Question

In a semiconductor device, the Fermi-energy level is $0.35$ eV above the valence band energy. The effective density of states in the valence band at $T=300$ K is $1\times10^{19}\ \text{cm}^{-3}$. The thermal equilibrium hole concentration in silicon at $400$ K is ___________ $\times 10^{13}\ \text{cm}^{-3}$ (rounded off to two decimal places).
Given \(kT\) at \(300\) K is \(0.026\) eV.

Hint:

Remember $N_v\propto T^{3/2}$ and $kT\propto T$. For holes: $p_0=N_v\,e^{-(E_F-E_v)/kT}$.

Answer 1

For non-degenerate semiconductors, $p_0 = N_v\,\exp\!\left(-\dfrac{E_F-E_v}{kT}\right)$.
Temperature scalings: $N_v(T)=N_v(300)\left(\dfrac{T}{300}\right)^{3/2}$ and $kT(400)=0.026\left(\dfrac{400}{300}\right)=0.03467\ \text{eV}$.
Thus $N_v(400)=1\times10^{19}\left(\dfrac{400}{300}\right)^{3/2}=1.5396\times10^{19}\ \text{cm}^{-3}$.
With $E_F-E_v=0.35$ eV,
\[ p_0=1.5396\times10^{19}\exp\!\left(-\frac{0.35}{0.03467}\right) =6.349\times10^{14}\ \text{cm}^{-3}. \]
Expressed as $\times 10^{13}\ \text{cm}^{-3}$: $p_0=63.49\times10^{13}\ \text{cm}^{-3}$.
\[ \boxed{63.49} \]