Thank you! Now I know how to measure internal resistance. I cannot trust intelligent chargers in showing me the internal resistance, because they fluctuate too much. This method is the best until now. You were so simple and clear. Good job!
One of the best informative videos on UA-cam, easy and simple language. Thanks mate for sharing this secret with us :) Wish you all the best. All the way from Arabian desert,,,
Still a valuable video 5 years later. However, you said back the charge off a bit to 4.1 instead of 4.2V, but didn't suggest a method for doing this. I am mortally wounded (like Nelson!) Thank you very much. I have saved this video!
Hi Tom, thank you very much for this very valuable explanation of this calculation, so simple yet so hard if one does not understand it, once again a great big thank you from Namibia. I do not understand why there are 23 dislikes on this video when it explains exactly what to do and how to do it, I do suppose though that there will always be a keyboard worrier out there somewhere to break down what we YT'ers try to help others with.....
Good explanation of something that most people don't even know about, this is very important when recycling used lithium cells, You should mark your cells with their capacity and I. R. to match them up to build packs, this helps prevent one of them dying prematurely
You Brits not only speak with a funny accent (no offense meant) but you also talk very fast. At the crucial moment after 8'00" I could not understand a thing. Congratulations, thanks!
Great clip, To be accurate you should account for your wire resistance. I estimate 4 to 7mOhm is your wire resistance , so your internal resistance is 8 to 11mOhm.
Very clear and helpful - thanks. A safety note though: at 5:27 you pick the cell up, with the croc' clips still attached and "popped it over there". If those clips should touch each other or touch a piece of metal,. they will short the cell which will cause large currents to flow, lots of heat to be produced and burst batteries/fire. Always disconnect at the battery end first.
Thank you ! Really good info on the slight under-charge to 4.1v, rather than 4.22v. I had been under the impression I would have to only charge to 80% to significantly extend cell life, but I much prefer the lesser tradeoff of taking it to a 97% charge and gaining 17% more capacity while still extending battery longevity. Much appreicated ! :-)
@@jamesrowlands8971Yes but you have to modify the electronics of your charger the only solution could be to check with multimetre every so often that the voltage does not rise above 19,5-20 volts
Nice video, also use thick wires (and use the same wires over) to limit the voltage drop across the wires (with that 4Amps current draw). Very nice battery holder.
Only 15m Ohm! that's a great battery you have there. My best battery is 133 m Ohm. Low internal resistance is a must for high current applications (>1.5 A). Thanks for the video, very enlighting.
Considering this problem with some details still yielded the same answer. When measuring voltage without a load resistor, we are assuming that Vo is approximately equal to emf since the internal resistance of the battery, r, is much smaller than the internal resistance of the voltmeter, R. Thus, R+r~=R and Vo=emf/(r+R)*R~=emf/R*R~=emf. Now on to measuring r part. It is important to select a load resistor RL of a small value not too much greater than r (which I think 1 ohm with proper POWER rating is very reasonable for RL such as in this video). Since r can no longer be ignored in this step, it must be included in the final computation as part of determining the current term as follows: let VL be the voltage across the RL, then VL=Vo/(r+RL)*RL, solving for r we obatin r=Vo/VL*RL-RL. So using your measurements r=4.10/4.04*1-1= 0.01485 ohms. Good video. Thank you.
Very helpful, been using my chargers measurement of internal resistance, but it keeps varying by quite a lot, I'll use you method as a base line, thanks.
that's nice video, but, i think it'll much better if you measure the open circuit voltage later. I mean, it's possible that the open circuit voltage being lower when you wait for the stable measurement of resistance voltage.
i should have paid attention in algebra! thank you for simplifying it. this will come in handy for ever. gotta exercise that gray matter every once in a while.
Great video thanks! Just a note: to add some precision you could have measured the resistor with the multimeter and used that instead of the number taken from the resistor.
Nice video but a little confusing because of the 1 ohm resistor. Two or 5 ohm resistor would have made the calculation a little harder but the video would have been easier to follow. Thanks again!
I've been looking at the various methodologies used to calculate internal resistance of a cell and this seems about as straight forward as it gets, so thank you. I've got a 4-wire battery box, not unlike the one you use here. I'll be interested to see how the answers derived using this method differ from the more complex approach espoused by other UA-camrs, using multiple measurements and analysis using a graph. Well thought out and well-delivered vid. :-)
crashed on my 1st flight of my new plane....lost all control at the 5 min mark and was in close range, but near the ground , and there was no chance to recover...........checked voltage on my very old 3S 2200mah lipo right after the crash and it read 3.80 per cell......this seemed OK but all indications before the crash were a loss of control ( brown out due to a LVC ) ....now searching for a simple way to check IR on all my LiPos .....thanks for the video..
It would be good to provide some guidance as to what current range is acceptable for obtaining an accurate/meaningful internal resistance for different cells/batteries. For example, you could tell folks why you are using a 1ohm resistor instead of a 100ohm or higher resistor. That would then let people decide if they could use an Xohm resistor for a given type of battery and/or battery voltage and still obtain reasonable information. Or...you could provide more disclaimers as to exactly what batteries/cells your 1ohm resistor can test accurately. For example, are you limiting your testing to just 1S1P 18650 lithium-ion cells?
nice video thank you, people should not mix this numbers with resistor meters we use, they use ac internal impedance to get the meter reading number you should divide the result with 0.0022 ohms, the meter should show 67ohm which is pretty bad for a battery, good 18650 should read around 20ohm on ac (resistor meter), dc resistor should be 44ohm or the way u say 4.5 ohm.
Good explanation of the matter. Here are couple of my notices on the measurement part: that power resistor has 5% precision, so it is slightly different than 1 Ohm. Little change in resistance will significantly affect the current running over that circuit. Another note: the wires used in the experiment has small diameter. It also creates a resistance when 4 A current runs. I would increase the wire diameter first and measure the resistance at the connection of the battery holder then run the measurement. In this case, the internal resistance of a battery would be a bit precise.
I got a question in terms of factoring in the resistance of the wires - would shorting out the the multimeter by touching both probes together give the resistance value of the internal circuitry and wires of the measuring device? So whatever value it shows when doing so - one would subtract this from the value on what you get with the resistance of the battery?
i think you should have compared your measure with the one given by the imax v2 , to test if they are equal and can trust the charger functionality,, right?
Great vid. So from what I can tell one can only calc the internal resistance of a battery using values obtained under load conditions? Is my observation accurate...
I have tested with 1 Ohm and 10 Ohms load, and the difference was only 11 mOhm more for the 1 Ohm load, I'm doing two sets of test for better accuracy. With a lower resistance battery, the difference was negligible, so am thinking the precision goes down with bad cells at high loads... edit: I wrote 100 Ohms, but it's 10 Ohms
Great vid.So we always need a load to determine the internal resistance of the battery? Secondly you mentioned the SCI enjoying high voltages , can you explain what you ment there to prolong the life of a battery. What does that mean in simpler terms? Cheers
You said "just make sure the power rating is correct" how do you determine that the power rating of the resistor is correct? I would love to use this method to get an idea of internal resistance of some Yinlong 40ah LTO cells. They are much larger in capacity than these vape cells but the nominal voltage is lower. How do I figure out what resistor I need to test them? I plan to have the cells charged at 2.3v for the test, and I don't care about the number itself as much as the results being similar (I just want to make sure none of them have a much higher result of the equation) Any help is much appreciated!! They are 66160 cells 2.3v
does the internal resistance consume power? Because I have 4pcs 32650 6A lifepo4 cells and after fully charging all of them, after a day, two goes down to 3.33v and one to 3.41v and the last one to 3.54v. Great video btw
Hey, I thought that you were going to show us an IR camera in high detail at different C rates, so we can see the resistance as thermal output from the caps and sides, so that we can have some idea of how thermal output an efficiency at different C rates. :)
lovely video nice and easy to understand thank you. i gotsa a question: on the resistor does the 20W rating mean anything over that would mean it starts to burn up? like 5A -->> Ploss = (I^2)*R so 25W it starts to get Hot? also the 5% uncertainty is concerning the resistance right? so youd have to calculate the extreme "edge" case value for both resistance cases of the load resistor to get the uncertainty in the internal resistance of the cell, you cant just take that 5% and apply it to the result you calculated right? many thanks
The resistor is used slightly outside the rating but it will survive a short test like this without problems. The uncertainty on the measurement is dominated by the +/-5% of the resistor plus +/-0.5% for the multimeter inaccuracy. So it is easy to improve the accuracy by measuring the real resistance of the 1ohm resistor plus its wires, and using that measuremed value instead of the nominal 1 Ohm. Doing it like that, the error reduces from +/-0.83 mOhm to +/-0.075 mOhm.
This is great and very comprehensive but I should add the resistance should be measured more accurately, not all resistances are made the same, but at the same time small resistance resistors are a pain to measure
for better precision, it would be better to do various measurements, with various set of resistors values, like 20, 10 and 1 Ohm, I use 10 and 1 Ohm and I get very close results
I think you need to measure the voltage after load , not before load, to see how much the voltage drop was. The voltage just before stopping the current is v0; and the voltage measured after a while ( about 30 secs.) you stop the current is v1.
if I don't know the resistor value, but I know that the discharge rate is at 1 amp this becomes simply subtracting Vr from Vo to get the internal resistance? I found an old e-bike pack and want to quickly figure out which cells are better and which are worse. I have a power bank case which accepts these cells. I test voltage across the leads with no load and with 1 amp load. Can it really be that simple?
Tom how much will the 5 % tolerance on your 1 ohm 20 watt +- 5% ceramic resistor effect the final internal resistance reading, i could do the math but was wondering if it would be worth the small extra cost for a precision wire wound audio 1 ohms 20 watt +- 1% ceramic resistor.
I think when the resistence we are looking for is too low, we need to consider the resistence of the wires, olligators pressure, contacts, etc. Maybe mesuring the voltage on the wires too. What do you think about ?
question: any resistor can be used instead of your 1.0 ohm 20w resistor you said, but if i use your Vo and Vr numbers and replace the 1,0 ohm 20w resistor with a 7,2 ohm 20w resistor i get a realy differen result, why is your calculation correct and not mine then?
with no load my cell was 3.77v with load it was 3.43v after the test without load it was 3.77v I checked the ampere before hand and it was 0.69 ampere so 0.34÷4.8 resistor = internal resistance of 0.51 ohm is this cell good for 1a load?
With the YR 1035 milliometer tester I measured the internal resistance and voltage of numerous rechargeable batteries that I had finished recharging. I wanted these conditions to see the parameters at full charge of all the batteries. I noticed that some of the same make and model have almost the same voltage (varies by a few millivolts) but different values of internal resistance (sometimes even double and triple). This is because some batteries are old, others partially new and others new (an older battery has a higher internal resistance). So the data I have are: internal resistance and voltage at full charge but without load connected to the battery. With these data with which calculation is it possible to know the real charge in Ampere available to the battery? To give an example: two identical batteries measure like this: batt1: 1,2806 volt and 0,0227 Ohm (or 22,7 mOhm) batt 2: 1.3204 volt and 0.0216 (or 21.6 mOhm) How many amps does one and the other battery have inside? Thank you
Hy, I got lifepo4 battery pack from all in one solar panel it's rating is 12.8 15Ah it has 4s 3P configuration but the problem is it's bms auto cut off its charging current when battery reached around 13 volt but it should cut of at 14 . 4 volt because fully charged li fepo4 cell voltage is 3.6 volt so 3.6 x 4 = 14.4 but why it cut of at 13 volt because if you do the maths opposite, for each cell there is just 3.2 volt average which is known as discharge cell but, bms thinks it's charged why?
@@crazytom and I noticed one more thing , it also pops when this 5000mah cell deep discharge below 3 volt at very light load . This is weard for me because at 3p cell of 5000mah rating I was drawing just 1amp dc motor as load but when it deep discharged it pops without being hot it happened 2 times
can you use this to tell if the batter is still good without using a a device that discharged the battery? meaning could we use the internal resistance to figure out current using ohns law with Vo=i Ro. where Ro is internal resistance?
I'm attempting this process right now to make a 3S balanced battery pack from 18650's, using Nichrome wire (3 ohms worth) as the resistor, but the voltage I get on my multimeter changes constantly making it hard to get any useful data. Is there a problem with the concept of using nichrome to do this? it's not coiled up, and it's not noticeably heating up.
Currently I hv a project for lion battery performance where I need to calculate IR and DCIR. is it correct procedure to test IR? Can you please make a video for DCIR testing. By the way thanks for the video
I just came back almost a year later because i forgot the formula. One of the most useful things i learned about batteries, thanks!
Thank you! Now I know how to measure internal resistance. I cannot trust intelligent chargers in showing me the internal resistance, because they fluctuate too much. This method is the best until now. You were so simple and clear. Good job!
4 year old video and best explanation IMO Thank You
dude that is the best internal resistance video ever
thank you so much
respect from Turkey
One of the best informative videos on UA-cam, easy and simple language.
Thanks mate for sharing this secret with us :)
Wish you all the best.
All the way from Arabian desert,,,
Thank you!!!! Using your example, i've managed to test my UPS batteries without an expensive battery tester!
Still a valuable video 5 years later. However, you said back the charge off a bit to 4.1 instead of 4.2V, but didn't suggest a method for doing this. I am mortally wounded (like Nelson!) Thank you very much. I have saved this video!
Gold dust explanation of internal resistance and bonus dont charge to 4.2 stop at 4.1v. Just what Ive been looking for thank you.
brilliant explenation plus...the calculator is the onky one which had been allowed using during my Master studies back then.
Hi Tom, thank you very much for this very valuable explanation of this calculation, so simple yet so hard if one does not understand it, once again a great big thank you from Namibia. I do not understand why there are 23 dislikes on this video when it explains exactly what to do and how to do it, I do suppose though that there will always be a keyboard worrier out there somewhere to break down what we YT'ers try to help others with.....
Maybe the 23 down thumbs were from guys who couldn't follow his super clear and precise explanation.
Good explanation of something that most people don't even know about, this is very important when recycling used lithium cells,
You should mark your cells with their capacity and I. R. to match them up to build packs, this helps prevent one of them dying prematurely
Couldn't agree more. I ensure that cells are within +/- 2mOhm for Radenite batteries.
radenite.com
You Brits not only speak with a funny accent (no offense meant) but you also talk very fast. At the crucial moment after 8'00" I could not understand a thing. Congratulations, thanks!
You could turn closed caption or reduce the playback speed.
Great clip,
To be accurate you should account for your wire resistance. I estimate 4 to 7mOhm is your wire resistance , so your internal resistance is 8 to 11mOhm.
yeah, those wires will show their resistance at those 4 amps of current
Very clear and helpful - thanks. A safety note though: at 5:27 you pick the cell up, with the croc' clips still attached and "popped it over there". If those clips should touch each other or touch a piece of metal,. they will short the cell which will cause large currents to flow, lots of heat to be produced and burst batteries/fire. Always disconnect at the battery end first.
+Chris Wesley. Yep. Set fire to my sleeve twice in the last few months when that happened!
@@crazytom was it like instant heat up , boiling and fire or explosions ? Want to know whats the first indicators :D
Thank you ! Really good info on the slight under-charge to 4.1v, rather than 4.22v. I had been under the impression I would have to only charge to 80% to significantly extend cell life, but I much prefer the lesser tradeoff of taking it to a 97% charge and gaining 17% more capacity while still extending battery longevity. Much appreicated ! :-)
With a charge range of about 1.4V, a .1V shortfall is about 93%.
@@crazytom is there a way to jerry rig existing chargers to not charge fully? For example for a power tool battery?
@@jamesrowlands8971Yes but you have to modify the electronics of your charger the only solution could be to check with multimetre every so often that the voltage does not rise above 19,5-20 volts
Wow, it really is that simple. And, helped solidify the need for the 4 wire measurement system. Very well explained!
Where can you get a 4 wire measurement system as shown in the video?
Nice video, also use thick wires (and use the same wires over) to limit the voltage drop across the wires (with that 4Amps current draw).
Very nice battery holder.
Only 15m Ohm! that's a great battery you have there. My best battery is 133 m Ohm. Low internal resistance is a must for high current applications (>1.5 A). Thanks for the video, very enlighting.
30A
Brilliant video - straight to the point and very well explained, thanks!
Exactly what I needed and simple to understand. Thanks.
Excellent clear and understandable explanation. Thank you.
viewing this in July 2021. brilliant! thank you.
Brilliantly explained. Thanks
Thanks from India 🇮🇳
"Probably health and safety nowadays."
"That's the measurement over with. I _told_ you it wasn't gonna be difficult."
I want to be your friend.
This video is golden even I little confused with the formula 😛
Considering this problem with some details still yielded the same answer. When measuring voltage without a load resistor, we are assuming that Vo is approximately equal to emf since the internal resistance of the battery, r, is much smaller than the internal resistance of the voltmeter, R. Thus, R+r~=R and Vo=emf/(r+R)*R~=emf/R*R~=emf. Now on to measuring r part. It is important to select a load resistor RL of a small value not too much greater than r (which I think 1 ohm with proper POWER rating is very reasonable for RL such as in this video). Since r can no longer be ignored in this step, it must be included in the final computation as part of determining the current term as follows: let VL be the voltage across the RL, then VL=Vo/(r+RL)*RL, solving for r we obatin r=Vo/VL*RL-RL. So using your measurements r=4.10/4.04*1-1= 0.01485 ohms. Good video. Thank you.
Very helpful, been using my chargers measurement of internal resistance, but it keeps varying by quite a lot, I'll use you method as a base line, thanks.
omg thx so much this explained exactly what i was trying to figure out
Excellent video I understand more completely now 👍🏻
Thank you. A little bit quick of a talking for a non-native english speaker but the most is understandable.
that's nice video, but, i think it'll much better if you measure the open circuit voltage later. I mean, it's possible that the open circuit voltage being lower when you wait for the stable measurement of resistance voltage.
i should have paid attention in algebra! thank you for simplifying it. this will come in handy for ever. gotta exercise that gray matter every once in a while.
Great video thanks! Just a note: to add some precision you could have measured the resistor with the multimeter and used that instead of the number taken from the resistor.
Good point.
I understand your position, but keep in mind that many meters people commonly use aren't very accurate at such a low resistance.
Nice video but a little confusing because of the 1 ohm resistor. Two or 5 ohm resistor would have made the calculation a little harder but the video would have been easier to follow. Thanks again!
That was incredibly easy to understand. Thank you!
Very useful information. Thank you!
Very good explanation
I've been looking at the various methodologies used to calculate internal resistance of a cell and this seems about as straight forward as it gets, so thank you. I've got a 4-wire battery box, not unlike the one you use here.
I'll be interested to see how the answers derived using this method differ from the more complex approach espoused by other UA-camrs, using multiple measurements and analysis using a graph.
Well thought out and well-delivered vid. :-)
once again a brilliant explanation....well done you! AG xx
Thanks AG. (Fiver's in the post)
Thank you for this, really helpful!
Great video. I have a calculator similar to that from 1977.
I remember that calculator its was my pride and joy at school in the 80's. I wish i'd learnt to use it properly :-(
Nicely explained thanks, and so easy and quick using a spreadsheet...thanks
They all play together. Voltage, Current, and Resistance are related. Good video. (+1)
I would have measured the actual resistance (±5%) and used 3+ digits for better accuracy, nice video
Good advice to check my resistor, but as far as i know they are quite acurate
crashed on my 1st flight of my new plane....lost all control at the 5 min mark and was in close range, but near the ground , and there was no chance to recover...........checked voltage on my very old 3S 2200mah lipo right after the crash and it read 3.80 per cell......this seemed OK but all indications before the crash were a loss of control ( brown out due to a LVC ) ....now searching for a simple way to check IR on all my LiPos .....thanks for the video..
Awesome video, man! Keep up the awesome work.
It would be good to provide some guidance as to what current range is acceptable for obtaining an accurate/meaningful internal resistance for different cells/batteries. For example, you could tell folks why you are using a 1ohm resistor instead of a 100ohm or higher resistor. That would then let people decide if they could use an Xohm resistor for a given type of battery and/or battery voltage and still obtain reasonable information.
Or...you could provide more disclaimers as to exactly what batteries/cells your 1ohm resistor can test accurately. For example, are you limiting your testing to just 1S1P 18650 lithium-ion cells?
Excellent tute, excellent presentation, excellent tips!
nice video thank you, people should not mix this numbers with resistor meters we use, they use ac internal impedance to get the meter reading number you should divide the result with 0.0022 ohms, the meter should show 67ohm which is pretty bad for a battery, good 18650 should read around 20ohm on ac (resistor meter), dc resistor should be 44ohm or the way u say 4.5 ohm.
Good explanation of the matter. Here are couple of my notices on the measurement part: that power resistor has 5% precision, so it is slightly different than 1 Ohm. Little change in resistance will significantly affect the current running over that circuit. Another note: the wires used in the experiment has small diameter. It also creates a resistance when 4 A current runs. I would increase the wire diameter first and measure the resistance at the connection of the battery holder then run the measurement. In this case, the internal resistance of a battery would be a bit precise.
Yes good points but it's illustrative.
Simple explanation. Thank you
Very well explained!
Thanks for making this so easy to understand. :)
Thanks you for sharing
I got a question in terms of factoring in the resistance of the wires - would shorting out the the multimeter by touching both probes together give the resistance value of the internal circuitry and wires of the measuring device? So whatever value it shows when doing so - one would subtract this from the value on what you get with the resistance of the battery?
What I didnt get from this was is after you measure the resistance will it get higher or lower if you battery is knackered?
i think you should have compared your measure with the one given by the imax v2 , to test if they are equal and can trust the charger functionality,, right?
Great vid. So from what I can tell one can only calc the internal resistance of a battery using values obtained under load conditions? Is my observation accurate...
Nice video. Very clear and well explained. Muchas gracias.
So what is a good resistance ?
Doesnt it also heavily depend on what load you use ? because im getting different results with the exact same battery.
I have tested with 1 Ohm and 10 Ohms load, and the difference was only 11 mOhm more for the 1 Ohm load, I'm doing two sets of test for better accuracy. With a lower resistance battery, the difference was negligible, so am thinking the precision goes down with bad cells at high loads...
edit: I wrote 100 Ohms, but it's 10 Ohms
Great vid.So we always need a load to determine the internal resistance of the battery? Secondly you mentioned the SCI enjoying high voltages , can you explain what you ment there to prolong the life of a battery. What does that mean in simpler terms? Cheers
0:59 C/2 is half C, not double C. Double C is called 2C. but otherwise a great video with good info.
Thanks. I didn't know that.
You said "just make sure the power rating is correct" how do you determine that the power rating of the resistor is correct? I would love to use this method to get an idea of internal resistance of some Yinlong 40ah LTO cells. They are much larger in capacity than these vape cells but the nominal voltage is lower. How do I figure out what resistor I need to test them? I plan to have the cells charged at 2.3v for the test, and I don't care about the number itself as much as the results being similar (I just want to make sure none of them have a much higher result of the equation) Any help is much appreciated!! They are 66160 cells 2.3v
does the internal resistance consume power? Because I have 4pcs 32650 6A lifepo4 cells and after fully charging all of them, after a day, two goes down to 3.33v and one to 3.41v and the last one to 3.54v.
Great video btw
"some of my friends in extreme vaping community are on quest for a cumulonimbus of vaping clouds"
@Rob Hunter Wow, way to be a prick. You sound quite ignorant.
Highly deserved like and sub
Hey, I thought that you were going to show us an IR camera in high detail at different C rates, so we can see the resistance as thermal output from the caps and sides, so that we can have some idea of how thermal output an efficiency at different C rates. :)
lovely video nice and easy to understand thank you.
i gotsa a question: on the resistor does the 20W rating mean anything over that would mean it starts to burn up? like 5A -->> Ploss = (I^2)*R so 25W it starts to get Hot?
also the 5% uncertainty is concerning the resistance right? so youd have to calculate the extreme "edge" case value for both resistance cases of the load resistor to get the uncertainty in the internal resistance of the cell, you cant just take that 5% and apply it to the result you calculated right?
many thanks
The resistor is used slightly outside the rating but it will survive a short test like this without problems.
The uncertainty on the measurement is dominated by the +/-5% of the resistor plus +/-0.5% for the multimeter inaccuracy. So it is easy to improve the accuracy by measuring the real resistance of the 1ohm resistor plus its wires, and using that measuremed value instead of the nominal 1 Ohm. Doing it like that, the error reduces from +/-0.83 mOhm to +/-0.075 mOhm.
What's its working theory electronically can you plz explain ?
Really helpful thank you
This is great and very comprehensive but I should add the resistance should be measured more accurately, not all resistances are made the same, but at the same time small resistance resistors are a pain to measure
for better precision, it would be better to do various measurements, with various set of resistors values, like 20, 10 and 1 Ohm, I use 10 and 1 Ohm and I get very close results
I think you need to measure the voltage after load , not before load, to see how much the voltage drop was. The voltage just before stopping the current is v0; and the voltage measured after a while ( about 30 secs.) you stop the current is v1.
Excellent video! Thanks for explaining that issue.
GOOD EXPAINATION !!!! THANK YOU.
thank you
and thank you
and thank you
and thank you
if I don't know the resistor value, but I know that the discharge rate is at 1 amp this becomes simply subtracting Vr from Vo to get the internal resistance?
I found an old e-bike pack and want to quickly figure out which cells are better and which are worse.
I have a power bank case which accepts these cells. I test voltage across the leads with no load and with 1 amp load.
Can it really be that simple?
Tom how much will the 5 % tolerance on your 1 ohm 20 watt +- 5% ceramic resistor effect the final internal resistance reading, i could do the math but was wondering if it would be worth the small extra cost for a precision wire wound audio 1 ohms 20 watt +- 1% ceramic resistor.
Awesome video, thank you!
I think when the resistence we are looking for is too low, we need to consider the resistence of the wires, olligators pressure, contacts, etc. Maybe mesuring the voltage on the wires too. What do you think about ?
Do I need to use a 20W resistor or will any resistor do the job? I have a very large battery, does that make a difference?
Compute the Wattage dissipated by the resistor and then use a resistor rated, at least, double that ( for safety ). The resistor will get HOT !!!
Since resistor will be used for short period of time you can use any, i use 5w and it only gets warm when testing
@@Boz1211111 What size battery?
@@yehudagoldberg6400 18650
@@Boz1211111 I want to test 100 AH LFP cells.
question: any resistor can be used instead of your 1.0 ohm 20w resistor you said, but if i use your Vo and Vr numbers and replace the 1,0 ohm 20w resistor with a 7,2 ohm 20w resistor i get a realy differen result, why is your calculation correct and not mine then?
with no load my cell was 3.77v with load it was 3.43v after the test without load it was 3.77v I checked the ampere before hand and it was 0.69 ampere
so 0.34÷4.8 resistor = internal resistance of 0.51 ohm is this cell good for 1a load?
what is this battery holder called and where i can find online in India
excellent, thank you very much..
Isnt 15mΩ really low? I'm testing some recycled 18650 and I'm getting results between 115 and 200 using a LiPo charger IR mesurement
Yeah he tested good high drain cell
You meant "milli Ohms" at the end of the video right? Not "milli amps".
Nice video.
With the YR 1035 milliometer tester I measured the internal resistance and voltage of numerous rechargeable batteries that I had finished recharging. I wanted these conditions to see the parameters at full charge of all the batteries. I noticed that some of the same make and model have almost the same voltage (varies by a few millivolts) but different values of internal resistance (sometimes even double and triple). This is because some batteries are old, others partially new and others new (an older battery has a higher internal resistance).
So the data I have are: internal resistance and voltage at full charge but without load connected to the battery. With these data with which calculation is it possible to know the real charge in Ampere available to the battery? To give an example: two identical batteries measure like this:
batt1: 1,2806 volt and 0,0227 Ohm (or 22,7 mOhm)
batt 2: 1.3204 volt and 0.0216 (or 21.6 mOhm)
How many amps does one and the other battery have inside? Thank you
Thanks for the video
Thank you for your video sir, i was thinking on doing the same and then i saw this and made it even more clear
Hy, I got lifepo4 battery pack from all in one solar panel it's rating is 12.8 15Ah it has 4s 3P configuration but the problem is it's bms auto cut off its charging current when battery reached around 13 volt but it should cut of at 14 . 4 volt because fully charged li fepo4 cell voltage is 3.6 volt so 3.6 x 4 = 14.4 but why it cut of at 13 volt because if you do the maths opposite, for each cell there is just 3.2 volt average which is known as discharge cell but, bms thinks it's charged why?
One or more cells over-charged.
@@crazytom and I noticed one more thing , it also pops when this 5000mah cell deep discharge below 3 volt at very light load . This is weard for me because at 3p cell of 5000mah rating I was drawing just 1amp dc motor as load but when it deep discharged it pops without being hot it happened 2 times
how to test 18650 battery total capacity without lowering the cut off voltage too much?
Hi nice video,i bought some 18650 online and ahen i tested the IR on my Liitokala most of them were between 38 and 46 can i still these?
Sure. They will most likely be 2C cells meaning the max current draw is 2 x Ampere Capacity/hour.
Im using those for some time and they seem like very decent cells, tho mine are 60mR 3Ah
can you use this to tell if the batter is still good without using a a device that discharged the battery? meaning could we use the internal resistance to figure out current using ohns law with Vo=i Ro. where Ro is internal resistance?
thank you. very helpful
thanks maaan helped a lot
I'm attempting this process right now to make a 3S balanced battery pack from 18650's, using Nichrome wire (3 ohms worth) as the resistor, but the voltage I get on my multimeter changes constantly making it hard to get any useful data. Is there a problem with the concept of using nichrome to do this? it's not coiled up, and it's not noticeably heating up.
Nevermind I was just getting inconsistent contact with my multimeter probe against the battery! battery holder makes all the difference!
Currently I hv a project for lion battery performance where I need to calculate IR and DCIR. is it correct procedure to test IR? Can you please make a video for DCIR testing. By the way thanks for the video
15 mOhm it's too low for 18650 Cells mean more than 200Amps can pass have you tested and compared it with ir Meter for real test ?