Troubleshooting Battery Internal Resistance Measurements

[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.

 

[sophiecentaur]

It's all old and used,

So are we all. But you are clearly a capable experimenter and that helps a lot when finding a pathway through to good results.
The secret of this is not to try and see how a top of the range cell works but to use the cheapest and cheerfullest 'market stall' cell.

 

[neilparker62]

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?

The test 'set-up' is about as simple as it gets. You measure open circuit volts on the meter's DC volts setting as usual. Then you change to the "battery test" setting and measure again. The voltage reading is lower than the open circuit volts and I'm assuming that's because the meter is (internally) applying a load across the battery terminals. Based on the manual indicating a current draw of "approximately 50 milli amps", I'm presuming the load is 30 ohms. Have sent an email to the meter manufacturer asking about that so will wait to hear.

 

[sophiecentaur]

You measure open circuit volts on the meter's DC volts setting as usual.

@DaveE 's question still needs to be answered, though. Where, in the circuit, do you actually measure the Volts? You have to have the Voltmeter connections directly across the battery to eliminate any other drops.

 

[neilparker62]

Using the battery test facility on the meter does not require any external circuit. You just put the probes directly onto the battery terminals.

 

[sophiecentaur]

Using the battery test facility on the meter does not require any external circuit. You just put the probes directly onto the battery terminals.

The system uses only two terminals. ( I assume?) That means the Volt measurement will include an IR drop across contacts and leads. So you won’t do as well as a proper four terminal method.
Very convenient though so we can’t knock it completely.

 

[DaveE]

Using the battery test facility on the meter does not require any external circuit. You just put the probes directly onto the battery terminals.

OK then, try this: Measure the battery with one probe on each terminal. Then measure an offset, or "zero", with each probe on the same terminal. Subtract those two measurements. I'm not sure your meter will do this well, but it's worth a shot.

 

[sophiecentaur]

with each probe on the same terminal.

Can this work? The battery is no longer a source of current so you'd just be measuring 0/0. No? or have I got the idea wrong?

 

[DaveE]

Can this work? The battery is no longer a source of current so you'd just be measuring 0/0. No? or have I got the idea wrong?

Yes, you're probably correct that the EMF comes from the battery. Duh... LOL.

 

[neilparker62]

The system uses only two terminals. ( I assume?) That means the Volt measurement will include an IR drop across contacts and leads. So you won’t do as well as a proper four terminal method.
Very convenient though so we can’t knock it completely.

From what I've read online, the key point about 4-wire measurement is not about "4 wires" per se but rather about ensuring that the voltage measurement probes are placed directly across the resistance being measured.

You are correct in pointing out that this is NOT the case when using battery test mode since the probes are carrying the load current going through the (presumed) 30 ohm load resistance within the meter. And hence there's a $2IR_{lead}$ voltage drop.

However it does now beg the question, was there anything particularly wrong with my previous "one meter method" of measuring first battery emf , then load voltage and finally load current? Since here (for the voltage readings) I was indeed placing the probes "directly across the resistance being measured" ie directly onto the battery terminals. And load current went through a separate path to the external resistance.

 

[sophiecentaur]

And load current went through a separate path to the external resistance.

If, as you suggest, you were swapping the same meter for volt and current measurements, the resistance of the meter would be affecting the current. There's more to this than meets the eye.

 

[neilparker62]

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.

This advice has taken some time to 'register' properly!

I would say the ammeter resistance is the major component of the resistance 'offset' (y-intercept) in my graph of $R_{calc}\;vs\;R_{measured}$. But I don't think either leads nor ammeter resistance matter too much unless I attempt obtaining load volts by using IR rather than the measured value.

However I probably do indeed need to "beg / borrow /steal" (as you put it) another meter to ensure that load voltage is measured concurrently with load current. Load volts measured in a circuit without the ammeter could be significantly different from load volts measured in a circuit with the ammeter on account of ammeter resistance ( +- 2 ohms). All the more so if load resistance values are of the order of $10 \Omega$.

I look forward to better measurements - if you hear of burglaries at DVM shops you'll know who is the prime suspect

 

[sophiecentaur]

I would say the ammeter resistance is the major component of the resistance 'offset' (y-intercept) in my graph of RcalcvsRmeasured.

If you could find the value of the internal Ammeter shunt resistance then you could replace the meter with a resistor of that value when measuring the volts and more or less eliminate the problem. I would suspect that, on the A, mA ranges, the display will be showing directly measured Volts, mV etc. so no scaling of what you read; i.e to measure 200mA, if the meter uses the 200mV range then the shunt resistor would be 1Ω . This random link I found shows the sort of internal shunt resistor found for a particular DMM. You could be dealing with 100mΩ for a 1A current range but 2Ω for lower currents. It would all depend on the number of digits and accuracy of the meter.

HAha - catch 22 says that you could measure that resistance if only you had a second meter!

 

[sophiecentaur]

@neilparker62 I have two decent (mid range cost) DMMs and I measured the resistance of one with the other. For mA current measurement, the resistance was 2 Ohms and for higher currents, 0.1 Ohms. For cheaper meters the resistance would be greater to bring the voltage drop to something measurable with a few sig digs.

 

[neilparker62]

@neilparker62 I have two decent (mid range cost) DMMs and I measured the resistance of one with the other. For mA current measurement, the resistance was 2 Ohms and for higher currents, 0.1 Ohms. For cheaper meters the resistance would be greater to bring the voltage drop to something measurable with a few sig digs.

Thanks - that lines up with the regression analysis on $R_{calc}$ vs $R_{measured}$. Indicated about 1.9 Ohms. I also measured ammeter resistance using my brother's fluke and the result was again about 2 Ohms.

 

[neilparker62]

So after all the helpful discussion in this thread, I am finally starting to get some (hopefully) sensible data and results from this experiment. Setup is as per following pic. The first pencil supplies a minimum load of about 13 ohms whilst the second exposed lead enables a set of readings (volts / amps) to be taken simply by changing the point of attachment of the red crocodile clip.

I hope my understanding of 4-wire measurement has improved somewhat. I could not understand the reason for Dave's insulated copper strips but having initially connected negative lead of load circuit to the top of the crocodile clip attaching voltmeter to battery negative , I finally realized what Dave was on about and reconnected as follows:

The circuit is completed by touching the ammeter positive probe onto the terminal at right. I hope it is clear that load leads and voltmeter leads are insulated from each other right up to point of mutual contact on the battery terminals.

Result (from one of many datasets)

Many thanks for all the tips - have learned a hugh amount about the importance of taking into account various small (usually ignored) resistances such as ammeter resistance, lead resistance etc. And of course about 4-wire measurement in general.

 

[Tom.G]

One other variable that has not yet been mentioned:

Carbon has a Negative temperature coefficient of resistance, -4.8x10^-4, not enough to worry about in this test unless it starts glowing red!

Cheers,
Tom

 

[neilparker62]

One other variable that has not yet been mentioned:

Carbon has a Negative temperature coefficient of resistance, -4.8x10^-4, not enough to worry about in this test unless it starts glowing red!

Cheers,
Tom

Actually I did look at that coefficient and concluded it would indeed not be a factor. Not so sure about the temperature coefficient of battery internal resistance though!

 

[hutchphd]

One more weird variable in this setup is that carbon in a matrix (as it is an a pencil lead) can be sensitive to pressure. They make big carbon pile variable resistors which are effective. I think this is due to the connectivity of the carbon matrix. Don't know if this is significant for your rig.

 

[neilparker62]

An interesting slant on this experiment is to apply the binomial theorem to obtain an expression for current in terms of conductance: $$I=\frac{V}{R+r_i}=\frac{V}{R(1+\frac{r_i}{R})}\approx\frac{V}{R}\left(1-\frac{r_i}{R}\right)=VG-Vr_iG^2$$ The one term binomial expansion assumes $r_i<<R$ but we can assess the error even with very modestly chosen values of $R=4\Omega$ , $r_i=0.5\Omega$ and $V=1.5\; V$. Accurately determined current is $\frac{1}{3}\;A$ and using the expression on the right we obtain $I\approx\frac{21}{64}$. The difference is just $\frac{1}{192}$ or about 1.6% (of $\frac{1}{3}$).

For the data set graphed above with R ranging from about 13 to 20 ohms, the approximation will be even better. Accordingly we can plot current vs conductance (calculated as $\frac{I}{V_{load}}$) and obtain the relevant coefficients (emf and internal resistance) from quadratic regression. The plot obtained is also an illustration of Ohm's law since the dominant term in $VG-Vr_iG^2$ is of course $VG=\frac{V}{R}$.

The statistical 'caveat' of this method is that in applying quadratic regression (using Excel's trendline facility), we force a zero intercept based on the equation having a theoretical (0,0) intercept. I am not sure what statistical 'gurus' would have to say about doing that!

I am a bit concerned about $R^2=1$ (too good to be true ??) but otherwise the coefficients obtained are in good correspondence with the previous graph: $emf = 1581 \;mV$ and $r_i=\frac{458.56}{1581}\approx 0.29\Omega.$

 

[sophiecentaur]

we force a zero intercept based on the equation having a theoretical (0,0) intercept.

Hellfire. Why ever would you use a zero intercept? Did anyone actually measure that point??