12v Battery Internal Resistance Test
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- čas přidán 30. 07. 2024
- Here I'm testing a bank of used 12v batteries to find out the internal resistance and if any are worth keeping or taking to the recycle centre.
Generally the lower the internal resistance a battery has the healthier it is, although that didn't seem the case with one particular battery I tested.. - Zábava
He did the math super fast in the vid. It was 0.1 amps then multiplied by 12v to get the wattage of 1.2. He uses that formula shown in the vid to get his readings. From what I've seen mostly used, Ohms(R) =(v1-v2) / (v1/ohms used for test) using the same drain method to get V2. The results should be in the milliohms for what I have read. The result should match closely to the battery specs, when new of course.
It does thx! I used your formula and tested my batteries and they came out at less than .5 ohms. I definitely feel confident in their potential! 👍Cheers
Thank you, a very informative and well laid out video, regards Doc Cox.
Wow never seen a better demonstration. Awesome
dude this is the best multimeter measuring internal resistance video ever!
thanks so much with every tiny details
great master
respect from Turkey
Very educational and informative, thanks.
Good video, interesting information.
You could also measure the current with the resistor in place. (R1 + R2) = V/I. You measure I, you measure V, R1 is the resistor value, and R2 is the unknown internal resistance. Solve for R2.
Yes, probably more accurate? but you need to do two measurements at the same time.... so need two meters.
Hi
Is a hi ressist is bad ? How mucu sold be a resistens of a 12 v battery ? Thenks.
Great video
nice video. What if we connect some batteries in series like two in series? Can you calculate the internal resistance of each cell for the series combination?
Thanks man for right point. So, is there any advice for use correct resistor-load compare measuring voltage ? Think etc. 12V lead-acid car batery, 3,8V Li-po, 6V SLA or 12V SLA . I see you use 120 Ohm .
battery #16 maybe have dead/shorted cell ?
Thank you 🙏 ✔️
Try measuring voltage from battery terminals. Not from resistor terminals! (your wires is thin)
i know I'm kinda randomly asking but do anybody know of a good place to watch new tv shows online?
@Kellen Lionel i would suggest FlixZone. Just google for it =)
2 ohms seems very high internal resistance for a sealed lead acid battery. I have used your method, but I could never work out the time at which I should take the measurement.
Yes you’re right 2ohms is a high reading for a lead acid battery. A brand new battery should have a reading of milli ohms. Those batteries were actually nearing their end of life and I was trying to filter out the ones to recycle by using the simple formula and a resistor. The time I took the measurement at was a bit guesswork I admit, but there should be a point at which the reading slows up dropping or flicking display between two figures e.g 12.3/ 12.4 say.
The ESR tends to get higher as the batteries age.
Thank You so much
Thank you
Nice video. So the important here is to choose the right watt not the ohm range. For example, 10w or higher could work. I wonder, in order to make an acurate voltage measurament, if could be better to use a ceramic resistor. I see you are using one with alluminium coat. Perhaps this kind of resistor are more efficiently so the voltage is more unestable and decrease constanly due the cooling coat.
It seems you got it wrong...Ohms and Watts of the loading resistor are interdependent. The point is that the load resistor should withstand the test conditions...
Is this possible to do with a car battery as well? What power resistor to use then? And where did that 120ohm number come from?
For car batteries you'd need a much lower resistance probably less than 8 ohms and rated for tens of watts (resistor will get very hot if left connected for long). But cars are outside and experience wider temperature swings than indoors, cold in particular will reduce the cold cranking amps so a cold night may be required to get a better picture (let the battery sit out there all night in the freezing cold first).
Very nice job on the video, Thanks.
Do you need to put a resistor on the battery to calculate the internal resistance. Can't you just put the meter in Ohms scale and find the internal resistance of the battery ?
Simply NO! Your meter can blow up! (Measuring resistance on a live circuit is prohibited)...
And internal resistance cannot be measured in this way...It looks like the ESR of capacitors...
Why you choose120ohm for 12v.for 6v battery tell me resistor value? .to take v2 reading is it fixed waiting time,because reading will continuously decrease
Hi, I chose the 120ohm resister as it was the one I had at the time. It’s not really that you have an ideal resister for a particular volt battery, more that you know the resister value to do the equation. It would work equally well with 100ohm etc, just be mindful of the wattage so it can handle the current. It could get quite hot otherwise. Yes you can be a lot more scientific about measuring the internal resistance but here I was just showing an easy way to highlight a bad battery. Basically any load (resister, bulb, motor etc) can be used as the load and you observe how much the voltage drops. An old, bad battery for instance will drop in working voltage more than another of the same voltage/capacity. Makes it easy to compare. Hope that helps
Yes, the measurement will continue to drop, as current is being used but just use a fixed amount of time, say 10seconds to measure each battery so it all stays constant.
TheLastPost thanks for quick response and good answer
newbie here, great vid, not familiar with all of this but i have 12V 7ah batteries I'd like to test however I only found those little wired resistors 1K and 5.5K (used on home alarm systems), can I use these to test the batteries - I quickly tried it but the Voltage did NOT drop at all with these resistors
Needs to be lower resistance, ideally a few hundred milliamps to an amp for batteries of this size to get any useful resolution. Try a 22 ohm 10 watt resistor or maybe a 12 ohm 15 watt, cheap ones from ebay are fine. What do you intend on using the battery for? A worn out battery may still be perfectly fine in low power applications or bench use but unsuitable for others.
Voltage should be measured at the battery terminals, The resistance for the load should factor in the leads that were used. The resistance of the load will change slightly due to heating. Outside of that close enough to get reference measurements compared to the others.
I don't understand why your formual and method is correct, when, in fact, commercial battery testers require a value for 'CA' ( ie, 'cranking amps ) for a particular battery, obtained via the 'data sheet' .. Also, would you explain where you got the value of your 120 ohm resister ?? is that 'industry standard'? (Eg: the Ancell BST200
tester )
Hi, I’m not sure on the commercial battery testers and how they work. I guess the need for the CA value and data sheets is because cranking a cold engine will have a huge draw on the battery.
These were only old 12v 2Ah batteries. I used the 120ohm power resistor as it was one I had to hand. As this was a rudimentary test I could have used a 12v car light bulb or similar to put a load on the battery. It’s easier to use a known value resistor as it makes the calculation easier
I use V1 - V2 = V3 V2 / R = I V3 / I = Ohms IR Ohms * 1000 = milliOhms .
12.66V1 - 12.36V2 = 0.30V3 voltage drop 12.36 v2 divided by 120ohms resistance or load = 0.10 I amp draw. 0.30V3 voltage drop divided by 0.10 I amp draw = 2.8691 Ohms internal resistance or 2869.13milliohms. I couldn't find the internal resistance of your exact battery but similar models like the UB1222 have an IR rating of 80milliOhms. Those batteries are showing there age.
Interesting.
Interchanging current, watts and voltage all in one video...glad I know the difference or it might've really confused me🤣
Great video and simple equations
I guess this is best for a constantly measuring of a batteries life or as you have done a number of the same batteries
I tried it with a 10w 63ohm resistor on some car batteries, they where all new and measure milli-ohms
You do know that the manufacturer's specified internal resistance is 0.17 Ohms? Quite the difference from 2.91 Ohms. A dead short on a fully charged battery should yield a 73AH drain.
Should if been in milliohms..
Assuming you don't have one shorted cell in that battery, the reason your good battery limits to 12.03V when charging is probably because it is trying to draw a high initial charge current and your converter/regulator has gone into current limit. Typically a good lead acid / SLA, battery with the low internal resistance will pass a high initial charge current and then increase in voltage slowly as it absorbs the charge. A bad battery gets up to voltage too quickly with a low current. I am surprised how high the resistance is for all of your good batteries, did you say about 2.5 ohms(?) I guess you are using them at very light load as with an internal resistance of about 2.5 ohms you be loosing 6v in the battery for a load of only 2.4 amps.
what is this, why is this confusing ?
We measure the value for V1 is the source or original value of battery with DMM, and then for V2 we measure through connected leads of DMM with battery to the ends of resistor and got V2 value. and R is 120 for all cells.
But you also write the Original Value of voltage in chart and then V1 is different. Why like that ?
V1 is actually the battery first value. Isn't it ?
yes
some time parameter to consider this measurement other wise this resistance will increase with time
Sir can I use electronic load for the same measurements instead of resistor
Of course you can!
No. An electronic load will vary its resistance to maintain the desired current that you set. You need a constant, known resistance to do the calcs.
@@paulpkae You can always use the formula δV/δI in order to calculate the internal resistance. This means that you can use two different values of resistance in order to define two different currents and then calculate the difference of the voltages over the difference of the currents. The battery, one way or another, doesn’t care if the load is purely resistive or electronic semiconductor. Where do you see any mistake?
@@user-pz6cx8zf2y Do you think the battery should be loaded for both measurements? e.g. measure at 0.1A and at 0.2A.
@@jamesbrown99991 If you use the classic formula (Ri=dV/dI), yes, this will give you a more accurate result. Most people use the non loaded minus the loaded voltage difference, which also works, but just because the non loaded voltage gives zero amps in the denominator of the fragment, the formula puts out a more inaccurate results (compared to two specific currents for the delta value of the denominator)...
What would your benchmark resistance be? 10? 7.5? Would all 12v batteries be measured on the same scale?
Well I was just comparing those used batteries against each other, and even figures of 2ohms is high tbh. For example a new 12v battery (of the type I showed here, 12v 2amp) will have a resistance that measures in milli ohms, say around 0.11mOhms. I was using this formula and a simple resistor to figure out the worst ones ready for recycling. Hope that helps
I use a car battery tester which ises this principal
In order to measure the internal resistance of these batteries, they must be fully charged before that measurement. Their internal resistance ranges between 25-35mOhm. An usual car 12V battery, when healthy, should range between 7 to 15mOhm.
12V/7.2AH lead acid battery with 12.9V open circuit voltage that I suspect to be gone bad, when connected to a 10W 1R resistor showed 4.3V till I can see on multi meter, the connecting wire to resistor and resistor were very hot, I had to remove the wire. So it was passing 8.9A. It shouldn't be right. Too much of current passing through it.
A 1Ohm resistor connected across a 12V battery draws 12A from the battery, or consumes 144W of power. In order for it to withstand such power, it should be made of at least three times that power, that is, nearly a 450W one, instead of the 10W you used.
The feeding cables should have been accordingly calculated for proper diameter or (better) cross section, securing low voltage drop for that current magnitude instead of their inadequacy causing overheating. Apart of all these above, there is nothing wrong with your test. In fact, using high current, makes it much more reliable as regards the health of the battery under test...
The point is to do it correctly and safely...
@@user-pz6cx8zf2y Wrong, the terminal voltage dropped from 12.9v to 4.3v the moment he connected the load. The high internal series resistance of the battery meant the actual combined load resistance was higher than 1 ohm. Also you don't need such a high wattage resistor for tests like this, three times higher is for product design where the resistor will be dissipating such power reliably for long periods of time, the 10w resistor getting overwatted for a few seconds during tests isn't the end of the world as long as it's resistance doesn't change too much, if in doubt put it in distilled water for cooling. Although I'd probably have used a 10 ohm 15w instead.
@@CamelCasee Where exactly am I wrong? My comment is based on the hypothesis that the battery under test is healthy and the calculations are accordingly done. Apart from that, it is obvious that the result is due to high internal resistance. What do you think that would happen during testing a healthy battery under the conditions you suggest? Have you ever seen the load resistors used inside the test instruments of the market which use the same method of dynamic testing of batteries? Do you know their wattage? They also are used for a few seconds per test as you say...but take a look at one of them before putting such comments...
@@user-pz6cx8zf2y He used a 120 ohm resistor.
its too high IR for a SLA batt.. i think something is wrong...
use over charge to de-sulfate them use current limiting resistor or light bulb 12 volt on one leg of input voltage from charger then 5 times run to 11 volts recharging as normal every time ,,its a way work best on flooded as one just add water but if they do not dry out being sealed one can crack sulfation away and reform themn
@Dodai Dibri just charge battery then run it dead but only to 10.5 --10 volts recharge it should get better capacity on say 5th time then charge it 24 hour to 2 weeks yep a bout it
THELASTPOST, 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.
1.2 amps = 1.2 watts ?
He did the math super fast in the vid. It was 0.1 amps then multiplied by 12v to get the wattage of 1.2. He uses that formula shown in the vid to get his readings. From what I've seen mostly used, Ohms(R) =(v1-v2) / (v1/ohms used for test) using the same drain method to get V2. The results should be in the milliohms for what I have read. The result should match closely to the battery specs, when new of course.
👍
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this is not a proper way to test internal resistor of battery
Yes, it is. Theoretically there is nothing wrong with it.
❤️🌹❤️
If the internal resistance was actually 3 Ohms all the time then it would be pulling 4A internally and be flat within 30 minutes when on the bench.
The internal resistance is more likely to be
Internal resistance appears in series with the cells, not parallel.
Nice vid but very sloppy with terminology and fine detail. This could be extremely confusing to "new to electronics" people and sight impaired people, otherwise well presented.
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