Which type of conductors don't follow Ohm's law? [closed]

Which type of conductors don't follow Ohm's law?

1. I know that semiconductors and superconductors don't follow them, but why?

2. And what about ideal conductors. What are they? Do they follow ohm's law? There are so dubious and contrasting answers in quora. Please clear this doubt.

• Try to stick to one specific question in your post. – M. Enns Jul 6 '19 at 2:09

To be sure, Ohm's 'law' is not a law and, in fact, the current-voltage relationship for any physical conductor can only approximate Ohm's law over some limited region of operation.

Note that Ohm's law $$V = IR$$ defines the ideal resistor of resistance $$R$$. The ideal conductor is essentially an ideal resistor in the limit that $$R\rightarrow 0$$. For an ideal conductor, there is zero voltage across for any finite current through.

An ideal conductor is an abstraction that is useful for simplifying the analysis of, e.g., circuits and deriving results in good agreement with reality (in the appropriate region of operation).

For example, consider a series circuit consisting of a cell, two wires, and a resistor. As long as the resistance of the wires is much less than the resistance of the resistor, we model them as ideal conductors and say that the voltage across the resistor equals the voltage across the cell. In reality, the voltage across the resistor is slightly less since there is a small (and generally insignificant) voltage drop across the wires.

I know that semiconductors and superconductors don't follow them, but why?

An appropriately doped n(p) type semiconductor can be a good conductor since the doping results in an excess of mobile electrons (holes) that can participate in a current. A p-n semiconductor junction, however, is an entirely different story and is beyond the scope of this answer.

A superconductor (over some region of operation) has, like an ideal conductor, zero resistance.

• Can we say from it that Ideal conductor doesn't follow Ohm's law then? – Ritwik Bhattacharyya Jun 29 '19 at 8:49
• @RitwikBhattacharyya, as I wrote above, an ideal conductor can be thought of as a resistor with resistance $R = 0$, i.e., the voltage across is zero for any finite value of current. Imagine instead, a conductor with non-zero resistance $R = \epsilon$ that is so small that the voltage across is effectively zero for any applicable value of current. Surely such a conductor follows Ohm's law and yet is effectively an ideal conductor. – Alfred Centauri Jun 29 '19 at 11:30

Ohm's Law simply states that for a given resistor, the current and the voltage are directly proportional. So if you increase the voltage by 10%, the current goes up by 10%. If you halve the voltage, the current is halved.

This matches up quite well with the old water-in-a-pipe analogy that you often hear about when discussing electricity. If you take a section of pipe (resistor) and apply water pressure at one end (voltage), you will get a flow of water (current). If you increase the pressure, you get more flow, with a roughly proportional relationship - double the pressure, you get double the flow.

The reason Ohm's Law is so simple is that when electrons are flowing in a conductor, the underlying physical process is rather like water flowing in a pipe - electrons can move from atom to atom with some resistance that is (roughly) independent of the voltage.

However, Ohm's Law has its limits. One problem is that the current heats up the resistor. This increases the resistance, which means you don't get the exact current increase that Ohm's Law predicts.

Semi-conductors don't obey Ohm's Law because there is a different physical mechanism at play. The electrons, instead of shuffling past atoms that are in in a continuous lattice, have to jump between quantum energy levels.

Superconductors don't obey it either because again there is a different physical process - it involves Bose-Einstein condensates, which is straying a bit out-of-scope..

Ohm's Law should be called Ohm's Rough Guideline.

In a superconductor, there is a critical current density. When the current goes above that, the resistance goes drastically (perhaps ~10^10 orders of magnitude!!) up. For most superconductors, the current limit isn’t that great either. This limits the practical application of ohm's law in superconductors.

The resistance of a semiconductor changes depending on the applied current or voltage. As such, you can not simply look up the resistance of a semiconductor and use "Ohm's Law" to determine the relationship between voltage and current by the good old V=IR formula like you can with a resistor. And thus semiconductors do not follow Ohm's law.

I do not have much knowledge of ideal conductors but I know that they have very high conducting ability and thus resistance offered by them is zero. So, I think they are not supposed to follow the Ohm's law. Though, I am not really sure for the last one.

I would suggest the following approach.

Take any piece of material, now apply potetial difference, i.e. voltage $$V$$, accross it. Given fixed size of the piece, the applied voltage will be proportional to the force on all the charged particles within the material (let us ignore screening for now). In some cases this force will be sufficient to get electrons (or ions in liquids) to move, so you get the current $$I$$.

Now all of this can be encapsulated by a simple relationship: $$f\left(V, I\right)=0$$ for some function $$f$$. In some cases this function is analytic so for small currents and voltages we can write:

$$\left(\frac{\partial f}{\partial V}\right)_I V + \left(\frac{\partial f}{\partial I}\right)_V I = 0$$ and you get the Ohm's law with resistance $$R=-\frac{\left(\partial f/\partial V\right)_I}{\left(\partial f/\partial I\right)_V}=const$$. As long as material stays within this regime we call it conductor. Once it is outside this regime we call it something else.

Semiconductors. Stay within this regime for VERY small currents/charges, then $$f$$ becomes nonlinear and you need to include more terms in the expnsion.

Superconductors. Function $$f$$ is not analytic at point $$V=0$$ so the linear expansion fails even for smallest voltages.

Metals. Linear expansion is vaolid for quite a wide range of currents and voltages, but then fails since $$f$$ becomes nonlinear, e.g. due to heating.

Now the three statements above should either be taken as experimental observations, or you should be prepared to read several chapters from solid state physics book to get a better picture.

I agree with the other answers. I would just add that the degree to which Ohms law applies depends on the degree to which the relationship between voltage and current is linear. In some cases it approximately holds for a device over a limited range of voltage and current if not over the entire range.

Hope this helps.