Here $E_c$ is the energy of an electron at the bottom of the conduction band and $E_v$ is the energy of an electron at the top of the valence band.
Observe that because we are describing particles in the valence band as holes, $m_h > 0$ and $E_h > -E_v$.
Lastly, it is important to point out the notation that we adopt here.
In the above plot, $m_h \neq -m_e$ because it refers to two **different** bands: valance band for holes and conduction band for electrons.
In most literature, the band indices are neglected such that $m_{h,v} \to $m_{h}$ and $m_{e,c} \to m_e$.
In most literature, the band indices are neglected such that $m_{h,v} \to m_{h}$ and $m_{e,c} \to m_e$.
??? question "a photon gives a single electron enough energy to move from the valence band to the conduction band. How many particles does this process create?"
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@@ -221,7 +222,7 @@ Since the hole energy is opposite $E_h = -E$, we replace the Fermi energy $E_F \
In the third step, we need to solve the equation under charge balance $n_e = n_h$.
The equation is not a pleasant one and cannot be solved analytically unless an approximation is made.
Therfore, the fourth step assumes that the Fermi level is far from both bands $E_F-E_v \gg kT$ and $E_c - E_F \gg kT$.
Therefore, the fourth step assumes that the Fermi level is far from both bands $E_F-E_v \gg kT$ and $E_c - E_F \gg kT$.
As a result, the Fermi-Dirac distribution is approximately similar to Boltzmann distribution:
