Q2. What is the effect of the GLAD Ti film on the electron mean free path?
Since a very high porous architecture is especially favored for grazing incident angles of the particle flux (α > 70°), the number of electrical pathways available for the free carriers are reduced leading to a decrease of the electron mean free path.
Q3. Why is scattering particularly favored at the column interfaces?
Due to the high porous structure of the GLAD films [11], scattering is particularly favored at the column interfaces reducing the electronic transport properties.
Q4. What is the effect of the shadowing effect on the GLAD metallic films?
In addition, GLAD deposition of columnar films with incident angles close or higher than 60° gives rise to a significant structural and uniaxial anisotropy in the substrate plane induced by the shadowing effect [19].
Q5. How much resistivity is there at room temperature?
For the first cycle, the return to the room temperature gives rise to a linear evolution of ρ versus T and the resistivity is down to ρ298 = 7.24×10-6
Q6. What is the effect of the oxidation on the surface of the Ti film?
The oxidation only occurs on the surface of the conventional Ti film (α = 0°) whereas it becomes significant through the thickness of the GLAD Ti film due to its high porous structure [15].
Q7. What is the reason for the increase in resistivity?
In the end, the concentration of hopping sites falls below the threshold for percolation and the resistivity becomes too high to be measured by the vdP technique.
Q8. How can the van der Pauw method be applied to anisotropic materials?
Price [6] showed that the method may be applied to anisotropic materials provided that one of the principal axes of the resistivity tensor is perpendicular to the plane of the sample.
Q9. What is the resistance of a TiO1+-like compound?
It supports the formation of a TiO1+ε-like compound since the resistivity ρyy remains in the order of magnitude of metallic compounds.
Q10. How does the resistivity of the GLAD film change?
Further annealing treatments (2nd to 5th cycles) reduce the metallic-like character of the film since the resistivity reaches ρ298 = 1.83×10-5
Q11. How does the resistivity of titanium change with temperature?
A further increase of the temperature up to 473 K leads to a non linear evolution of the resistivity versus temperature since ρ473K reaches 8.11×10-6
Q12. What is the resistivity of the GLAD Ti film?
it should be reminded that their system is composed of Ti stacks where two parallel strips of Ti deposited by conventional sputtering (α = 0°) cover the GLAD Ti film (α = 80°).
Q13. What is the important jump of the resistivity?
This last annealing cycle leads to the most important jump of the electrical resistivity since 3 orders of magnitude separate resistivity values measured between the 8th and 9th cycle (axis break in Fig. 2).
Q14. What is the effective anisotropy of the vdP?
The anisotropy in their vdP configuration AvdP is defined as [7,9]:( )[ ] ( )[ ]∑ ∑ + + = − = −− × × == oddn eff oddn eff xx yy vdP nAn nAn R R A 1 1 1 sinh sinh π π(4)This equation (4) is used to calculate the effective anisotropy Aeff.
Q15. What is the effect of the van der Pauw method on the GLAD Ti film?
An extension ofthe van der Pauw method has been applied to get resistivities ρxx and ρyy in the mutually perpendicular directions x,y of the two-dimensional plane and the anisotropic coefficient during incremental annealing cycles in air (from 298 to873K).