Effect of Burden Voltage of a Multimeter for battery capacity measurement - Page 1
Forum section: electronics / beginners

author: tonythetonguc
author_rank: Newbie
timestamp: January 28, 2023, 11:38:55 am
content: Hi Everybody, as a newbie i have a small question/confirmation that may sound funny for most of you. I am not an technical electrician nor electrical engineer, I am just an engineer who has tiny electric knowlage from High-School lectures. Let's get in to the question:First of all, I am planing to measure battery capacities (AA rechargable or non-rechargebles on the market) with using 2 multimeters with computer logging capability (UT61+ series). I will add an electric load and I will measure the voltage and current simultaneously with 2 multimeters and finally i will calculate the capacity.When I was searching for multimeters, I heard about "Burden Voltage" issues of MMs which I was suspect previously cuz' I knew that a DMM measures the current by adding a resistance on the circuit and measures the potential difference from this shunt resistance. Some of comments, it was told that U61E has too much of burden voltage, so i digged a bit more and in an individual comment it was told that this would cause wrong current measurement cuz DMM would dismiss (not take in to account) this shunt resistance and will effect your measurement (also the total Current flow). So briefly in that forum it was told that especially low current measurements will be faulty. And i asked this user with Direct Message he answered that my measurements will be falutly if use about 100 - 200 mA current drain from batteries. I am here to clarify this point, vuz according to my knowledge he/she was wrong.As far as I know, multimeters add a resistance on the circuit but measures the final current which is a valid value for the total current flow throught the circuit. Let's analyse it with an example:So when I add an electric load to the circuit (For example 1.2 Ohms) and If let's say I have a 1.2 Volt Ni-MH source. In this curciut without a MM i will have 1.0 Ampere of current. When i add a multimeter in seires with 0,1 ohm shunt resistance at mA measurements (I know they are lower but i gave a value for easier calculation) the total resistance in the circuit will be 1,3 Ohm, So my final and real current will be 0,923 A through the circuit and the MM will measure this value (0,923 A), So in this case my measurement will be correct, what is currrent drained from the battery will be measured by the MM correctly, only the difference the Current that is drained will not be as I planned in the original circuit.But the guy i mentioned above told that The MM will not take take in to account this shunt resistance drop and measure 1A of current in this circuit. When i said how it is possible, he added "Actually it measures the real current (0,923A) but removes the effect of shunt by calculation and shows you 1A to obtain the circuits real current without the multimeter. And this confused me actually.If he is right my measurements will be false (The total circuit will drain 0,923 A but the Multimeter will show 1 A )So, what is the reality?

author: wasedadoc
author_rank: Super Contributor
timestamp: January 28, 2023, 12:04:25 pm
content: Quote from: tonythetonguc on January 28, 2023, 11:38:55 amHe is not right. You are correct that the meter will ideally show 0.923. (In practice it is likely to be slightly different due to meter tolerance error.)The difference between 0.923 and 1.000 depends on the other items (the voltage source and the load) in the circuit. The meter has no way of knowing what parameters those other items have so it it impossible for the meter to calculate a correction.

author: BeBuLamar
author_rank: Super Contributor
timestamp: January 28, 2023, 02:42:12 pm
content: Your meter should show 0.923 A if there is no error (that is if the current measurement is specified as no error). The meter would show that actual current flowing thru the meter. If you remove the meter the current would be 1.0 A and that is the effect of burden voltage. When you insert the meter the current is less that it would have if you don't have the meter but the meter display is the actual current.Yes the guy is wrong. There is no way the meter can compensate for the burden voltage.

author: nightfire
author_rank: Frequent Contributor
timestamp: January 28, 2023, 03:39:53 pm
content: Your multimeter is adding a certain value of resistance to that simple circuit, therefore causing some voltage drop. (Principle of the burden)So there are the following options you have in checking stuff out:1) You go with your resistor you already have, maybe a special model with high precision or thermal value, so you do not want to buy a new one. Then you get the 0.923 Amps in your circuit as described. You then measure the battery voltage at the end of the battery and everything is ok.2) If you want to get an initial current of 1 A, buy another resistor to whether replace the one you (already or maybe hypothetically) got or put a higher value in paralell. But due to declining voltage over the course of discharge of the battery, the voltage will go down and therefore also the current, if resistance remains constant3) You use an electronic load that does constant current discharge and therefore changes resistance accordingly to the voltage presented to achieve a constant current flow.

author: rstofer
author_rank: Super Contributor
timestamp: January 28, 2023, 05:53:21 pm
content: The shunt resistor also changes value with heating.$10 for a purpose built tester for various chemistries:https://www.amazon.com/Digital-Battery-Capacity-Voltage-Controller/dp/B07RNPHJN6/ref=asc_df_B07RNPHJN6I don't have one so I don't know how well it works.

author: wasedadoc
author_rank: Super Contributor
timestamp: January 28, 2023, 05:59:59 pm
content: Quote from: rstofer on January 28, 2023, 05:53:21 pmThat gives anestimateofpercentagecapacity simply based on cell voltage. It doesnotmeasure or even estimateabsolutecapacity. The only parameter it measures is voltage. Not current. Not time.

author: tonythetonguc
author_rank: Newbie
timestamp: January 28, 2023, 06:36:01 pm
content: @wasedadoc@BeBuLamarThanks, I wrote the values for a theoritical example, in order to describe the situation. Values don't matter, so my DMM will read the real current and i will have the correct result for the battery capacity...@nightfireThanks, I gave resistor sample to have a theoritical, My actual plan is to use a CC electric load as you recommended on 3rd item of your post. I will adjust the current rating depending on the battery specs. for rechargeble batteries I will try in various current values like 0.2 - 0.3 - 0.5A etc. Like it was done in web site "http://lygte-info.dk/"@rstoferActually i want to log the data to computer, this device is too simple for my purpose. I saw some electronic loads in local resellers which can log to computer but they are all out of stock and expensive. There are some in aliexpress there are some but seems untrusty, if you can recommend one which has computer logging feature with a reasonable price, I would be happy.

author: wasedadoc
author_rank: Super Contributor
timestamp: January 28, 2023, 06:43:06 pm
content: Quote from: tonythetonguc on January 28, 2023, 06:36:01 pmYou do not require logging ability on the electronic load. Rather, you want to log the battery voltage and that can be done with a multimeter with logging ability. Usually just some output to a PC which does the actual logging.

author: nightfire
author_rank: Frequent Contributor
timestamp: January 28, 2023, 06:49:47 pm
content: Logging: If you have t wo Multimeters that can be interfaced to PC they can do the logging. The Software test Controller in a separate threat mentioned here in the forum might help here, or some stuff like "Sigrok" as a different project.I myself use a Korad KEL102 clone, that has been rebadged by Stamos soldering and sold for 30$ cheaper. Don't know if this fits your budget, though.Verifying if a DMM calculates current correctly: Measure with a second DMM the resistance of the shunt of the first multimeter, then let it measure the discharge current, and with DMM#2 you check the voltage drop across the shunt and can do your own calculation with U=RI.

author: tonythetonguc
author_rank: Newbie
timestamp: January 28, 2023, 07:43:38 pm
content: Quote from: wasedadoc on January 28, 2023, 06:43:06 pmI know. I already have an electronic load which cannot log data."Rather, you want to log the battery voltage and that can be done with a multimeter with logging ability."This is what i will do exactly as i wrote on the first post"First of all, I am planing to measure battery capacities (AA rechargable or non-rechargebles on the market) with using 2 multimeters with computer logging capability (UT61+ series)"Also my existing electronic load is able to measure capacity but very untrusty. Even it doesn't have additional inlet (or probes / or cables) for Voltage measurement. You can only connect the battery to the load and it measures the battery voltage from the cables that carry the circuit current. They have made an update in the newer models of this chineese load but i don't want to pay more fot this untrsuty and shitty things.I have an E. load similar to this one:https://aliexpress.com/item/32821877897.htmlNewer versions have separate wires for measuring and current flow.https://aliexpress.com/item/4001153199791.html

author: tonythetonguc
author_rank: Newbie
timestamp: January 28, 2023, 07:59:12 pm
content: Quote from: nightfire on January 28, 2023, 06:49:47 pmThis is what i would do exactly but with the default manufacturer's software, also thanks for advice, I was searching exactly about Sigrok and also the test controller software by lygte-info.dk.As far as i see, Sigrok has no support for 61 PLUS series yet, but it supports the older versions (without PLUS). Hope will get support for plus series soon.