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NCERT Class XII Chapter
Semiconductor Electronics
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Question : 13 of 19
Marks: +1, -0
In an intrinsic semiconductor the energy gap EgE_g is 1.2 eV. Its hole mobility is much smaller than electron mobility and independent of temperature. What is the ratio between conductivity at 600 K and that at 300 K? Assume that the temperature dependence of intrinsic carrier concentration nin_i is given by nin_i = n0n_0 exp (−Eg2kBT)\left(-\frac{E_g}{2k_BT}\right) , where n0n_0 is a constant.
kBk_B = 8.62 × 10−510^{-5} eV K−1K^{-1}
Solution:  
Conductivity is given by
σ = e (neμe+nhμhn_e \mu_e + n_h \mu_h)
For intrinsic semiconductor nen_e = nhn_h = nin_i
Also mobility of holes (μh\mu_h) << mobility of electrons (μe\mu_e)
So, conductivity σ = eneμeen_e\mu_e
Temperature dependence of intrinsic carrier concentration
nin_i = n0e−Eg2kBTn_0 e^{-\frac{E_g}{2k_BT}}
So, conductivity, σ = eniμeen_i\mu_e = eμen0e−Eg2kBTe\mu_en_0 e^{-\frac{E_g}{2k_BT}}
where eμen0e\mu_en_0 = constant σ0\sigma_0
Hence , σ = σ0e−Eg2kBT\sigma_0 e^{-\frac{E_g}{2k_BT}}
Conductivity at 600 K, σ1\sigma_1 = σ0e−1.2 eV2kB(600)\sigma_0 e^{\frac{-1.2\,\mathrm{eV}}{2k_B(600)}} ... (i)
Conductivity at 300 K, σ2\sigma_2 = σ0e−1.2 eV2kB(600)\sigma_0 e^{\frac{-1.2\,\mathrm{eV}}{2k_B(600)}} ... (ii)
Dividing (i) and (ii)
σ1σ2\frac{\sigma_1}{\sigma_2} = e−[0.6 eV600kB−0.6 eV300kB]e^{- \left[ \frac{0.6\,\mathrm{eV}}{600k_B} - \frac{0.6\,\mathrm{eV}}{300k_B} \right]} = e+0.68.62×105[1600]e^{+ \frac{0.6}{8.62 \times 10^5} \left[ \frac{1}{600} \right]} or σ1σ2\frac{\sigma_1}{\sigma_2} = e11.6e^{11.6} = 1 × 10510^5
So, σ600 K\sigma_{600\,\mathrm{K}} = 10510^5 σ300 K\sigma_{300\,\mathrm{K}}
Conductivity increases rapidly with the rise of temperature.
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