Troubleshooting Battery Internal Resistance Measurements

[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.

 

[sophiecentaur]

The purpose of the experiment I am doing has not changed at all.

OK that's all fine and you are clearly getting somewhere useful. It was only the 'aside' that I didn't understand.
My question still remains about what that graph in post #89 is all about. What 'component' do the values of Conductance represent? Are they the load conductance? What did you measure that with? Did you actually need to - if V across the cell and I are all that's used?

Also, people (not just me) really appreciate proper schematic diagrams for describing electrical experiments. Photos are a bit like verbal versions of equations in that they're open to misinterpretation. For simple circuits they are very easy to produce with an object based drawing app (not shakey-hand-drawn as in Paint). PowerPoint and other similar apps are available for most people. Even Excel has the basic Microsoft system for building items from shapes. Once learned, never forgotten.

 

[neilparker62]

OK that's all fine and you are clearly getting somewhere useful. It was only the 'aside' that I didn't understand.
My question still remains about what that graph in post #89 is all about. What 'component' do the values of Conductance represent? Are they the load conductance? What did you measure that with? Did you actually need to - if V across the cell and I are all that's used?

Also, people (not just me) really appreciate proper schematic diagrams for describing electrical experiments. Photos are a bit like verbal versions of equations in that they're open to misinterpretation. For simple circuits they are very easy to produce with an object based drawing app (not shakey-hand-drawn as in Paint). PowerPoint and other similar apps are available for most people. Even Excel has the basic Microsoft system for building items from shapes. Once learned, never forgotten.

Concerning the graph in post #89.

I derived the following equation: $$I \approx E_{cell} \times G - E_{cell} \times r_i G^2 $$. This equation is of the form $y=ax + bx^2$ with y=I and x=G. Quadratic regression on the x,y pairs will yield the coefficients a $=E_{cell}\;$ the cell emf and b $=-E_{cell} \times r_i$. "b" can be divided by the already obtained value of "a" to determine $r_i$ the internal resistance of the cell.

In this equation G is indeed the load conductance - the reciprocal of load resistance R. It is obtained from the measured load volts and amps simply by dividing measured load amps by measured load volts.

The graph in post #89 looks very linear and indeed it is because the above quadratic expression is dominated by the term $E_{cell} \times G$ which would represent ideal battery behaviour in the case $r_i=0$. Remarkably however the quadratic regression fit does yield quite convincing values for both cell emf and internal resistance. A parabola can look very linear over certain limited domains.

Concerning circuit schematic

Please see following summary of the experimental procedure for the "A level" physics prac. The circuit diagram does not show it but they do mention including a second resistor to limit current. In my case the pencil showing in my photo acted as that second resistor providing a minimum load of about $13\;\Omega$. Perhaps $15 \Omega$ would be better since we need to limit current draw to values that do not significantly alter cell emf (open circuit voltage) over the course of the experiment. The rheostat shown in the circuit diagram is my exposed pencil lead which enables one to change load resistance/conductance simply by changing the point of attachment of the crocodile clip.

In the circuit diagram referred to above, there are misleading 'wires' apparently connecting battery terminals to voltmeter leads. There should be no such 'parasitic' resistance between battery terminals and voltmeter leads - that is the exact point of Dave's 4-wire measurement setup (if I understand it correctly ?).

 

[neilparker62]

So here's my setup to measure internal resistance using battery test mode on my multimeter. The meter's manual indicates that when testing a 1.5V battery, current draw will be about 50 milli amps. From which I infer that the "battery test" setting is supplying the battery under test with a 30 ohm load as per circuit diagram above. If there was no lead resistance we could directly determine the battery's internal resistance from $$r_i=\frac{\Delta V}{I}$$ where $\Delta V=E_{cell}-V_1$ and $I=\frac{V_1}{30}$.

If both meters above are set to read open circuit volts, then the readings will be the same since there will be no 30 ohm resistor and no current in the circuit. However if the meter measuring $V_1$ is now set to "battery test" , the 30 ohm resistor is connected and current flows. There is now a small terminal voltage drop for both $V_1$ and $V_2$ but $V_1$ drops marginally more - by measurement this difference is 7 or 8 millivolts. The difference is due to lead resistance and can be divided by load current to obtain a value for lead resistance. Given $I_{load} \approx 50 mA$ we can determine lead resistance as $\approx \frac {7.5}{50}=0.15 \Omega$.

Having determined lead resistance, we may now dispense with $V_2$ and move back to the simple method described in paragraph 1 above. And just subtract $0.15 \Omega$ from the calculated $r_i$ value obtained accordingly. Here are some results I have obtained using this technique. The dataset is ordered on the last column. Voltage readings are in milli volts.

 

[neilparker62]

Data below is for Energiser Rechargeables with $r_i$ between 30 and 40 milli-ohms. We obtain about 33 milli ohms. A bit different using regression statistics - 37 milli ohms based on $\frac{V_L}{E}\approx 1- r_i G.$ Exceptional AA batteries - they hardly 'bend' at all when drawing currents between 200 and 400 milli-amps.

 

[neilparker62]

I've written an Insights article on the above and quoted or linked to a number of contributing posts - I hope the relevant contributors won't mind me doing that. I'm about to finish up with it now and will be posting "for review" - would greatly appreciate your comments/suggestions on the article. Will definitely be acknowledging all assistance.