The contribution in the total current flowing | vedclass

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(A) The total current $I$ in a semiconductor is the sum of electron current $I_e$ and hole current $I_h$.Given that $\frac{I_e}{I} = \frac{3}{4}$ and

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$6 : 5$

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(A) The total current $I$ in a semiconductor is the sum of electron current $I_e$ and hole current $I_h$.Given that $\frac{I_e}{I} = \frac{3}{4}$ and $\frac{I_h}{I} = \frac{1}{4}$,the ratio of currents is $\frac{I_e}{I_h} = \frac{3/4}{1/4} = 3$.We know that current $I = n A q v_d$,where $n$ is the concentration,$A$ is the area,$q$ is the charge,and $v_d$ is the drift velocity.Thus,$\frac{I_e}{I_h} = \frac{n_e q A v_e}{n_h q A v_h} = \frac{n_e}{n_h} 	imes \frac{v_e}{v_h}$.Given $\frac{v_e}{v_h} = \frac{5}{2}$,we substitute the values: $3 = \frac{n_e}{n_h} 	imes \frac{5}{2}$.Solving for the ratio of concentrations: $\frac{n_e}{n_h} = 3 	imes \frac{2}{5} = \frac{6}{5}$.

The contribution in the total current flowing through a semiconductor due to electrons and holes are $\frac{3}{4}$ and $\frac{1}{4}$ respectively. If the drift velocity of electrons is $\frac{5}{2}$ times that of holes at this temperature,then the ratio of concentration of electrons and holes is

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