In aluminum, how does electricity travel through the surface oxide layer?

Question

Suppose I connect a conductive wire (cross section 1 mm$^2$) to an aluminum object. Since aluminum is highly conductive, electricity will flow smoothly inside the object with little resistance. However, since aluminum is also very reactive, there is a thin layer of highly resistive aluminum oxide on the surface of the object. Wikipedia says this layer is about 4 nm thick (citing this paper). Naively, we can calculate the resistance of the oxide layer using the resistivity of alumina, which is about $10^{14} \,\Omega \cdot \text{cm}$:

$$R=\rho \frac{l}{A} \approx 4\times 10^9 \,\Omega$$

Of course, we don't actually measure such a large resistance. But why not? How exactly does an electric current pass through the oxide layer?

The obvious answer is that electrons simply tunnel through the oxide layer. So let's calculate the tunneling probability. According to this document from MIT OpenCourseWare, the aluminum oxide layer presents a potential barrier of 10 eV. Then the transmission coefficient across a 4 nm layer is given by

$$T \approx e^{-2\left( \sqrt{2 m_e / \hbar^2 \cdot (10\text{ eV})} \right) (4\text{ nm})} = 5.16 \times 10^{-57}$$

This is an extremely small number. In principle, we could now find the actual rate from the density of states and Fermi's golden rule, but it seems likely the result will be a very small current.

It's possible that the parameters that I am using might be incorrect. I checked a few other sources and found widely varying values for the potential barrier and oxide thickness. However, the fact that slightly anodized aluminum with a thicker oxide layer (for example, a few tens of nm) still conducts electricity makes me think that tunneling is not a full explanation, since the tunneling rate decreases exponentially with oxide layer thickness.

Another possible explanation might be electrical breakdown or some other change in the oxide crystal structure, such as melting. But if this is the correct answer, what exactly changes in the oxide layer to make it electrically conductive? Normally, oxides are not conductive because the oxygen atoms scavenge free electrons. Does this stop happening for some reason?

I am willing to accept a good theoretical answer, but I am hoping for experimental evidence if possible.

Answer 1

The native oxide that coats aluminum is slightly porous, and the pores tend to trap tiny amounts of moisture in them. This renders them electrochemically active and ever so slightly conductive. (In fact, for the aluminum oxide layer to grow in thickness in hot environments requires that both the aluminum atoms be capable of diffusing up through the existing oxide to reach oxygen in the atmosphere and the oxygen atoms be capable of diffusion down through the oxide to reach unreacted aluminum underneath the oxide.)

In order to render an oxidized aluminum surface pore-free, the aluminum piece must be baked in an oven with an oxygen atmosphere in it, to close off those pores.

In the absence of porosity in the oxide, the conduction mechanism is Frenkel-Poole emission, where a random thermal fluctuation will occasionally promote a bound electron into the conduction band, where it can then drift under the influence of an external field.

Answer 2

The engineering answer is that current doesn't pass through the oxide layer very well at all, and if you want to make a good contact to an aluminum object you have to be very careful about how you do it.

If you just press a copper wire (for example) against an aluminum one, you will get a very high resistance contact. Probably not 4 gigohms, but maybe on the order of hundreds or thousands of ohms, so there may be something to the earlier answers which suggest the oxide layer is fragile enough to break away and allow some contact.

But, for example, you can abrade the oxide away with sandpaper, and then make an air-tight connection of (or solder or weld) the other object to the aluminum before the oxide has time to re-form.

For small enough geometries (like the bond-wires used to connect integrated circuit chips to their lead-frames) you can pressure weld aluminum directly to other materials like gold or silver. This tends to deform the aluminum wire substantially, which must spread the oxide out enough to prevent it interfering with the contact.

Or you can use a chemical "coating" or surface treatment on the aluminum to keep the oxide from forming. One of these treatments goes by several names such as "Alodine", "chromate conversion" or "chem film". (Note: traditional Alodine treatment is not usable for products to be sold in Europe due the RoHS directive, but newer chemical treatments are available that are acceptable under RoHS)

Or you can use a very aggressive flux to displace the oxide during soldering. But this flux must be cleaned very thoroughly to avoid continued corrosion of your parts.

Answer 3

I believe @Maxim Umansky is correct in his comment: the breakdown potential of the aluminum oxide layer is just a few volts (see, e.g., Fig.8 in http://citeseerx.ist.psu.edu/viewdoc/download?doi=10.1.1.877.5366&rep=rep1&type=pdf (J. Electrochem. Soc., Solid-State science and technology, October 1976, p. 1479). For a layer thickness of about 4 nm we get a breakdown voltage of a few volts.

Answer 4

My hypothesis is that the layer of aluminum oxide on the surface of a piece of aluminum is so thin and malleable that when a copper (say) conductor is pressed against it, the aluminum oxide is easily pushed aside so that electrical contact is made. I believe that QM tunneling could not account for the large current flows such as we observe in such cases.

Correction: while the alumina layer is very thin indeed (due the high oxygen reactivity of aluminum combined with the ability of aluminum oxide to prevent further oxidation of the aluminum surface, which is why aluminum can stay shiny), it is not malleable as compared with aluminum. It bonds very tightly and is hard. As the reference in my comment says, the resistance of the alumina layer is low just because of its thinness.

I have successfully soldered a copper multistrand wire directly to an aluminum chassis by simply scratching the aluminum surface with a steel wire brush repeatedly while applying the solder (60/40 lead/tin around rosin flux) and the soldering iron (gun type), then soldering in the wire. The resulting joint looked good and conducted electricity without measurable resistance.

Answer 5

When aluminium oxide is formed naturally, it will inevitably contain defects like dust particles, metal contamination, trapped moisture etc. In addition, the pressure applied to the mechanical connection is enough to break the oxide layer. As a result, if the contact area is large enough, the effective thickness of the oxide layer will be much less than the expected 4nm, essentially, it will be zero. As a result, there will be no measurable breakdown voltage under typical conditions, and when you connect wires to an aluminium object, it simply acts as a conductor.

By the way, oxide layer is not a unique feature of aluminium. What is notable is how fast it oxidises (preventing soldering in most cases) and how bad the galvanic corrosion is when it's connected to a dissimilar metal such as copper (which caused many house fires back in the day). But essentially, when you connect two copper wires, the same reasoning about oxide layers apply.

Answer 6

The resistance of aluminum oxide is 1x10^14 /cm ohms. It has good thermal conductivity and can reduce thermal shock resistance. Alumina is very useful in that it is available in a variety of purity ranges from 94% to 99.9%. It is usually white, but is sometimes pink (88% alumina) and brown (96% alumina). The composition of aluminum oxide can be easily changed to enhance certain desirable material characteristics such as the hardness or color. Aluminum oxide is an electrically insulating material with high resistivity that increases with purity.

While it is a good insulator, it is not a pure insulator so current will travel through it.