Troubleshooting Battery Internal Resistance Measurements

[Tom.G]

The "LOAD VOLTS" graph in post #14 shows an error 0.07%! You can get that much variation just by wiggling an alligator clip on a contact, and I doubt your meter is that accurate. If the meter has 4-digit readout that is the ±1 digit uncertainty inherent in all digital measurements.

For the graph in post #1 R_CALC vs R_MEAS, two possibilities come to mind:

• the meter (or its leads) need calibration and/or repair
• the battery holder is defective

Try the experiment with one battery (cell), making contact directly to the battery without the battery holder.

You mentioned that you used the same meter to measure the current. The Current range of a digital ammeter has some low, non-zero, internal resistance, which would add to your effective load resistance. Has this been taken into account?

Beg, borrow, or steal a second meter to measure the current at the same time as the voltage measurement.

Please let us know what you find.

Cheers,
Tom

 

[sophiecentaur]

If you are actually doing the experiment yourself, in advance of the class, (you should) then I would recommend doing away with crocodile clips and probes. Replace them with screw terminals ('chocolate block' type connectors) and a proper switch to the battery (in a proper holder). If you compare that with the results of your croc clips and probes you may find sufficient difference to justify giving the class proper connectors.
There are several comments higher up about the time factor of an AA cell under high current. You need to find a 'sweet spot' for a current that's low enough to protect the cell but high enough to give a measurable Vr_i drop. Such an apparently simple class practical needs to be designed better than one might think and preparation is essential. I have seen (and suffered from) the results of winging it when the equipment / setup is just not adequate for the exercise. ("These experiments never work Sir") It goes with the territory.

 

[DaveE]

If you are actually doing the experiment yourself, in advance of the class, (you should) then I would recommend doing away with crocodile clips and probes. Replace them with screw terminals ('chocolate block' type connectors) and a proper switch to the battery (in a proper holder). If you compare that with the results of your croc clips and probes you may find sufficient difference to justify giving the class proper connectors.
There are several comments higher up about the time factor of an AA cell under high current. You need to find a 'sweet spot' for a current that's low enough to protect the cell but high enough to give a measurable Vr_i drop. Such an apparently simple class practical needs to be designed better than one might think and preparation is essential. I have seen (and suffered from) the results of winging it when the equipment / setup is just not adequate for the exercise. ("These experiments never work Sir") It goes with the territory.

Yes, this. Plus you should review "kelvin contacts" or "4-wire sensing" measurements. You don't want the load current flowing through any contacts that would create an extra voltage drop for your meter to measure. The data you are collecting is what your meter sees (i.e. inside the meter). This is critical for sensitive measurements.

 

[neilparker62]

If you are actually doing the experiment yourself, in advance of the class, (you should) then I would recommend doing away with crocodile clips and probes. Replace them with screw terminals ('chocolate block' type connectors) and a proper switch to the battery (in a proper holder). If you compare that with the results of your croc clips and probes you may find sufficient difference to justify giving the class proper connectors.
There are several comments higher up about the time factor of an AA cell under high current. You need to find a 'sweet spot' for a current that's low enough to protect the cell but high enough to give a measurable Vr_i drop. Such an apparently simple class practical needs to be designed better than one might think and preparation is essential. I have seen (and suffered from) the results of winging it when the equipment / setup is just not adequate for the exercise. ("These experiments never work Sir") It goes with the territory.

Thankyou - exactly my experience trying to do the prac myself ! You'd think it would be a relatively straight forward exercise but the numerous sources of error (due to poor contacts as well as the actual chemistry of the battery) make it quite a challenge to find the 'sweet spot' you mention.

Following is my latest attempt. I am using 2 AA cells in parallel and that is stabilising the readings somewhat at the expense of a more limited range of readings. Some will no doubt quibble that 3 of the 5 readings are 1501, 1503 and 1505 millivolts so right at the edge of the meter's resolution. I take Tom G's point about the battery holder but it's new, the springs are strong and overall it enables more solid connections than probes direct onto battery terminals.

As part of the brief is using a "home-made" resistor one pencil lead supplies 8.8 ohms while the other is used as a variable resistor (depending on where you attach the crocodile clip) enabling the 4 other readings.

 

[neilparker62]

 

[nasu]

Why are you using two batteries in parallel when the goal seems to be finding the internal resistance of one battery?

 

[neilparker62]

Why are you using two batteries in parallel when the goal seems to be finding the internal resistance of one battery?

To try and stabilise the load voltage readings. I am working on the assumption that if the batteries are identical I will end up with 1/2 the value of r (internal) for a single battery.

 

[neilparker62]

The "LOAD VOLTS" graph in post #14 shows an error 0.07%! You can get that much variation just by wiggling an alligator clip on a contact, and I doubt your meter is that accurate. If the meter has 4-digit readout that is the ±1 digit uncertainty inherent in all digital measurements.

You mentioned that you used the same meter to measure the current. The Current range of a digital ammeter has some low, non-zero, internal resistance, which would add to your effective load resistance. Has this been taken into account?

I

Yes - I'm aware the range of load voltages is very small - only about 25 millivolts in the last graph I posted.

I think the systematic error in resistance is indeed mainly due to ammeter resistance - many thanks for pointing that out. Will use the regression obtained offset value (y-intercept) to correct the measured resistance value.

 

[sophiecentaur]

Yes, this. Plus you should review "kelvin contacts" or "4-wire sensing" measurements.

Basically, any method is better than the crude method that the OP is suggesting. For a start, if you want to measure internal battery resistance than it is far better to use a low grade, cheap battery that actually has a poor internal resistance. A PP3 (9V) battery will give a voltage drop with lower, more sensible current values - matching the test to the available equipment.
This is one way to start getting some decent results. Basically, this sort of measurement should involve the DMM working over a range that uses all of its sig digits. If you work on the principle of a Bridge circuit, your DMM can be measuring the difference between a reference AA cell that's unloaded and which will keep its volts for days on end. You connect your DMM ( at maximum or appropriate sensitivity range) between the + terminals of the Test and Reference cells and see how that difference varies as you alter the current through the test cell. All that's necessary is to have the negative terminals of the two cells well connected.

Of course, if its necessary to be using a single meter for current and volts measurements then so be it and there's no way of being sure that the currents through the test cell is the same with and without the meter. But schools and colleges usually have more than just one instrument to use.

Digital meters are very handy but they actually tend to play hell with people's understanding of the circuits they are using. We just tick off the item on the syllabus and move on etc. etc.

 

[neilparker62]

EDIT:
Yeah, they are one and the same, I didn't catch that. If you rearrange $r = 2 \left( \frac{\mathcal{E}}{I} - R \right)$ with the substitution that $V_t = I R$ you get right back to the first equation...

Sorry for the detour.

No worries - we are all learning!

 

[DaveE]

To try and stabilise the load voltage readings. I am working on the assumption that if the batteries are identical I will end up with 1/2 the value of r (internal) for a single battery.

OK, it's really the parallel combination of the source resistances. Which, as you say, is 1/2 for 2 identical batteries.

But, the cell voltage should be stable. Any instability is most likely a problem with your test fixture. Fix that first, adding a second cell probably isn't addressing the root cause and might show up as variation in your impedance data. In a good test setup, I don't see how two cells is more stable than one.

 

[neilparker62]

OK, it's really the parallel combination of the source resistances. Which, as you say, is 1/2 for 2 identical batteries.

But, the cell voltage should be stable. Any instability is most likely a problem with your test fixture. Fix that first, adding a second cell probably isn't addressing the root cause and might show up as variation in your impedance data. In a good test setup, I don't see how two cells is more stable than one.

It's more stable because the current draw relative to the charge capacity (amp hours) of the pair of batteries is 1/2 as much. Ideally open circuit voltage does not change or only changes (decreases) very gradually. But if you are drawing a relatively large current "gradual" becomes tangibly "rapid". Try experimenting with a 4 ohm resistor and an AA cell and you'll see that.

We face a classic measurement conundrum. You want to drop more volts across the internal resistance so you increase the current. But in increasing the current, you lose stability of open circuit voltage. That's why an earlier post mentioned trying to find the 'sweet spot' where you are dropping reasonably measurable millivolts across the internal resistance but not affecting the open cct voltage too much.

 

[neilparker62]

Basically, any method is better than the crude method that the OP is suggesting. For a start, if you want to measure internal battery resistance than it is far better to use a low grade, cheap battery that actually has a poor internal resistance. A PP3 (9V) battery will give a voltage drop with lower, more sensible current values - matching the test to the available equipment.
This is one way to start getting some decent results. Basically, this sort of measurement should involve the DMM working over a range that uses all of its sig digits. If you work on the principle of a Bridge circuit, your DMM can be measuring the difference between a reference AA cell that's unloaded and which will keep its volts for days on end. You connect your DMM ( at maximum or appropriate sensitivity range) between the + terminals of the Test and Reference cells and see how that difference varies as you alter the current through the test cell. All that's necessary is to have the negative terminals of the two cells well connected.

Of course, if its necessary to be using a single meter for current and volts measurements then so be it and there's no way of being sure that the currents through the test cell is the same with and without the meter. But schools and colleges usually have more than just one instrument to use.

Digital meters are very handy but they actually tend to play hell with people's understanding of the circuits they are using. We just tick off the item on the syllabus and move on etc. etc.

I tried using a reference voltage and I can't say it made much difference. We're trying to measure millivolt changes and are up against the resolution limits of the meter. Whether I measure one voltage as 1500 millivolts and the other as 1490 or 0 and -10 makes no odds. It is what it is. Also if the instability I have mentioned is at the millivolt level, having extra digits (even if the meter was capable of it) would not make any difference.

Once again I'm going to recommend trying out the measurement and seeing what happens. I can post a video here if you like - my poor student was doing his best with a 10 ohm resistor and the load volts was drifting around all over the place. And I have every sympathy because I had exactly the same problem. Current readings also drifted but less drastically.

You hit the nail on the head with your observation in an earlier post that the trick is to find the "sweet spot" in terms of load resistance and hence current draw.

 

[neilparker62]

If you are actually doing the experiment yourself, in advance of the class, (you should) then I would recommend doing away with crocodile clips and probes. Replace them with screw terminals ('chocolate block' type connectors) and a proper switch to the battery (in a proper holder). If you compare that with the results of your croc clips and probes you may find sufficient difference to justify giving the class proper connectors.
There are several comments higher up about the time factor of an AA cell under high current. You need to find a 'sweet spot' for a current that's low enough to protect the cell but high enough to give a measurable Vr_i drop. Such an apparently simple class practical needs to be designed better than one might think and preparation is essential. I have seen (and suffered from) the results of winging it when the equipment / setup is just not adequate for the exercise. ("These experiments never work Sir") It goes with the territory.

I'll try to improve on the connections as per your suggestions above.

 

[DaveE]

It's more stable because the current draw relative to the charge capacity (amp hours) of the pair of batteries is 1/2 as much. Ideally open circuit voltage does not change or only changes (decreases) very gradually. But if you are drawing a relatively large current "gradual" becomes tangibly "rapid". Try experimenting with a 4 ohm resistor and an AA cell and you'll see that.

We face a classic measurement conundrum. You want to drop more volts across the internal resistance so you increase the current. But in increasing the current, you lose stability of open circuit voltage. That's why an earlier post mentioned trying to find the 'sweet spot' where you are dropping reasonably measurable millivolts across the internal resistance but not affecting the open cct voltage too much.

Oh, I see. That kind of stability!
But with two you are measuring half as much resistance, so to get the same voltage drop you'll need twice the load, right? Then the two batteries should discharge as fast as a single one. Or, if you prefer, each battery only sees half the load current.

 

[neilparker62]

Then you want to avoid significant discharge of the battery or temperature change. One way to do that is to keep elapsed time of the experiment short. Can you turn it on and do all the measurements in a few seconds?

I've tried to do this by setting up such that the load only connects at the same time as the multimeter probes are put on the battery terminals. Then the idea is it's a touch/release measurement.

An idea that's occurred to me (which is probably way 'overkill' admittedly) is to program a raspberry pi to operate a relay or transistor switch which switches in the load. Then use an analogue input pin to measure the voltage immediately and switch off again. I guess this would be a crude 'pulse test'.

 

[sophiecentaur]

There is a well known saying that the speed of a convoy is the speed of the slowest ship. There is a direct analogue of this, concerning experimental errors.
The parameters you seem to have chosen do not help with errors. Taking twice the current through two cells seems completely pointless.

An idea that's occurred to me (which is probably way 'overkill' admittedly) is to program a raspberry pi to operate a relay or transistor switch which switches in the load.

In the dim and distant nineteen sixties, I did the internal resistance experiment and it was fine. We used old fashioned Zinc Carbon cells and analogue lab volt and amp meters. Also we used a nice hefty Rheostat with good contacts. I don't have my old exercise book available but I remember drawing a graph, getting a straight (enough) line which yielded a 'good' answer; much the same as the rest of the class. The suggestions in my post higher up should give you a good chance of very believable and consistent results. Stick to tried and tested methods and leave the Raspberry Pi until you want to want to data-log the whole thing (and to give the students something more to manage not to understand).

 

[neilparker62]

In the dim and distant nineteen sixties, I did the internal resistance experiment and it was fine. We used old fashioned Zinc Carbon cells and analogue lab volt and amp meters. Also we used a nice hefty Rheostat with good contacts. I don't have my old exercise book available but I remember drawing a graph, getting a straight (enough) line which yielded a 'good' answer; much the same as the rest of the class. The suggestions in my post higher up should give you a good chance of very believable and consistent results. Stick to tried and tested methods and leave the Raspberry Pi until you want to want to data-log the whole thing (and to give the students something more to manage not to understand).

Yup - nothing much has changed it seems (other than the meters!). I would say they used a very flat battery if emf was 1.415 volts and internal resistance > 2 ohms though!

 

[sophiecentaur]

I can imagine all the contributors to this thread running through their memories of this sort of experiment, looking for a credibility gap to explain where the problem is coming from. Basic EE is often outside the comfort zone of teachers (and most normal human beings, aamof).

I can't help feeling that the likely value of r for an AA cell will be around 1Ohm. So the voltage drop should be about 1mv per mA. If the current load is between 0 and 100mA (very representative for an AA cell) then the PD would drop by up to 100mV. That's an ideal set of values for 'school' level equipment. Even the tackiest DMM should be able to measure that with four digits. 0.1V in 1.5V shouldn't be using the least sig digit so what is happening with the DMM?

There's something going on that's messing up the experiment, I'm sure.

The use of two cells just has to be a red herring. I can't remember ever using two cells (and the video doesn't either). Are the cells identical and in states of equal discharge? Chuck one away - or, as I suggested higher up, use one cell loaded and one unloaded and compare the PDs between them to give a voltage measurement of between Zero and 100mV. That's going to give really sharp resolution of V - much better than you actually need.

 

[neilparker62]

I think this prac needs a lot more careful investigation. For example in the video , they found an emf of 1.415 volts. I wouldn't mind betting that if you actually measured the cell emf it would be much closer to 1.5 volts. I have a whole lot of used Duracell AA batteries here and they still measure well over 1.5 volts.

Then they tried using a button battery (with not very good connections!) and guess what - the voltage readings fluctuated like crazy. Good winging it to say that it would be a good exercise for students to have to think about "how they would cope with the fluctuation."

My data in posts #39 and #40 gave a regression determined emf of 1518 mV whereas measured was 1521 mV or so. But right 'ball park' anyway. The regression determined internal resistance was 0.23(4) ohms so for a single cell about 0.47 ohms. I think 0.5 ohms is about the right 'ball park' number for typical internal resistance of an AA battery. Probably lower for "high drain" more expensive AA cells.

If I were to do (as indeed I attempted) the same experiment with just one AA cell, I would expand the range of measured load voltage values from 20mV to about 40mV. But then measurement becomes difficult because the load volts are changing continuously. That's why I tried using 2 AA cells in parallel. As I explained in a previous post , that stabilised the load voltage readings at the expense of a narrower range of values.

Other data and measurements I have taken suggest that unfortunately internal resistance of two cells in parallel is not 1/2 that of one cell. You can't win it seems!

 

[Tom.G]

How about:
1) use one cell
2) connect the meter to cell
3) document the reading
4) connect the load
5) after a timed 5 (or 10) seconds, read the meter and document it

 

[sophiecentaur]

As I explained in a previous post , that stabilised the load voltage readings at the expense of a narrower range of values.

If connecting two in parallel made a difference to the 'stability' of the volts then something is wrong with either the batteries (not identical) or the holder (dodgy contact). What technical reason could affect the stability if the setup was OK? If the cells in parallel don't show half the value of r then they cannot be matched; you need to investigate that and clear the problem before moving on.
Connecting the two - terminals together and then putting the meter across the two + terminals should show near zero V. Does it vary over time?

 

[neilparker62]

I have been experimenting on various Duracell Plus AA batteries. In light of the following information from that particular cell's datasheet, I rather wonder if I'm not trying to measure something that doesn't exist or is too small to be measurable ??

The relevant datasheet can be found here and it states 65 milli ohms "impedance" at 1 kHz. ie 0.065 ohms!

DC resistance varies with DOD (depth of discharge) as per graph below but seemingly does not exceed 0.15 ohms.

 

[DaveE]

I rather wonder if I'm not trying to measure something that doesn't exist

It definitely does exist.

or is too small to be measurable

That depends on your test setup. It is measurable if you try hard enough and spend enough time and money on good instruments and test fixtures.

 

[hutchphd]

In light of the following information from that particular cell's datasheet, I rather wonder if I'm not trying to measure something that doesn't exist

On one level it does not exist. The model of a battery as a perfect voltage source in series with an immutable resister certainly does not exist in real life.
But as an hueristic and at least semiquantitative tool it is obviously useful, particularly if you don't suck on the battery too hard. Consider it the Thevenin equivalent of a horrible bunch of temperature-dependent and memory-laden wetware.
For precision measure the optimal is nearly always some form null measure using a bridge. I am unaware but surely there is a standard "four wire" technique to do this measurement at the zero current level. Shall we invent one?

 

[neilparker62]

It definitely does exist.

That depends on your test setup. It is measurable if you try hard enough and spend enough time and money on good instruments and test fixtures.

Ok so here is my "fast fix" solution - please advise if you think it is valid.

1. Measure open cct volts
2. Set meter dial to battery test and measure closed cct volts

According to the meter manual , current draw for 1.5V battery test setting is approximately 50mA. 1.5/0.05 = 30 ohms. So I assume the battery test setting provides a 30 ohm load. Hence: $$ r_i = \frac{V_{oc} - V_{cc}}{I} = \Delta V \times \frac{30}{V_{cc}}. $$For a new Duracell AA battery I measure $V_{oc}=1596 \; mV$ and $V_{cc}=1565 \; mV$. Hence $\Delta V=31 \; mV$ and $r_i=\frac{30 \times 31}{1565} \approx 0.59 \; \Omega.$

So now if my assumption about the meter's battery test setting is correct, how does this result tie in (if at all) with the impedance data supplied on the Duracell datasheet ?

 

[DaveE]

Ok so here is my "fast fix" solution - please advise if you think it is valid.

1. Measure open cct volts
2. Set meter dial to battery test and measure closed cct volts

According to the meter manual , current draw for 1.5V battery test setting is approximately 50mA. 1.5/0.05 = 30 ohms. So I assume the battery test setting provides a 30 ohm load. Hence: $$ r_i = \frac{V_{oc} - V_{cc}}{I} = \Delta V \times \frac{30}{V_{cc}}. $$For a new Duracell AA battery I measure $V_{oc}=1596 \; mV$ and $V_{cc}=1565 \; mV$. Hence $\Delta V=31 \; mV$ and $r_i=\frac{30 \times 31}{1565} \approx 0.59 \; \Omega.$

So now if my assumption about the meter's battery test setting is correct, how does this result tie in (if at all) with the impedance data supplied on the Duracell datasheet ?

I can't assess this measurement for a very small resistance without seeing the details of your test set-up. The devil is in the details for this sort of measurement. My initial impression is that you are mostly measuring the resistance of your test leads and similar test fixture issues. Have you read and complied with the 4-wire sensing setup I referred to earlier?

 

[DaveE]

OK, I threw something together to demonstrate a 4-wire measurement.
1) A battery holder just for the mechanical part.
2) Scissors, Cu foil and a plastic insulator to make 2 "2-wire probes" Each strip of Cu is isolated from the other, but both can touch the battery terminal.

3) connect one set of terminals to the voltmeter and NOTHING ELSE. The other set is connected to a load resistor (~10Ω) and an ammeter all in series.

4) Because the load data drifts slowly, take a picture to collect the data easily.

My results:
No load, 1.52251V, 0A
w/ load, 1.49712V, 0.14442A
=> 176mΩ

PS: You'll want to take the w/ load data quickly or battery discharge will also lower the voltage and appear as greater resistance. This one was 2-3 seconds of discharge. You could follow up with another no load data point afterwards and do some creative compensation for that if the effect is big.

 

[Tom.G]

I have been experimenting on various Duracell Plus AA batteries.

Ahh! There is part of your problem. Duracell has one of (or THE) lowest internal impedances on the market... to the extent that many Smoke Alarms list it as one of the 2 or 3 preferred brands.

I strongly recommend that you try the experiment with a low cost, low quality Carbon/Zinc (LeLanche) cell. Much easier!

Cheers,
Tom

 

[neilparker62]

On one level it does not exist. The model of a battery as a perfect voltage source in series with an immutable resister certainly does not exist in real life.
But as an hueristic and at least semiquantitative tool it is obviously useful, particularly if you don't suck on the battery too hard. Consider it the Thevenin equivalent of a horrible bunch of temperature-dependent and memory-laden wetware.
For precision measure the optimal is nearly always some form null measure using a bridge. I am unaware but surely there is a standard "four wire" technique to do this measurement at the zero current level. Shall we invent one?

Happy to invent! As per post #54 we could perhaps use an op-amp to obtain a differential measurement between identical cells one unloaded and the other loaded.

 

[neilparker62]

My results:
No load, 1.52251V, 0A
w/ load, 1.49712V, 0.14442A
=> 176mΩ

PS: You'll want to take the w/ load data quickly or battery discharge will also lower the voltage and appear as greater resistance. This one was 2-3 seconds of discharge. You could follow up with another no load data point afterwards and do some creative compensation for that if the effect is big.

Cool - could you perhaps try with $\approx 30\;\Omega$ just for the sake of comparision with what I think the battery test setting on my multimeter is doing ?

 

[neilparker62]

Ahh! There is part of your problem. Duracell has one of (or THE) lowest internal impedances on the market... to the extent that many Smoke Alarms list it as one of the 2 or 3 preferred brands.

I strongly recommend that you try the experiment with a low cost, low quality Carbon/Zinc (LeLanche) cell. Much easier!

Cheers,
Tom

It looks to me like Duracell batteries manufactured here are not up to standard or we are getting cheap imports ?! Dave's measurement above indicates a terminal voltage drop of about 25mV (for a 10 ohm load) and that looks like it's for an old Duracell (could Dave confirm ?) since open cct voltage for a new battery is about 1.6 Volts. Meanwhile my new battery drops about 50 mV from 1595 mV down to 1545 mV across the same (or similar) load.

I would say these numbers are measurable enough - even with an ordinary multimeter such as I have.

Even Dave's measurement is out of spec against the data on the datsheet whereby internal resistance/impedance does not exceed 0,15 ohms. I wonder if they are mistaken in thinking that impedance measuring at 1kHz gives a 'fair' reflection of DC resistance ?

 

[DaveE]

that looks like it's for an old Duracell (could Dave confirm ?)

Yes, old and partly used, I think.

I wonder if they are mistaken in thinking that impedance measuring at 1kHz gives a 'fair' reflection of DC resistance

I would trust AC measurements more, not less. It removes some of the chemistry concerns that @Baluncore referred to by removing other DC and drift issues. AC synchronous detection at low frequency also gives you phase information to verify it's really resistance you are measuring. So just from a modelling point of view, the data is better. It can also be very sensitive. OTOH, it's harder to do.

Cool - could you perhaps try with ≈30Ω just for the sake of comparision with what I think the battery test setting on my multimeter is doing ?

Why don't you do it? My setup is easy to replicate.
My next step would be to compensate for discharge by doing this:
1) Measure no load.
2) Connect load and wait for 3 seconds.
3) Measure with load (take a photo) then wait for 3 seconds.
4) Disconnect load.
5) Measure no load.
6) Use the average value of the 2 no load measurements to compensate for discharge.

 

[sophiecentaur]

OK, I threw something together

I had to smile. What you 'threw together' was a lot more posh than the OP had available and the money you or your employer spent shows in the quality of results. Of course, it still takes skill to get the best out of good equipment - respect bro.

There are several other (cheaper) tricks to get more Sig Digs out of cheaper equipment - I would still go for my 'Bridge' arrangement for improving on the performance of a cheapo DMM by working with lower currents and measuring a few mV, rather than 1.5V.

 

[DaveE]

I had to smile. What you 'threw together' was a lot more posh than the OP had available and the money you or your employer spent shows in the quality of results. Of course, it still takes skill to get the best out of good equipment - respect bro.

There are several other (cheaper) tricks to get more Sig Digs out of cheaper equipment - I would still go for my 'Bridge' arrangement for improving on the performance of a cheapo DMM by working with lower currents and measuring a few mV, rather than 1.5V.

That was literally about 1/2 hour, not counting writing the post.

Employer? I haven't had one of those in at least 10 years. That's my backyard workshop.

That test equipment wasn't expensive. It's all old and used, rescued from my employers trash, bought from auctions, flea markets, or eBay. At least 1/3 of it was broken and repaired by me. That's partly why it's all older stuff. Back in the day the good equipment manufacturers would publish real service manuals with schematics, BOMs, test procedures, etc. I wouldn't buy anything cheap or broken if I couldn't first see a manual online. BTW, that's why I don't have a decent spectrum analyzer, they're really hard to find cheap and usually hard to repair. The really good ones came after HP stopped doing the service manuals.

Yes, I agree, there are better ways of doing this other than a simple DC ohms measurement. But maybe not in 1/2 hour. I was really just trying to show a 4-wire measurement setup, since I thought that message might not be sinking in. That part isn't rocket science. Also, I really don't care about the answer, I'm good with "small"; the battery resistance is small.