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The width of a depletion region is 400 nm. The intensity of the electric field at the depletion region is 5xx10^(5)V/m. Then calculate the following quantities: (1) The value of the potential barrier. (2) The minimum energy required by an electron to move from the n-type to the p-type region of the diode. |
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Answer» SOLUTION :`d=400 nm = 400xx10^(-9)m` `E=5xx10^(5)V//m` `1eV=1.6xx10^(-19)J` (1) `E=(V)/(d)` `THEREFORE V=E.d=5xx10^(5)xx400xx10^(-9)` `therefore V = 0.2V` (2) Kinetic energy`(1)/(2)mv^(2)=Ve` `=0.2xx1.6xx10^(-19)` `=3.2xx10^(-20)J` ` =(3.2xx10^(-20))/(1.6xx10^(-19))` Kinetic energy `=0.2eV` |
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