The resistivity of a pure semiconductor is 0.5 Ωm. If the electron and hole mobility be 0.39 m²/V-s and 0.19 m²/V-s respectively then calculate the intrinsic carrier concentration.
No solution provided for this question.
The resistivity of a pure semiconductor is 0.5 Ωm. If the electron and hole mobility be 0.39 m²/V-s and 0.19 m²/V-s respectively then calculate the intrinsic carrier concentration.
No solution provided for this question.
A Ge specimen is doped with Al. The concentration of acceptor atoms is \[\sim 10^{21} atom/m^{3}\]. Given that the intrinsic concentration of electron hole pairs is \[\sim 10^{19}/m^{3}\], the concentration of electrons in the specimen is :
According to mass action law
\[ n_{e}\times n_{h}= n_{i}^{2} \]
\[ n_{h}=10^{21} ; n_{i}= 10^{19}; n_{e}=n_{i}^{2}/n_{h} = n_{e}= 10^{38}/10^{21}= 10^{17} \]
Intrinsic semiconductor is electrically neutral. Extrinsic semiconductor having large number of current carriers would be :
An extrinsic semiconductor with a large number of current carriers is still electrically neutral overall because the total number of positive and negative charges remains balanced.
The contribution in the total current flowing through a semiconductor due to electrons and holes are 3/4 and 1/4 respectively. If the drift velocity of electrons is 5/2 times that of holes at this temperature, then the ratio of concentration of electrons and holes is :
Current i = neAv
\[ \frac{I_{1}}{I_{2}}= \frac{n_{1}}{n_{2}} * \frac{v_{d1}}{v_{d2}} \]
\[ \frac{3}{1}= \frac{n_{1}}{n_{2}} * \frac{5}{2} \]
\[ \frac{n_{1}}{n_{2}} = \frac{6}{5} \]