3.1.3 The Hydrodynamic Transport Model

In the hydrodynamic transport model, carrier temperatures are assumed to be different from the lattice temperature. The basic equations (3.2) through (3.4) are augmented by energy balance equations which determine the carrier temperatures. The current relations take the form

$$\mu_{n}^{} \left(\vphantom{\mathrm{grad}\left(\frac{E_{C}}{\mathrm{q}}-\psi\right) + \fra... ...N_{C,0}}{n}\cdot\mathrm{grad}\left(\frac{n\cdot T_{n}}{N_{C,0}}\right) }\right. \left(\vphantom{\frac{E_{C}}{\mathrm{q}}-\psi}\right. {\frac{E_{C}}{\mathrm{q}}} \psi \left.\vphantom{\frac{E_{C}}{\mathrm{q}}-\psi}\right) {\frac{\mathrm{k_{B}}}{\mathrm{q}}} {\frac{N_{C,0}}{n}} \left(\vphantom{\frac{n\cdot T_{n}}{N_{C,0}}}\right. {\frac{n\cdot T_{n}}{N_{C,0}}} \left.\vphantom{\frac{n\cdot T_{n}}{N_{C,0}}}\right) \left.\vphantom{\mathrm{grad}\left(\frac{E_{C}}{\mathrm{q}}-\psi\right) + \fra... ...N_{C,0}}{n}\cdot\mathrm{grad}\left(\frac{n\cdot T_{n}}{N_{C,0}}\right) }\right)$$ (3.7)

$$\mu_{p}^{} \left(\vphantom{\mathrm{grad}\left(\frac{E_{V}}{\mathrm{q}}-\psi\right) - \fra... ...N_{V,0}}{p}\cdot\mathrm{grad}\left(\frac{p\cdot T_{p}}{N_{V,0}}\right) }\right. \left(\vphantom{\frac{E_{V}}{\mathrm{q}}-\psi}\right. {\frac{E_{V}}{\mathrm{q}}} \psi \left.\vphantom{\frac{E_{V}}{\mathrm{q}}-\psi}\right) {\frac{\mathrm{k_{B}}}{\mathrm{q}}} {\frac{N_{V,0}}{p}} \left(\vphantom{\frac{p\cdot T_{p}}{N_{V,0}}}\right. {\frac{p\cdot T_{p}}{N_{V,0}}} \left.\vphantom{\frac{p\cdot T_{p}}{N_{V,0}}}\right) \left.\vphantom{\mathrm{grad}\left(\frac{E_{V}}{\mathrm{q}}-\psi\right) - \fra... ...N_{V,0}}{p}\cdot\mathrm{grad}\left(\frac{p\cdot T_{p}}{N_{V,0}}\right) }\right)$$ (3.8)

The energy balance equations state conservation of the average carrier energies. In terms of the carrier temperatures, T_n and T_p, they can be written as

$$\left(\vphantom{\frac{E_{C}}{\mathrm{q}} -\psi}\right. {\frac{E_{C}}{\mathrm{q}}} \psi \left.\vphantom{\frac{E_{C}}{\mathrm{q}} -\psi}\right) {\frac{3\cdot \mathrm{k_{B}}}{2}} \left(\vphantom{\frac{\partial \,(n\cdot T_{n})}{\partial t} + R\cdot T_{n} + n\cdot\frac{T_{n} -T_{L}}{\tau_{\epsilon,n}}}\right. {\frac{\partial \,(n\cdot T_{n})}{\partial t}} {\frac{T_{n} -T_{L}}{\tau_{\epsilon,n}}} \left.\vphantom{\frac{\partial \,(n\cdot T_{n})}{\partial t} + R\cdot T_{n} + n\cdot\frac{T_{n} -T_{L}}{\tau_{\epsilon,n}}}\right)$$ div S_n = (3.9)

$$\left(\vphantom{\frac{E_{V}}{\mathrm{q}} -\psi}\right. {\frac{E_{V}}{\mathrm{q}}} \psi \left.\vphantom{\frac{E_{V}}{\mathrm{q}} -\psi}\right) {\frac{3\cdot \mathrm{k_{B}}}{2}} \left(\vphantom{\frac{\partial \,(p\cdot T_{p})}{\partial t} + R\cdot T_{p} +p\cdot\frac{T_{p} -T_{L}}{\tau_{\epsilon,p}}}\right. {\frac{\partial \,(p\cdot T_{p})}{\partial t}} {\frac{T_{p} -T_{L}}{\tau_{\epsilon,p}}} \left.\vphantom{\frac{\partial \,(p\cdot T_{p})}{\partial t} + R\cdot T_{p} +p\cdot\frac{T_{p} -T_{L}}{\tau_{\epsilon,p}}}\right)$$ div S_p = (3.10)

Here, $\tau_{\epsilon,n}^{}$ and $\tau_{\epsilon,p}^{}$ denote the energy relaxation times, while S_n and S_p are the energy fluxes.

$$\kappa_{n}^{} {\textstyle\frac{5}{2}} {\frac{\mathrm{k_{B}}\cdot T_{n}}{\mathrm{q}}}$$ S_n = (3.11)

$$\kappa_{p}^{} {\textstyle\frac{5}{2}} {\frac{\mathrm{k_{B}}\cdot T_{p}}{\mathrm{q}}}$$ S_p = (3.12)

The thermal conductivities, $\kappa_{n}^{}$ and $\kappa_{p}^{}$, are assumed to obey a generalized Wiedemann-Franz law [54].

$$\kappa_{n}^{} \left(\vphantom{\frac{5}{2} + c_{n}}\right. {\textstyle\frac{5}{2}} \left.\vphantom{\frac{5}{2} + c_{n}}\right) {\frac{\mathrm{k_{B}}^{2}}{\mathrm{q}}} \mu_{n}^{}$$ = (3.13)

$$\kappa_{p}^{} \left(\vphantom{\frac{5}{2} + c_{p}}\right. {\textstyle\frac{5}{2}} \left.\vphantom{\frac{5}{2} + c_{p}}\right) {\frac{\mathrm{k_{B}}^{2}}{\mathrm{q}}} \mu_{p}^{}$$ = (3.14)