InterviewSolution
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InterviewSolution
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A semiconductor is known to have an electron concentration of 8 xx 10^(13) cm^(-3) and a hole concentration of 5 xx 10^(12) cm^(-3) . a . Is the semiconductor n- type of p-type ? b. What is the resistitivity of the simple , if the electron mobility is 23,000 cm^(2) V^(-1) s^(-1) and hole mobility is 100 cm^(2) V^(-1) s^(-1) ? Take charge on electron , e = 1.6 xx 10^(-19) C . |
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Answer» Solution :Data SUPPLIED , `n_(e) = 8 xx 10^(13) cm^(-3) = 8 xx 10^(19) m^(-3)` `n_h = 5 xx 10^2 cm^(-3) = 5 xx 10^(18) m^(-3)` `mu_(e) = 23,000 cm^(2) V^(-1) s^(-1) = 2.3 m^(2) V^(-1) s^(-1)` `mu_(H) = 100 cm^(2) V^(-1) s^(-1) = 0.01 m^(2) V^(-1) s^(-1)` a. Here semiconductor has GREATER electron concentration . So it is n- type semiconductor . b. `(1)/(rho) = e(n_(e) mu_(e)+ n_(h) mu_(h)) = 1.6 xx 10^(-19) (8 xx 10^(19) xx 2.3 +5 xx 10^(18) xx 0.1) = 30 m HO m^(-1)` `rho = (1)/(30) = 3.3 xx 10^(-2) Omega m` |
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