3.1.2. Smoothness *γ* = *k*B*T*0

Also in the case where the barrier gets smoother, such that *γ* equals the base temperature, *γ* = *k*B*T*0, we can perform an analytical investigation. Focusing on the linear-response regime, Δ*T*/*T*0 1, where the effect of the smoothness is most pronounced, we can write the power in a compact form as

$$P = -\frac{eV}{h} \left\{ \left[ -\mathcal{N}(E\_0) - E\_0 \frac{d\mathcal{N}(E\_0)}{dE\_0} \right] eV - \frac{1}{2} E\_0^2 \frac{d\mathcal{N}(E\_0)}{dE\_0} \cdot \frac{\Delta T}{T\_0} \right\}.\tag{27}$$

where N (*E*) is the Bose-Einstein distribution function, N (*E*) = 2 exp[*E*/(*k*B*T*0)] − 13−1. As discussed above, in the linear-response regime the maximizing voltage is *V*max = *V*s/2, where the stopping voltage *V*s is the voltage that makes the expression in the curly bracket in Equation (27) vanish. Further maximizing over *E*0 then gives *<sup>E</sup>*0,max = 1.6 *k*B*T*0, which inserted into the power expression gives

$$P\_{\text{max}}^{V,E\_0} \approx 0.5 \, P\_{\text{W}}.\tag{28}$$

From Figure 3 it is clear that both *<sup>E</sup>*0,max and *P*V,*E*<sup>0</sup> max are in good agreemen<sup>t</sup> with the numerical result.
