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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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

Vedclass · Accepted Answer

$0.4$

Vedclass · Answer

(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 $I = \sigma \cdot A \cdot E = 1.6 	imes 10^{-4} 	imes 2500 = 0.4 \,A$.

Similar Questions

If one indium atom is doped per $5 \times 10^7$ silicon atoms,and the number density of silicon atoms is $5 \times 10^{28} \, \text{atoms}/\text{m}^3$,find the number density of acceptor atoms in $\text{atoms}/\text{cm}^3$.

When the temperature of a semiconductor increases,then:

In an $n$-type semiconductor,the free electrons donated by the impurity atoms occupy energy levels in:

The valence of the impurity atom that is to be added to a germanium crystal to make it an $N$-type semiconductor is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module