High School Troubleshooting Battery Internal Resistance Measurements

[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.

 

[neilparker62]

Hi again. I've written a second article which is a sequel to the first one above. I hope I won't tax contributors patience if I again request review ? Techniques used in measurement of battery internal resistance are expanded to measure low resistance generally. By simple ratio of voltages in a series circuit.

The article is entitled "The Poor Man's Milli Ohm Meter" and has been posted to the Insights development forum.