A potential difference of 2.5 ,V is applied a | vedclass

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(B) The current $I$ in a semiconductor is given by the formula: $I = A \cdot e \cdot (n_e \mu_e + n_h \mu_h) \cdot E$,where $E = \frac{V}{d}$.Given:Po

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$0.4$

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(B) The current $I$ in a semiconductor is given by the formula: $I = A \cdot e \cdot (n_e \mu_e + n_h \mu_h) \cdot E$,where $E = \frac{V}{d}$.Given:Potential difference $V = 2.5 \,V$Area $A = 1 \,cm^2 = 10^{-4} \,m^2$Thickness $d = 1.0 \,mm = 10^{-3} \,m$Electron concentration $n_e = n_h = 2 	imes 10^{19} \,m^{-3}$Mobility of electrons $\mu_e = 0.33 \,m^2/V \cdot s$Mobility of holes $\mu_h = 0.17 \,m^2/V \cdot s$Elementary charge $e = 1.6 	imes 10^{-19} \,C$Electric field $E = \frac{V}{d} = \frac{2.5}{10^{-3}} = 2500 \,V/m$Conductivity $\sigma = e(n_e \mu_e + n_h \mu_h) = (1.6 	imes 10^{-19}) 	imes (2 	imes 10^{19}) 	imes (0.33 + 0.17) = 3.2 	imes 0.5 = 1.6 \,S/m$Current $I = \sigma \cdot A \cdot E = 1.6 	imes 10^{-4} 	imes 2500 = 0.4 \,A$.

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