B Troubleshooting Battery Internal Resistance Measurements

[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.

 

[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??

 

[neilparker62]

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

Well - yes. If I don't connect the circuit (zero conductance) the ammeter registers zero current.

1) Forcing 0 intercept is advisable if you know for a fact that it is 0. Anything you know a priori, you should use in your model. This model is rather crude, but uses 0 intercept as the consequence of the Big Bang Theory: at time 0 all the matter is in one place.09 Jun 2014

 

[neilparker62]

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.

Do you mean sensitive to ambient air pressure ? I think I could reasonably assume that over the relatively short period over which measurements are taken, that would not vary much.

 

[hutchphd]

I was thinking more of clamp pressure and perhaps bending moments on the pencil lead. I have no idea whether this is really an issue

 

[sophiecentaur]

Well - yes. If I don't connect the circuit (zero conductance) the ammeter registers zero current.

1) Forcing 0 intercept is advisable if you know for a fact that it is 0. Anything you know a priori, you should use in your model. This model is rather crude, but uses 0 intercept as the consequence of the Big Bang Theory: at time 0 all the matter is in one place.09 Jun 2014

But you can’t assume linear regression, can you? At some value of I the straight line will curve towards the origin. You have to fit the line to the actual measured data.

 

[neilparker62]

I was thinking more of clamp pressure and perhaps bending moments on the pencil lead. I have no idea whether this is really an issue

Well I tried applying some pressure/bending and the lead simply snapped. So I would say that's a binary 1 or 0

 

[neilparker62]

But you can’t assume linear regression, can you? At some value of I the straight line will curve towards the origin. You have to fit the line to the actual measured data.

I'm not assuming linear regression - I'm assuming a quadratic regression fit of the form ax^2 + bx.

 

[sophiecentaur]

I'm not assuming linear regression - I'm assuming a quadratic regression fit of the form ax^2 + bx.

Have I missed something? You have a device under test and you are measuring resistance / conductance for various currents. Passing zero current doesn't make the conductance zero. Or am I off beam about what you're doing?

 

[neilparker62]

Have I missed something? You have a device under test and you are measuring resistance / conductance for various currents. Passing zero current doesn't make the conductance zero. Or am I off beam about what you're doing?

Open circuit: $R=\infty \implies G=0$.

 

[sophiecentaur]

That's a correct definition. I think my problem is the lack of schematic diagrams and actual experimental description. What does that graph actually show?
If you are studying the properties of the carbon then don't you want to know how the conductance is affected by the current? I may not have read your account properly but no current for an open circuit is not the same as a zero intercept for G as a function of I for the pencil lead. With no current flowing, G will not be zero - G is a property of the carbon.
The thread has migrated from measuring the characteristics of a cell to the characteristics of a variable carbon resistor? A pencil lead being used as a variable resistor is a good idea but do you need to calibrate it if you are measuring the current with a meter?
Could you clear this up for me?

 

[neilparker62]

That's a correct definition. I think my problem is the lack of schematic diagrams and actual experimental description. What does that graph actually show?
If you are studying the properties of the carbon then don't you want to know how the conductance is affected by the current? I may not have read your account properly but no current for an open circuit is not the same as a zero intercept for G as a function of I for the pencil lead. With no current flowing, G will not be zero - G is a property of the carbon.
The thread has migrated from measuring the characteristics of a cell to the characteristics of a variable carbon resistor? A pencil lead being used as a variable resistor is a good idea but do you need to calibrate it if you are measuring the current with a meter?
Could you clear this up for me?

The purpose of the experiment I am doing has not changed at all. It is to measure the internal resistance of an AA cell. Characteristics of carbon resistors only came up in the context of being possible sources of error in the experimental setup. But I think we may fairly conclude that neither temperature nor pressure vs resistivity coefficients for graphite are such that they might change the resistance of the pencil leads in use. Even if they did it would not matter so long as (per Tom G's advice) readings of load voltage and current are taken simultaneously. So that we can be as certain as reasonably possible that the current which is measured flows through the battery's $r_i$ and is solely responsible for the voltage drop which is the difference between cell emf and the measured load voltage.

Complications arise because cell emf is supposed to be a fixed value according to the model we are using. In practise this turns out not to be the case. Cell emf falls as the battery discharges. And the greater the current draw, the more rapidly that happens.

Dave's post (#63) I think sets the "gold standard" for this measurement. In particular he goes to great lengths to ensure that the load voltage is measured right at the point of contact on the AA cell terminals. And that current carrying load leads are kept insulated from voltage probe leads right up to that point of contact. For example I had an experimental setup in which I attached the negative load lead to the top of the crocodile clip being used as a voltage probe. Whereby I am now measuring $r_i$ plus resistance of crocodile clip. When the resistance value being measured is $< 0.5\Omega$ you have to take into account every small resistance that could possibly add to that being measured. The only quibble I might have with that method is that the measured 144 milli amp current would discharge the AA cell rapidly and perhaps drop the cell emf (by a few milli volts) even as the measurement is taken (at least in my experience).

What Dave showed us is - in essence - the brief of the experiment my students were given. The 'elaboration' leading to the graphs I have posted above came from looking at the "A level" practical described in the video I posted. Whereby a set of readings of load volts and load current are taken and $r_i$ is determined from regression analysis according to the equation $V_{load}=Emf - r_i I_{load}$. Note that Emf is assumed constant in this equation and should emerge as one of the two constants obtained from linear regression. Within the bounds of experimental error, the regression obtained Emf should agree with measured open circuit voltage.

I also described a method for manipulating the equation such that the same two values (Emf and $r_i$) could be obtained from quadratic regression of current against conductance. Since conductance is determined as $\frac{I_{load}}{V_{load}}$, this method uses the same set of readings as for the linear regression.

Finally -in post #61 - I suggested a simple method for determining $r_i$ by using the battery test facility on a typical multimeter. However this fails because the voltage probes are carrying load current and thus the resistance obtained will be the sum of $r_i$ and the resistance of the probes and probe leads. This method would work reasonably well if one knew - or could measure - the actual resistance of the probes and probe leads.