# [Physics] Why does the measured I-V curve for a film of aluminum suggest high resistance

conductorselectric-currentelectrical-resistancevoltage

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.

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:

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.