Module 6 : PHYSICS OF SEMICONDUCTOR DEVICES
Lecture 34 : p- type Semiconductors
Example 10
  Pure germanium has a band gap of 0.67 eV. It is doped with $ 3\times 10^{21}$ per m $ ^3$ of donor atoms. Find the densities of electrons and holes at 300 K. (effective masses $ m_e = 0.55m_0, \ m_h = 0.37m_0$)
  Solution
  For Ge, the intrinsic concentration is
 
$\displaystyle n_i = \sqrt{N_cN_v}e^{-\Delta/2kT}$
  Substituting given numerical values, $ n_i = 2.4\times 10^{19}\ \ /{\rm m}^3$. The density of donor atoms is $ N_d = 3\times 10^{21}$ /m $ ^3$. Thus the electron density $ n$ is given by
 
$\displaystyle n=\frac{N_d}{2}+\sqrt{\frac{(N_d)^2+4n_i^2}{4}} \simeq N_d\ \ \ {\rm as} N_d\gg n_i$
  Thus $ n = 3\times 10^{21}\ \ {/m}^3$. Using $ n_i^2 = np$, we get, for the density of holes $ p= 1.92\times 10^{17}\ \ /{\rm m}^3$.
 

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