Does the conductance of a semiconducter change when the Fermi level $E_F$ stays the same, but the distance to the valence band $E_V$ increases? When talking about n-type semiconductors and assuming roomtemperature: Does only the Fermi level determine the conductance of the semiconductor or also the energy difference between the Fermi level and the valence band? So, is there a difference in conductance, when you compare situation (a) to situation (b)?

 A: In general, in a semiconductor you have always both electrons in the conduction band and holes in the valence bands. Their densities are related by $$n·p=n_i^2$$ where the intrinsic density $n_i$ (which is the density of both electrons and holes when the Fermi level is about half the band gap) is related to the band gap $E_G$ by $$n_i^2=N_C\exp{-\frac{E_C-E_F}{kT}}N_V \exp{-\frac{E_F-E_V}{kT}}=N_c N_V \exp{-\frac{E_G}{kT}}$$ where $N_c$ and $N_V$ are the effective densities of state of the conduction and valence band, and $E_C$ and $E_V$ the conduction and valence band edges, respectively. Thus, the closer the Fermi level $E_F$ is to the conduction band, the higher the density of electrons there, and the lower the density of holes in the valence band. For usual n-type semiconductors like silicon with a not too small band gap, the Fermi level is very close to the conduction band, so that the hole density in the valence band is extremely small and thus can be neglected in the conductance. For example, at room temperature silicon has an intrinsic carrier density of about $$n_i=1·10^{10}cm^{-3}$$ When it has an n-type doping of $N_D=1·10^{16}cm^{-3}$, the electron concentration in the conduction band is $n≈1·10^{16}cm^{-3}$ and the hole density in the valence band is $p≈1·10^{4}cm^{-3}$ which is 12 orders of magnitude smaller than $n$.
