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P002: Restructure SPB (RSPB) Model (J. Zhu et al., 2021)

• J. Zhu, X. Zhang, M. Guo, J. Li, J. Hu, S. Cai, W. Cai, Y. Zhang, J. Sui, Restructured single parabolic band model for quick analysis in thermoelectricity, npj computational materials 7 (1) (2021) 1-8. https://doi.org/10.1038/s41524-021-00587-5

• GitHub rSPB respository. https://github.com/JianboHIT/rSPB

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$$ \left( {\frac{m^{\ast}}{m_{e}} \frac {T}{300 K}} \right) ^{3/2} = \frac {n}{n_{m,0}} \left[ {\exp \left( {\frac{S}{S_{0}} - 2} \right) - 0.1455} \right] $$

$$ \mu _{0} = \mu \cdot \left[ {1 + \frac {1}{2 \exp \left( {S / S_{0} - 2} \right) - 0.2910}} \right] ^{1/3} = \mu \cdot \left[ {1 + \left( {\frac{m^{\ast}}{m_{e}} \frac {T}{300 K}} \right) ^{-3/2} \frac{n}{2 n_{m,0}}} \right] ^{1/3} $$

$$ \mu _{WT} = \mu _{0} \left( {\frac{m^{\ast}}{m_{e}} \frac {T}{300 K}} \right) ^{3/2} $$

$$ +------------------------------------------+ $$

$$ PF_{opt}[\mu W / (cm \cdot K^{2})] = 0.1212 \mu _{WT}{ }[cm^{2} /(V \cdot s)] $$

$$ n_{opt}[10^{19}cm^{-3}]= 3.15 \left( {\frac{m^{\ast}}{m_{e}} \frac{T}{300 K}} \right) ^{3/2} $$

$$ +------------------------------------------+ $$

$$ n_{m,0} = \frac {2(2 \pi m_{e} k_{B} \cdot 300 K) ^{3/2}}{h^{3}} = 2.5094 \times 10^{19} cm^{-3} $$

$$ n = n_{m,0} \left( \frac{m^{\ast}}{m_{e}} \frac{T}{300 K} \right) ^{3/2} n_{r} $$

$$ n_{r} = \frac {2}{\sqrt {\pi}} F_{1/2} (\eta) $$

$$ +------------------------------------------+ $$

$$ S = S_{0} S_{r} $$

$$ S_{0} = \frac {k_{B}}{q} = 86.1733 \mu V/K $$

$$ S_{r} = \frac {2 F_{1} ( \eta ) }{F_{0} (\eta)} - \eta = \ln \left( 1.075 + \frac{e^{2}}{n_{r}} \right) $$

$$ +------------------------------------------+ $$

$$ \mu = \mu _{0} \mu _{r} $$

$$ \mu _{0} = \frac {(8 \pi) ^{1/2} \hbar ^{4} q C_{ii}}{3 m_{I}^{\ast} (m_b^{\ast} k_{B} T) ^{3/2} \Xi ^{2}} $$

$$ \mu _{r} = \frac {\sqrt {\pi}}{2} \frac {F_{0} (\eta)}{F_{1/2} (\eta)} = \left( 1 + \frac {n_{r}}{2} \right) ^{-1/3} $$

$$ +------------------------------------------+ $$

$$ L = L_{0} L_{r} $$

$$ L_{0} = \left( \frac {k_{B}}{q} \right) ^{2} = 0.7426 \times 10^{-8} W \Omega / K^{2} $$

$$ L_{r} = 3 \frac {F_{2} (\eta)}{F_{0} (\eta)} - 4 \left( \frac {F_{1} (\eta)}{F_{0} (\eta)} \right) ^{2} = 2 + \frac {\pi ^{2} / 3 - 2}{\left[ {1+(n_{r} / 2 \pi) ^{-3/2}} \right] ^{3/2}} $$

$$ +------------------------------------------+ $$