As the concentration of donor atoms $n_{d}$ is varied the chemical potential (or Fermi level) $E_{F}$ changes and so the concentration of electrons $n_{e}$
in the conduction band changes. Equation 4.11 in your book is the formula for the concentration of electrons in the conduction band. It applies to an intrinsic semiconductor and also to an impurity semiconductor. However, the chemical potential $E_{F}$ changes in each situation.
For an n-doped material with a concentration $n_{d}$ of donors (Phosphorus atoms, say) the concentrations of electrons in the conduction band $n_{e}$ is your formula 4.11,
\begin{equation}
n_{e}=N_{c}\exp(-(E_{c}-E_{F})/k_{B}T) \ .
\end{equation}
Electrons from the neutral donor atoms move to the conduction band leaving positively ionized donors at concentration $n_{d+}$,
\begin{equation}
n_{d+}=\frac{n_{d}}{1+\exp(-(E_{d}-E_{F})/k_{B}T)}\sim n_{d}
\end{equation}
At $T\sim300K$ the donors are almost all positively ionized. Some electrons from the valence band also move to the conduction band leaving holes in the valence band at concentration $n_{h}$.
\begin{equation}
n_{h}=N_{v}\exp(-(E_{F}-E_{v})/k_{B}T)
\end{equation}
The crystal is electrically neutral so that the concentration of holes (positively charged) in the valence band and positively ionized donors
balances the negative charge of the electrons in the conduction band.
\begin{equation}
n_{h}+n_{d+}=n_{e}
\end{equation}
Substituting for the concentrations, but neglecting the contribution from the holes since $n_{h}<<n_{d}$,
\begin{equation}
n_{d}\sim N_{c}\exp(-(E_{c}-E_{F})/k_{B}T)
\end{equation}
the solution for the chemical potential is,
\begin{equation}
E_{F}=E_{C}+k_{B}T\ln(n_{d}/N_{c})
\end{equation}
As an example, take the zero of energy as the bottom of the conduction band $E_{C}=0$. Consider Si at $T=300K$. Assume the donors are phosphorus atoms. The donor level $E_{d}=-0.044$eV. $N_{c}=2.4\times 10^{19}$ cm$^{-3}$ , $N_{v}=1.3\times 10^{19}$ cm$^{-3}$. Let the donor concentration be $n_{d}=10^{16}$cm$^{-3}$. The chemical potential is $E_{F}=-0.201$eV. The concentration of positively ionized donors is $n_{d+}=9.977\times 10^{15}$cm$^{-3}$.The concentration of electrons in the conduction band is $n_{e}=2.092\times 10^{17}$cm$^{-3}$.
Now increase the donor concentration to $n_{d}=10^{17}$cm$^{-3}$. The chemical potential is now $E_{F}=-0.142$eV.The concentration of positively ionized donors is $n_{d+}=9.777\times 10^{16}$cm$^{-3}$.The concentration of electrons in the conduction band is $n_{e}=6.615\times 10^{17}$cm$^{-3}$.
