$\Lambda$CDM and Holographic Dark Energy Holographic dark energy is usually written $$\rho_{\rm DE} = 3M_p^2\frac{c^2}{L^2}$$ where $c$ is a dimensionless HDE parameter and $L$ is an infrared cutoff. In Li's 2004 model, $$L=L_e \equiv a(t)\int_t^\infty\frac{dt'}{a(t')}$$ is the future event horizon. Li assumed constant $c$, giving $$w_{\rm DE} = -\frac13 \left( 1+\frac{2\sqrt{\Omega_{\rm DE}}}{c} \right).$$ However, with $c=1$ and $\Omega_{{\rm DE},0}=0.6889$, $w_0\simeq-0.887$, rather than $w=-1$. Does not match observations! Also, constant-$c$ observational fits often give $c<1$, leading to phantom evolution, null-energy-condition violation and a turning point in $H(z)$. That conclusion assumes $c'=0$. What about if $c$ is not constant? Then: $$c<1, \qquad c'=c-\sqrt{\Omega_\Lambda}>0, \qquad w=-1.$$ Therefore $\dot H = -4\pi G(\rho_m+\frac43\rho_r) \le0$, so there is no finite-redshift turning point and no phantom phase. Thus ...
Yes! Also, it is not the 'Planck Power' (despite what you might have read in Misner, Thorne and Wheeler, P.980). The existence of black hole horizons implies a maximum luminosity (power) limit in General Relativity. Not even gravitational waves can escape a black hole . Consider an (almost) black hole made of light (this is called a Kugelblitz ) sphere of radius \begin{equation} \notag R \geq \frac{2Gp}{c^3} \end{equation} which is filled with photons with a total mass-energy of momentum $p$ times speed of light $c$ \begin{equation} \notag E=p \ c \end{equation} that leave after a time \begin{equation} \notag t=R/c \end{equation} with average power (luminosity) \begin{equation} \notag P_{max} = \frac{E}{t}=\frac{p \ c^2}{R}=\frac{c^5}{2G} \approx 1.8\times10^{52} \ W \end{equation} This is maximum power in GR , regardless of the nature of the system. You might be tempted to call this half a 'Planck Power' but there is no $\h...