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प्रश्न
The drift current in a reverse-biased p-n junction is increased in magnitude if the temperature of the junction is increased. Explain this on the basis of creation of hole-electron pairs.
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उत्तर
When the temperature of a reverse-biassed p‒n junction is increased, the breaking of bonds takes place because of the increase in the thermal energy of the charge carriers. Drift current is due to the flow of the minority carriers across the junction. So, when a p‒n junction is reverse biassed, the applied voltage supports the flow of minority charge carriers across the junction. Thus, the drift current increases with increase in temperature in a reverse-biassed p‒n junction.
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संबंधित प्रश्न
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I = `"I"_0 exp ("eV"/(2"k"_"BT") - 1)`
where I0 is called the reverse saturation current, V is the voltage across the diode and is positive for forward bias and negative for reverse bias, and I is the current through the diode, kBis the Boltzmann constant (8.6×10−5 eV/K) and T is the absolute temperature. If for a given diode I0 = 5 × 10−12 A and T = 300 K, then
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