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Draw V-I characteristicsof a p-n junction diode. Answer the following questions, giving reasons: (i)Why is the current under reverse bias almost independent of the applied potential upto a critical voltage ? (ii) Why does the reverse current show a sudden increase at the critical voltage ? Name any semiconductor device which operates under the reverse bias in the breakdown region. |
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Answer» Solution :The V-I characteristics of a p-n junction diode in forward bias and REVERSE bias arrangement have been shown in . (i) In reverse bias the junctionwidth increases. The higher junction potential restricts the flow of majority charge carriers. HOWEVER, such a field favours flow of minority carriers. Thus, reverse bias current is due to flow of minority carriers only. Since the NUMBER of minority carriers very small, the current is small and ALMOST independent of the applied potential upto a critical (before breakdown) voltage. (ii)At a critical (breakdown) voltage the reverse bias current shows a sudden increase. Under high reverse bias, the high junction field may strip an electron from the valence band, which can tunnel to the n-side through the thin depletion layer. This mechanism of emission of electrons after a critical applied voltage LEADS to a high reverse (breakdown) current. A zener diode operates under the reverse bias in the breakdown region. |
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