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I plotted the I-V curve between two points (few microns apart) on a thin aluminum film.

I expected this metal to be a conductor and have a much lower resistance, but the slope suggests that it is actually about 4Ω. Does this make sense? Is this an equipment limitation?

Here is the plot. The equipment is limited to 0.1A current.

enter image description here

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  • $\begingroup$ How thin was the film? Below a few tens of nanometres metal films tend to be reticulate rather than a smooth continuous film, and this increases their resistance. Can you give the details of the geometry? As it stands there's no way for us to calculate the resistivity. $\endgroup$ – John Rennie Nov 14 '14 at 6:59
  • $\begingroup$ You bring up a good point. The film was 100nm, but it was not smooth. It was created using e-beam evaporation. $\endgroup$ – sebo Nov 14 '14 at 7:03
  • $\begingroup$ And the geometry you used for the measurement? $\endgroup$ – John Rennie Nov 14 '14 at 7:09
  • $\begingroup$ I don't have any other relevant measurements. I just touched two points on the film which were fairly close together. This aluminum film was deposited on top of silicon substrate. $\endgroup$ – sebo Nov 14 '14 at 7:15
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    $\begingroup$ Did you use a 4-point probe (the correct way to measure sheet resistance)? If just two probes, did you use the 4-wire method to accurately measure the resistivity of just the film, and not the wires to the instrument? What kinds of probes did you use? $\endgroup$ – Jon Custer Nov 14 '14 at 14:34
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This isn't really an answer, because your question doesn't provide enough information for an answer. However it explains what you need to do. In fact this is exactly what I did (back in 1983!) to measure the resistivity of evaporated silver films.

If you have the film formed on some substrate you need to score two lines to leave a long narrow track like this:

Metal film

In the diagram I've exaggerated the width of the track. The width $t$ needs to be much smaller than the length $d$ so the track approximates a thin wire.

Now vary the distance between the electrodes, $d$, and measure the resistance. Graph the resistance against $d$ to get the resistance per unit length. You need to do this because your electrodes will have some contact resistance, and if you only make one measurement you can't separate the contact resistance from the resistance of the metal film. Your graph will have a non-zero $y$ intercept, and this will give you the value of the contact resistance. The gradient will give the resistance per unit length.

Once you have the resistance per unit length, $R/\ell$ the resistivity is given by:

$$ \rho = \frac{R}{\ell} t h $$

where $t$ is the track width and $h$ is the film thickness.

Now you can compare your measured resistivity with the resistivity of bulk aluminium. Films of around a micron or greater thickness should have a resitivity similar to the bulk metal. However thinner films will have a higher resistivity because they are not continuous but contain voids.

For evaporated silver films I found the resisitivity only deviated significantly from the bulk value below around 50nm, but this will depend on how much they anneal after striking the surface.

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  • $\begingroup$ This answer describes very well further experimentation needed to determine the resistivity of the material. However, I'm still looking for some possible reasons to explain the cause of the large resistance I'm seeing. I understand that creating a track should help, but does such a high resistance without a track make sense? $\endgroup$ – sebo Nov 14 '14 at 8:02
  • $\begingroup$ Maybe you had a high contact resistance. There's no way to tell. $\endgroup$ – John Rennie Nov 14 '14 at 8:15
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Another possibility is that you are having to break through the oxide film that all Aluminium has in air. The voltage needed to do that can vary considerably. You might need to use sharp steel electrodes that will punch through the oxide and contact the metal directly.

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Actually, I am not sure you should expect a much lower resistance value. For example, with aluminum resistivity of 2.65x10^(-8) Ohm-m, a sample with length of 5 micron and cross-section area of 100 nm x 1 micron has a resistance of about 1.3 Ohm. Of course, you can question the rather arbitrary effective width of the sample - 1 micron, but I guess this effective width depends on the dimensions of your electrodes, and I don't know them. Let me just note that, as far as I remember, for very small electrodes, the resistance between two points of an infinitely wide and long film can be extremely high (infinite, if you have point-like electrodes).

I'd also like to note that Dirk Bruere's answer can be relevant. A few months ago, I tried to measure the resistance of a thin (1.6 mm) and long (1 m) rod made of doped silicon. The measured resistance was infinite until I chucked the ends of the rods in collets, thus "breaking" the oxide film on the surface of the rods (the resistance was about 5 kOhm).

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  • $\begingroup$ I work with small signal electronics and the problems caused by oxide films on Al parts are sometimes a real PITA - so much so we use stainless steel as much as possible in our instruments. It is of special concern when Al casings have to be grounded. Quite often a screw and a ground wire are not good enough. $\endgroup$ – user56903 Nov 14 '14 at 9:00

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