What limits the maximum attainable Fermi Energy for a material experimentally?

Either through doping or gating. What are some good terms to search for if I'm looking for some experimentally obtained values for particular materials? I'm particularly interested in what the limit is for graphene, if anyone knows.

For example, the DOS for regular graphene has states between -3t and +3t, but what is the maximum fermi energy we can examine experimentally? Is there a breakdown of the sample at some limit of doping or voltage?

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if you don't mind could you please tell me what is still bothering you in the provided explanation? Alternatively, what is the key information missing here? –  NanoPhys Mar 20 '13 at 23:06

In the atomic ground state a carbon atom has the electronic configuration $1s^22s^22p^2$. In the sp$^2$ hybridization the $2s$, $2p_x$, and $2p_y$ participate in the formation of the three $\sigma$ bonds and the $2p_z$ orbital forms a $\pi$ bond. According to molecular orbital theory this $2p_z$ state would form the bonding ($\pi$) and anti-bonding orbitals ($\pi^*$). Now, extending this to a crystalline structure we would get the valence and conduction bands corresponding to the bonding and anti-bonding orbitals respectively.The dispersion of these two bands is given by the well-known tight-binding formula (considering only nearest-neighbor hopping) $$E(k_{x},k_{y})=\pm t\sqrt{1+4\cos\left(\frac{\sqrt{3}}{2}k_{y}a\right)\cos\left(\frac{1}{2}k_{x}a\right)+4\cos^{2}\left(\frac{1}{2}k_{x}a\right)}$$ where $a=0.142$ nm and $t=2.7$ eV. Using the value of the lattice constant and using the fact that this is a honeycomb lattice it can be found that the areal density of carbon atoms is $3.9\times 10^{15}$ cm$^{-2}$. So if you start $p$-doping your graphene sample then in principle you could tune the Fermi energy low enough to remove $3.9\times 10^{15}$ electrons (for a $1$ cm$^{-2}$ sheet) from the valence band. In other words, strip the graphene sheet of all $2p_z$ electrons. So the lowest Fermi energy would be $E_F=-3t$. If you continue to remove more electrons you would probably start compromising the structural integrity of the lattice. Maintaining structural integrity is precisely the role of $\sigma$ bonds.

On the other hand when you $n$-dope your graphene sheet the limiting factor would be the work function. By definition the work function is the amount of energy an electron needs to gain in order to escape the solid. In theory you could start considering higher orbitals ($3s$, $3p$, $3d$, $4s$, etc.) and try to compute the corresponding bands all the way up to infinite energy. However, the electrons lying in bands above the work function wouldn't really be considered part of the crystal. In the case of graphene the value of the work function ($4.5$ eV) is the difference in energy between the Dirac point and the energy at which the electron is no longer bound to graphene.

One way of experimentally determining the amount and type of doping graphene is by observing the field effect. If know the capacitance per unit area ($C$) of the gate and we pass a current through the graphene sheet while sweeping the gate voltage, then the voltage at which we observe the Dirac point ($V_{\rm{Dirac}}$) can tell us the doping in the sample. Using the simple capacitor formula we can find the density of doping as $CV_{\rm{Dirac}}$. The sign of $V_{\rm{Dirac}}$ will tell you if it is $n$- or $p$-doped. If the electric field due to the gate voltage is pointing toward the graphene sheet when we achieve the Dirac point then the sample is $p$-doped and $n$-doped otherwise. A voltage always depends on what you define as your reference (ground). So it's best describing it in terms of electric fields without much knowledge of system details.

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Structural integrity is compromised when you empty the $\sigma$ orbitals OR fill the $\sigma^*$ orbitals, right? Therefore, it seems to me that you can break the bonds by doping enough either n or p. –  Steve B Mar 20 '13 at 19:23
At the point where doping (say by some atomic/molecular dopants) can start breaking sigma bonds I would say graphene is participating in a chemical reaction with the intended dopant! But if you use a gate then I'm not sure how easy it would be to break sigma bonds –  NanoPhys Mar 20 '13 at 23:02

This answer is specific to graphene experimental technology. (I'm not an expert on this, sorry for any mistakes.)

For gating graphene, the state-of-the-art method is an ion gel gate. These can dope to around 1e14 electrons / cm2 n or p. Here's a reference with 6e13, but different formulations are slightly better or worse. This is not a fundamental limit; if a more chemically stable ion gel were found, the doping could increase.

For chemical doping, you can get many times higher values by putting on lithium, calcium, or similar atoms. For example, here's bilayer graphene where they added two electrons per unit cell (if I understand correctly). Does it really count as "graphene" with such a high density of foreign atoms? Maybe ... I'm sure you can find discussions about it in the literature.

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