I bumped into a few paper covering the topic of electrical conductivity in spinel NiCo2S4. However, the two seem to be arguing the opposite.
One read:
"In general, the electrical conductivity of the NiCo2S4 phase is closely dependent upon the mass of Co(II) and Ni(III), which can render extra holes as p-type doping and electrons as n-type doping, respectively "
The other read:
"Theoretically, the presence of Ni3+ will provide extra electrons as n-type doping while the presence of Co2+ will result in extra holes as p-type doping. Therefore, more Ni3+ and Co3+ will lead to higher electrical conductivity for electrons and better electrochemical performance "
Hence, my question relates to a material with high but equal amounts of p and n-type dopants - such a material would not provide any advantage in conductivity, correct?
Dear Yijie,
if you have 2 types of dopands (acceptor like states and donor like states), I expect the formation of donor-acceptor pairs. Such pairs don't contribute to enhancement of conductivity due to following reasens:
The conductivity follows from sigma = e*n(p)*µ (µ mobility, n,p remaining carrier concentration).
- The carrier concentration of one type (n) is neutralized by the other (p).
- The mobility is reduced remarkable by scattering at donor-acceptor pairs
Both effects reduce the conductivity. So we can conclude:
Your conclusion concerning the conductivity is fully correct.
With Regard
R. Mitdank
The simple expression of conductivity based on the compensation doping is valid for unary semiconductors like Si, Ge etc. I am doubtful about the validity of compensation doping in binary semiconductors such as oxides, sulfides etc. The change in the transition metal cation valence states significantly alters the M-X bonding, orbital interaction and associated energy levels. The charge transfer depends on the overlapping of the orbitals. Prediction should be based on the electronic structure calculations.
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