Module 6 : PHYSICS OF SEMICONDUCTOR DEVICES
Lecture 34 : Fermi Energy
Example 11
  Silicon crystal is doped with $ 5\times 10^{20}$ atoms per m $ ^3$. The donor level is 0.05 eV from the edge of the conduction band. Taking the band gap to be 1.12 eV, calculate the position of the Fermi level at 200 K.
  Solution
  The intrinsic carrier concentration can be obtained from the known carrier concentration in Si at 300 K. As the carrier concentration at 300 K is $ 1.5\times 10^{16}\ / {\rm m}^3$, the carrier concentration at 200 K is $ (200/300)^{3/2}\times 1.5\times 10^{16} = 0.82\times 10^{16}\ /{\rm m}^3$. As the doping concentration is much larger than $ n_i$, we can take $ n\simeq N_d= 5\times 10^{20}\ /{\rm m}^3$. Thus
 
$\displaystyle E_F^n-E_F^i = kT\ln\frac{n}{n_i} = 0.183\ \ {\rm eV}$
 

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