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Theoretical vs measured sheet resistance of conductivity paper

  1. Oct 3, 2015 #1
    1. The problem statement, all variables and given/known data
    After performing a lab experiment to determine the conductivity if Pasco conductivity paper, I found the conductivity to be 0.195 Ωm. We found conductivity by graphing measured values of voltage (y axis) and the length (x axis) between the probes of a voltmeter that are touching the conductivity paper. We know that the slope that equals ( rho * I ) / A, so using our measured values for current and cross sectional area, we can solve for resistivity. Doing so tells us that resistivity = 5.13 Ωm. Taking the inverse gives us conductivity. As a step in the lab in the lab report I must write, I must compare it to the manufacturer's value, which is given as 5000 Ω per square. I have read that the resistivity is equal to the sheet resistance multiplied by the thickness of the material. Thus, the manufacturer says that the resistivity = 62.5 Ωcm = 0.625 Ωm. The professor who supervised this experiment says that most of the class got around my value for resistivity and checked our units to make sure that units are not the issue. Thus, I am wondering what could could be going on that makes our value different by more or less 10 times.

    Thickness: t = 0.0125 cm
    Width: w = 2.12 cm
    Current: I = 1.05 * 10^-7 A / S
    Slope = 0.204

    2. Relevant equations
    A = width * thickness
    V = IR
    V = (resistivity * length / A) * I
    V = (rho * L / A ) * I
    V = (rho * I / A) * L ---> for graphing, since we do not change the current, cross sectional area, or resistivity, and we change the length between probes.
    Slope = ( rho * I ) / A
    rho / t = sheet resistance

    3. The attempt at a solution
    I have been attempting to figure out why our number is essentially off by a factor of ten by converting all measurements to centimeters in case the manufacturer did. I have also tried converting their 5000 Ω per square value (which is the resistance of a sheet w wide and L long) by adjusting the "square" that they use to fit the geometry of our sheets (we cut the original sheet into long strips). Originally, the sheets were 28 cm by 20 cm, with squares on the paper being 1 by 1 cm. After cutting, the sheets were 28 cm by 2.12 cm. I posting this to see if there is some piece of information that I am missing when comparing/converting sheet resistances.
     
  2. jcsd
  3. Oct 3, 2015 #2

    mfb

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    Staff: Mentor

    That has units of resistivity, not conductivity.
    Per square what?

    Some ideas and questions that could be relevant to find the origin of the difference:

    How did you apply a current, and how and where did you contact the paper?
    What is the total length where you had a current flow?
    What is S? Did you try other currents?
    Can you show your data points?
     
  4. Oct 3, 2015 #3
    0.195 Ω^-1 m^-1 is the value conductivity. I did not do units correctly.

    "Per square" as I understand it is 1 square of the material that is one slice along the dimension of the thickness. A picture can be seen in the Wiki link as to what the square is, but in words it is the square that the current flows parallel to. One side of this "square" is the width of the cross sectional area, the thickness "t" being the other dimension. I do not entirely understand if converting sheet resistance to resistivity is as simple as I think it is when we are taking the sheet resistance given for a big sheet and expecting it to work for a modified sheet. In theory, it sounds like to me that the sheet resistance would not change. This is the whole purpose of my post, to better understand sheet resistance and how manufacturer's use/define it. https://en.wikipedia.org/wiki/Sheet_resistance

    Current was supplied by a "power supply" box. That is what the professor called it. The power supply was attached to the paper by wires and alligator clips that contact thin copper sheets to ensure a more uniform current going through the paper. The paper and copper sheets where placed on top of a cardboard strip, though I cannot recall his reasoning for this. A The voltmeter probes touched the paper directly to measure voltage.

    "S" is meant to be seconds, as amps per second is the standard unit for current. Current was calculated by inserting a resistor into the circuit before we began the experiment. We measured the resistance of the resistance and then measured the voltage difference after applying a current. We then used V / R = I to solve for current.
    Current was kept uniform throughout the experiment so that for our graphing equation we would have a linear relationship between voltage and length, since every other factor should stay the same and thus be a constant, like the slope calculated.
    Data points are attached with other data and graph.
     

    Attached Files:

  5. Oct 3, 2015 #4

    mfb

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    Staff: Mentor

    It is not, Ampere is. 1 Ampere is 1 Coulomb per second: 1 A = 1C/s.
    Also, seconds are written with a small s.

    5000 Ohm for the whole sheet probably means (rho * L / A ) = 5000 Ω where L is the length of the strip.

    Sure, another current as cross-check would still be nice.

    Something went wrong with cm <-> m in your excel sheet in K15 and K17.

    Okay. The data points look nice, that eliminates many possible sources of error.
    What do you mean with modified sheet?

    How long was the total sheet?
     
  6. Oct 4, 2015 #5
    My mistake about the units for current. Are you saying then that the resistance for a sheet that is a fraction of the original width (assuming the length and thickness of the paper are unchanged) would be a fraction of 5000 Ω? If our new strips were ~2 cm wide, then our sheet is about 8.7% of the original width (23 cm is given by the manufacturer's website, though the actual marking on the paper give 20 x 28 cm) and thus the resistance of our sheet will by 8.7% of 5000 Ω, which is around 401.4 Ω. Are we considering this number to be the sheet resistance? If so, then the resistivity is 401.4 Ω * 0.0125 cm = 5.0175 Ωcm. The conductivity is then 0.199 Ω^-1 cm^-1. This is almost exactly our number, but the units are not correct (m vs cm). Quoting the exact wording for what value the manufacturer gives: "Its resistivity is approximately 5,000 ohms per square." Except that resistivity does not have those units, but the term sheet resistance does. It seems like they are mixing terminologies.


    The cm to meters in the excel sheet is a labeling error from when I had converted all to cm. I missed it when I was cleaning up the excel sheet to upload. Another current would be nice, but I know that another group that performed the experiment at the same time we did got 0.2 Ω^-1 m^-1 for conductivity, though I do not know the current they used, but I assume it would not be the exact same.

    When I say modified, I mean trimmed, cut into thinner sheets. So that instead of having a 30 x 23 cm sheet, we had thinner strips that were 30 x 2.12 cm sheets.
     
  7. Oct 4, 2015 #6

    mfb

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    Staff: Mentor

    If you reduce the width (and therefore the area), the resistance goes up.
    Apply the factor of 0.087 in the correct direction and it looks good.
     
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