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If I plan to measure the resistivity of a high resistivity material, would it not make sense to alter the geometry of the sensor to make A/l as small as possible and reduce the voltage to get a low resistance value to reduce the resistivity measured at conditions which are easy to fulfill experimentally. Because i'm a bit confused as i'm being told to increase the voltage and make sure A/l is large to measure the resistivity of this material. Thanks in advance for any help.

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When you do the experiment typically you will apply some known voltage $V$ and measure the current $I$. This is given by:

$$ I = \frac{V}{R} = \frac{V}{\rho} \frac{A}{\ell} $$

Your problem is that if the resitivity, $\rho$, is very high then $V/\rho$ will be very small and the current $I$ will be too small to measure accurately. If so you can increase the current back to measurable levels by increasing the value of $A/\ell$.

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  • $\begingroup$ Further, if the cross sectional area is too small your measured resistivity may be dominated by surface scattering, so you won't get an accurate measurement of the intrinsic material resistivity. This is often a problem in thin film measurements of low resistivity materials (e.g. copper in my experience), compounded by slight variations in film thickness. $\endgroup$
    – Jon Custer
    Commented Mar 20, 2019 at 12:53
  • $\begingroup$ to add to the answer of @John Rennie: the lowest dc currents that can be measured with the best (room-temperature) amplifiers today is about 10 fA, if the measurement bandwidth is 1 Hz. These amplifiers are based on commercial operational amplifiers with external circuitry comprising a negative feedback, which gives an I-V converter, whose conversion ratio is given by the feedback resistor. $\endgroup$
    – flaudemus
    Commented Mar 20, 2019 at 13:00
  • $\begingroup$ @JonCuster the material i'm trying to measure is a liquid, sorry didn't make that clear. I'm assuming that the surface scattering problem doesn't apply in that case. $\endgroup$
    – J. Doe
    Commented Mar 20, 2019 at 13:00
  • $\begingroup$ @flaudemus yh we'll be using AC current to avoid charge accumulation at the electrodes which apparantly causes errors. $\endgroup$
    – J. Doe
    Commented Mar 20, 2019 at 13:02
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    $\begingroup$ @Pieter yh comparing it with those liquids sounds like a really good idea, thanks a lot. will look into "guard electrodes" too, thanks. $\endgroup$
    – J. Doe
    Commented Mar 20, 2019 at 15:47

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