quality of a modern Li-ion battery, as it does not relate to its capacity to hold a charge. High C batteries can deliver a lot of power at high a current because of their low internal resistance, but that does not mean these batteries are able to hold a charge.

@MrSummitville

4 жыл бұрын

IR can be very useful ... When comparing cells of the same brand and model, any cells with extremely high IR should be excluded from the pack ...

So in conclusion: It is impossible to exactly measure the battery internal resistance 😁

*Hold up, say that again plz!!!*

battery IR is a bit like a dog chasing its own tail

You can't use this. By manufacturer specs internal resistance is measured at 1kHz and C/5 and C/10. So it actually is impedance no DC resistance

The true cost of salvaging batteries in time, life span and efficiency may be more than just buying new ones.

TOO COMPLICATED I DO MINE IN 2 LINES ALSO I THINK IT WRONG

what about leakage resistance. the resistance that makes the battery discharge just sitting there? please explain. America should make bodgy poor quality wrongly specked batteries and sell them cheap to china. tit for tat.

There's a whole field of EE over looked in this video. If you want to learn about battery performance start with a textbook not trinkets from China.

To see if the battery is good or not, you should have told us what the correct Internal Resistance should have been to compare it with the measured and calculated Internal Resistance.

I came to the praised tree with the bag, but the tree is at the level of a shrub. He knows a little more than an amateur.