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The concentration of hole-electron pairs in pure silicon at T=300K is 7xx10^(15)per cubic metre, Antimony is doped into silicon in a proportion of 1 atom in 10^7atons. Assuming that half of the impurity atons contribute electrons in the conduction band, calculate the foctor by which the number of silicon atoms per cubic metre is 5xx10^(28) |
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Answer» Solution :The number of charge carries beforedoping is equal to the number of holes plus the number of charge carriers per cubic metre before doping `2xx7xx10^(15)=14xx10^(15).` SINCE antimony is doped in a proportion of 1 in `10^(7)`,the number of antimony atoms per cubic metre is `10^(-7)xx5xx10^(28)=5xx10^(21)`.As HALF of these atoms contribute electrons to the conduction band,teh number of extra conduction electrons produced is `2.5xx10^(21)`per cubic metre after the doping is `2.5xx10^(21)+14xx10^(15)` `~~2.5xx10^(21)` The factor by which the number of charge is increased `=(2.5xx10^(21))/(14xx10^(15))=1.8xx10^(5).` In fact, as the n-type impurity is doped ,the number of holes and conduction electrons remains almost the same.However this does not AFFECT our result as the number of holes is anyway to small as compared to the number of conduction electrons. |
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