Measuring milliohms, and sheet resistance

[klugesmith]

It was the night before garbage pickup, so time for experiments with a take-out food container from local restaurant.
Aluminum foil pie dish, about 0.0025 inches (63 um) thick.  Cover is paper about 0.017 inches thick, coated with foil of unknown thickness.


Let's use electricity to find out how thick.

Sheets of conductive material can be characterized by sheet resistance, commonly stated as a number of ohms per square.
It's the resistance of any rectangular strip divided by the Length/Width ratio.  4 cm long and 1 cm wide has the same resistance as 4 feet long and 1 foot wide: four squares worth.   Sheet R is numerically the same as bulk resistivity (in ohm-cm or ohm-m) divided by the sheet thickness.   
In printed circuit boards made with 1-ounce copper foil, the sheet resistance is about 1/2 milliohm per square.

I cut half-inch wide strips of the dish and cover, and marked 7 inches (14 squares) on each part.


Next challenge: measuring small fractions of an ohm.  The foil-coated paper strip is less than 1/10 ohm, and aluminum foil strip is less than 1/100 ohm.   Ordinary multimeters in "ohms" mode use currents of 1 mA or less, which would make the voltage drop relatively small for inexpensive measurement.  The remedy: more power! If we set up a DC power supply to drive 1.00 ampere through the resistance under test, and measure voltage drop with a separate meter, each millivolt corresponds to resistance of 1 milliohm.


I found that the foil and foiled paper strips had 7.6 mV and 66 mV across 14 squares, so 0.54 and 4.73 milliohms per square.   The metal coating on paper is almost 9 times thinner than the pie plate foil (if they are made of the same stuff).


The literature value for resistivity of Al is 2.65e-8 ohm-m, which would make the metal thicknesses 49 and 5.6 um (1.9 and 0.22 thousandths of an inch).   

But measuring caliper tells us the thick foil is at least 2.5 thousandths thick.  Explanation: it must be an aluminum alloy.  As in copper alloys, there's a large resistance penalty for small amounts of other elements.  If the foil composition is like alloy 6061, with rho of 3.67e-8 ohm-m, then thickness of 68 um (2.7 thousandths) would match our resistance measurement.   Aluminum coating on paper lid would be 8 um (0.3 thousandths).

There was time for an independent check this morning, with the sounds of garbage trucks approaching.
Pie dish weighs 14.84 grams, and is formed from a 31-cm disk of foil.  Using the literature value for density of Al, the indicated metal thickness is 73 um (2.9 thousandths).

I hope somebody else learns something from this report.

[rikkitikkitavi]

Apart from using and mixing of  SI units and Imperial units which hurts my head a lot (MHz)  I would say this is really  a nice work of science done properly
Especially working under stress (garbage man is coming ...)

Very entertaining post.

I measured the primary and secondary resistance of a transformer windings in a similar way,it is a 3kVA 400V/20V transformer to deduct the approximate VA rating (based on an arbritrary full load efficiency rating of 95% ) or more exactly current capacity as I dont trust the old 3A/square mm for large transformers. I just need to get some proper graphite electrodes (arc gouging electrodes)  and we shall have some arc-fun.

[davekni]

Try standard rolled aluminum foil.  It's usually fairly pure.  Of course, so is aluminum electrical wire.  Most other "aluminum" materials are alloys tailored for mechanical properties, as I've found with my eddy-current levitation demonstrations.  Yes, nice technique.  I do similar tests.