Electron Doping and Permittivity

citw
I've read that electron doping in certain materials can lead to higher permittivity (for example, Nb in TiO2). This is a little confusing to me, as permittivity decreases with conductivity (see http://en.wikipedia.org/wiki/Permittivity#Lossy_medium). Can anyone explain this to me?

P.S.: Both the dielectric constant and dielectric loss appear to increase with the case I mentioned above. I can understand the loss term increasing, but not the overall dielectric constant.

Which frequency range do you have in mind?

citw
Which frequency range do you have in mind?

Any. I'd like to know how this would ever make sense.

Ok, so where did you read this:
I've read that electron doping in certain materials can lead to higher permittivity (for example, Nb in TiO2).

Now, as you refuse to specify the situation you are really interested in, I can't give you but a very general argument:
There are some sum rules which make statements about the integral of the permittivity over frequency. Typically, these sum rules are proportional to the total number of electrons, so if that number increases, the permittivity will do so on the mean, too.
See, e.g.: http://www.atlas.uni-wuppertal.de/FP/anleitungen/fpI-10/paper_FPII2/pr128b.PDF
Now adding just some electrons won't change the permittivity by much on average, but there may be certain frequency ranges where the change is larger.

I didn't mean it too seriously ;-)
But from that link I infer that you are interested in the low frequency region as they talk about capacitors, and not about e.g. optics.
But: Is the doped TiO2 really conductive? If the electrons are just quite weakly bound, it is clear that they are very polarizable and that this increases the dielectric constant.

citw
I think this is where I'm confused... if electrons are weakly bound, I'd take the material to be conductive, and isn't it true that increased conductivity reduces the dielectric constant (via the imaginary term, given for example in the Wikipedia article)?

No, the material won't be necessarily conductive. I guess that the electrons get a very large effective mass due to lattice distortions (polarons).
A criterion for electrons in a semiconductor to behave as a metal is that the Bohr radii calculated with the effective mass and dielectric constant of the background (TiO2) do overlapp. With a large mass, this requires quite some density.

citw
Just to make sure I'm following, polarizability is described by relative permittivity (dielectric constant). The dielectric constant decreases with increasing conductivity, specifically from the imaginary (loss) term. Conductivity is inversely proportional to effective mass, and therefore the polarizability and dielectric constant will increase with larger effective mass.

If I'm following what you're saying correctly, the argument for doping would be that the effective mass of the impurity is larger than that of the intrinsic material, hence increasing the permittivity. Correct?

The binding energy (in the hydrogenic approximation), however, is proportional to the effective mass... so wouldn't a larger effective mass imply a higher binding energy and therefore more tightly bound electrons?

Just to make sure I'm following, polarizability is described by relative permittivity (dielectric constant). The dielectric constant decreases with increasing conductivity, specifically from the imaginary (loss) term. Conductivity is inversely proportional to effective mass, and therefore the polarizability and dielectric constant will increase with larger effective mass.

If I'm following what you're saying correctly, the argument for doping would be that the effective mass of the impurity is larger than that of the intrinsic material, hence increasing the permittivity. Correct?

The binding energy (in the hydrogenic approximation), however, is proportional to the effective mass... so wouldn't a larger effective mass imply a higher binding energy and therefore more tightly bound electrons?

First I don't see why you thinkt that the conductivity decreases with increasing conductivity.
The conductivity makes an imaginary contribution to permittivity and therefore you cannot deduce from the minus sign that it decreases permittivity. On the contrary, its absolute value increases.

Second, this formula holds only for conductors and I was arguing that your doped system is probably not conductive.

At least at small doping levels, the extra electrons will remain bound to the niobium ions. These extra valence electron of the niobium will reside in an orbital whose size can be calculated approximately using Bohrs formula. Metallic behaviour requires that these orbitals overlapp strongly.
If you can't follow me, you should try to look out for "Wannier-Mott excitons" in a good book.

citw
At least at small doping levels, the extra electrons will remain bound to the niobium ions. These extra valence electron of the niobium will reside in an orbital whose size can be calculated approximately using Bohrs formula.

Basically, my understanding is that the effect of Nb doping is an increase in the relative permittivity, but also an increase in the dielectric loss due to delocalized electrons created by Nb.

I found a thesis (in German) on the properties of TiO2:Nb
http://www.gbv.de/dms/ilmenau/toc/685168492.PDF
As I suspected, the Mott criterion for metallic conductivity is fulfilled only for very high doping, at least in rutile (cf. Tab. 2.6).
Is the modification of TiO2 and doping level mentioned in the article you cited?

citw
Is the modification of TiO2 and doping level mentioned in the article you cited?

One case is VERY low - about 0.5%. Reportedly, the permittivity increased by an order of magnitude. However, dielectric loss was "mostly higher than 0.1".

In this case, the authors mention that "delocalized electron transport related interfacial polarization is probably responsible for the observed colossal permittivity behavior, given the obvious grain boundary effects..."

So with minor Nb doping, both permittivity and dielectric loss increase.

The authors also mention that "the electrons in such shallow donor energy levels are virtually delocalized, and can thus freely hop at room temperature, giving high electronic conductivity corresponding to the high dielectric loss observed..."

This sounds like what I was thinking with respect to the loss-conductivity relationship... can you explain what they're describing here if not what I was referring to earlier? It sounds like there is a competition between "interfacial polarization" increasing the permittivity and "high conductivity" increasing dielectric loss. Does this make sense?

citw
I did some reading and found the following article: