Recently I was doing calculations on the formation energies of perovskite oxides(Ba0.5Sr0.5Co0.8Fe0.2O3) containing transition metals. I have a problem that has puzzled me for many days.
I understand the formation energy calculation formula: ( Ef = Etotal - 0.5E(Ba) - 0.5E (Sr) - 0.8E(Co) - 0.2E(Fe) - 3E(O) )
In the energy calculation of Ba0.5Sr0.5Co0.8Fe0.2O3, DFT + U is required due to the existence of Co and Fe elements.
So when I calculate E(Fe), do I need to introduce DFT + U ? If I use vasp software to calculate, how should I generate the structure of Fe to calculate E(Fe)?
Hi Zhang,
This is known to be a difficult problem, applying the Hubbard U correction to metallic Fe will usually not give an accurate answer. The simplest method is to choose a Hubbard U value which replicates formation energies accurately for known metal oxides, and hope it still does a good job for new metal oxides. Here's a short discussion of the problem, and the Materials Project database will also be a good place to find structure files for all the metal phases.
https://docs.materialsproject.org/methodology/calculation-details/gga+u-calculations
For oxygen, the appropriate reference energy will be 1/2 the energy of an isolated O2 molecule (with the triplet spin state). This will give the formation energy under oxygen-rich conditions, but please note that DFT does not calculate O2 accurately, so many researchers use the experimental binding energy of O2 instead. Typically I would use experimental entropy data from the JANAF tables if I wanted oxygen-poor conditions.
Steven Hartman
Thank you very much for your reply.
This seems complicated and I don't want to get formability through pure components now. I use four terminal compounds like BaCoO3, BaFeO3, SrFeO3, SrCoO3 to get the formation energy of BaxSr1-xCoyFe1-yO3, so there is only a question of formula balancing, what do you think of this approach?
That should give you a reasonable approximation of the convex hull energy, which is a different quantity than formation energy, but in my opinion it's actually more useful. If you have the time and computational power available, I would recommend checking the binary oxides as well just to see if they are a more stable alternative.
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