A semiconducting material has a band gap of 1 eV. Acceptor impurities are doped into it which create acceptor levels 1 me V above the valence band. Assume that the transition from one energy level to the other is almost forbidden if kT is less than 1/50 of the energy gap. Also, if k T is more than twice the gap, the upper levels have maximum population. the temperature of the semiconductor is increased from 0 K. the concentration of the holes increases with temperature and after a certain temperature it becomes approximately constant. As the temperature is further increased, the hole concentration again starts increasing at a certain temperature. Find the order of the temperature range in which the hole concentration remains approximately constant.

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Solution

Given band gap = 1 eV

Net band gap after doping

$= (1 - 10^{- 3}) e V = 0.999 e V .$

According to the question,

$K T_{1} = \frac{0.999}{50}$

$T_{1} = \frac{0.999}{50 \times 8.62 \times 10^{- 5}}$

$= 231.78 = 231.0$

For the maximum limit,

$K T_{2} = 2 \times 0.999$

$T_{2} = \frac{2 \times 3 \times 10^{- 3}}{8.62 \times 10^{- 5}}$

$= \frac{2}{8.62} \times 10^{2} = 23.2$

Temperature range is (23.2-231.8).

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Q. Indium antimonide has a band gap of

0.23 eV

between the valence and the conduction band. Find the temperature at which the thermal energy

k T

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Given band gap = 1 eV Net band gap after doping =(1−10−3) eV=0.999 eV. According to the question, KT1=0.99950 T1=0.99950×8.62×10−5 =231.78=231.0 For the maximum limit, KT2=2×0.999 T2=2×3×10−38.62×10−5 =28.62×102=23.2 Temperature range is (23.2-231.8).