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The following three properties are related to current flow:

  1. Band gap energy
  2. Dielectric constant
  3. Resistance

I would expect them all to have the same trend (i.e. higher band gap energy would cause higher dielectric constant and higher resistance), but this is not the case. For example:

  • SiO2 is a very good insulator (band gap of 9eV), but its dielectric constant is very low ($\epsilon_r=3.9$), compared to many materials. Why so?
  • In this ref (page 8) a consistent inverse correlation between $\epsilon$ and $\rho$ for Si is presented. Why is it inverse? I would expect them to rise together.
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  • $\begingroup$ As you have discovered, they are not necessarily related. As another example, the resistivity of Si can be varied (by doping) over orders of magnitude, while the band gap and dielectric constant remain, well, constant. Any generalization you can come up with will fail on pretty trivial counter-examples. $\endgroup$ – Jon Custer Dec 5 '14 at 20:55
  • $\begingroup$ But why are the three parameters independent of each other? Why band gap is not representing permittivity? $\endgroup$ – Sparkler Dec 5 '14 at 21:09
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    $\begingroup$ Well, I'll turn it around: Why should the band gap determine the dielectric constant, a dc parameter? The band gap certainly will have influence on the optical properties near the band gap, but otherwise... You are looking for a simple relationship where none does (or should) exist. $\endgroup$ – Jon Custer Dec 5 '14 at 21:54
  • $\begingroup$ @JonCuster, I've updated the question with an additional example. $\endgroup$ – Sparkler Dec 5 '14 at 23:29
  • $\begingroup$ Again, all you are doing is finding additional, easy examples of where reality does not correspond to your intuition. As we move beyond Newtonian physics (which is intuitive) we encounter these types of things in quantum, solid state, relativity, ... All it means is that you have to abandon intuition, dive in and learn it, and build appropriate intuition. If you look you will find a huge number of cases where your desired correlation does not hold. It does not hold because there is no physical reason for it to hold. $\endgroup$ – Jon Custer Dec 5 '14 at 23:36
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The starting point for this question is the Drude model for conductivity in metals. One can be a lot more rigorous than the Drude model (e.g. Quantum Boltzmann transport etc.), but ends up with the same result under many circumstances.

Before getting into the model, consider the following definitions for the frequency dependent conductivity $\sigma(\omega)$, dielectric function $\epsilon(\omega)$, and resistivity $\rho(\omega)$:

$$\mathbf{J}(\omega) = \sigma(\omega) \mathbf{E}(\omega) \implies \mathbf{E}(\omega) = \frac{1}{\sigma(\omega)}\mathbf{J}(\omega) \equiv \rho(\omega) \mathbf{J}(\omega)$$ $$\mathbf{D}(\omega) = \epsilon(\omega) \mathbf{E}(\omega)$$ $$\epsilon(\omega) = \epsilon_0 + \frac{i \sigma(\omega)}{\omega} = \epsilon_0 + \frac{i}{\omega \rho(\omega)}$$

In the Drude model, which simply assumes there is a charge density $n$ that supplies a current responding to an electric field. This current then decays with a characteristic time $\tau$. Mathematically, one starts with Newton's law for an electron in an electric field:

$$\mathbf{F}=m\mathbf{a}= m \frac{d\mathbf{p}}{dt} = -ne \frac{d\mathbf{J}}{dt}$$

Because $\mathbf{J} = -ne \mathbf{p}$, where it is assumed there is a electron density of $n$.

The Drude model assumes a specific form for the force on the electron $$\mathbf{F} = -e \mathbf{E} -\frac{\mathbf{p}}{\tau}$$

The first term is from the electric field, while the second is a friction/resistance term that represents the current decaying with a time constant $\tau$. One can solve the above equations in frequency space (with a Fourier transform) to get the following:

$$\sigma(\omega) =\left(\frac{n e^2 \tau}{m}\right) \frac{1}{1+i\omega \tau} = \sigma_0 \frac{1}{1+i\omega \tau} = \sigma_0 \left(\frac{1}{1+\omega^2 \tau^2} - i\omega \tau \frac{1}{1+\omega^2 \tau^2}\right)$$

So in the DC limit, you have the following $$\sigma(0) = \sigma_0 = \frac{n e^2 \tau}{m}$$ $$\rho(0) = \rho_0 = \sigma_0^{-1}$$ $$\epsilon = \epsilon_0+\sigma_0 \tau= \epsilon_0+\frac{\tau}{\rho_0}= \epsilon_0(1+\frac{n e^2 \tau^2}{\epsilon_0 m})$$

Thus, for metals, you have that $\epsilon \sim \frac{1}{\rho}$, or that the dielectric constant is inversely proportional to the resistance.

However, materials with bandgaps are NOT metals, and have no carriers ($n=0$) at zero temperature! At finite temperature though, one can say that $n\sim e^{-\frac{E_g}{k_b T}}$, so then $\epsilon \sim n \sim e^{-\frac{E_g}{k_b T}}$. This means that the larger the bandgap, the less free carriers, and the lower the dielectric constant. So even if the bandgap is held the same, but temperature is varied you can get different resistances and dielectric constants. Considering that undoped Silicon has a carrier density of about 12-13 orders of magnitude lower than a metal like copper, you can see why metals are considered to have $\epsilon \approx 10^{12}\sim \infty$ for practical purposes.

Now you might think you would be done since the relationship between the dielectric constant, band gap, and resistance are all shown above; however, the constants $\tau,m$ are unknown! The former comes from how electrons scatter in the system, while the latter comes from the type of orbital the electrons come from (to a first approximation). These two parameters vary for materials significantly. If you keep the material the same, but simply dope it, then the parameters do not change as significantly, which is why you notice get those trends for the dielectric constant, band gap, resistivity, etc. for Silicon.

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Dielectric constant is based on relative capacitance to vacuum. And it is usually calculated by using static electric plates. You put two parallel plates in vacuum and measure the capacitance. Then you do the same, but put the medium you want to measure instead. Divide the new capacitance of the new medium with the capacitance of vacuum:

$ϵ = C_x/C_v$

Since capacitance is charge divided by voltage:

$C=q/V$

The charge will be the same on both setup. Though the voltage will not. It depends on the medium between the two plates. If the molecules in the medium are polarized it is easier to mediate static electric field across the medium. That means more energy is stored in each charge. If not it is less easy to mediate static electric field and less energy stored per charge. In this case the latter. So a medium that is easier to mediate electric field has a lower dielectric constant and vice versa.

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  • $\begingroup$ The dielectric constant is not defined for DC regime only and nit just for dielectric in capacitors. It is a general property and in general a function of frequency. $\endgroup$ – nasu Apr 7 '17 at 3:46

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