(3 pts) 1. In a reverse-biased ideal pn-junction diode, the current is small and

Question

(3 pts) 1. In a reverse-biased ideal pn-junction diode, the current is small and independent of the bias because A) the electric field in the depletion region stays constant B) of a limited supply of carriers to be drifted by the electric field C) of series resistance of the neutral regions D) of high-level injection (5 pts) 2. An abrupt Si p+n diode with No=2x1015 cm-3, ,-1 us, D,-25 crm'Ns has a breakdown voltage of 80 V (i) The breakdown mechanism is (2 pts) A) generation-recombination B) tunneling C) the Zener effect D) avalanching (i) If No is increased to 4x1015 cm,the new breakdown voltage will be (3pts) A) 80 V B) 40 V C) 160 V D) can't be determined (3 pts) 3. In a forward biased rectifying metal to n-type semiconductor, the dominant current flow A) is electrons flowing from the metal to the semiconductor B) is electrons flowing from the semiconductor to the metal C) is recombination of electron-hole-pairs in the depletion region of the semiconductor D) none of the above (3 pts) 4. In a pn-junction diode, the absolute value of the electric field is largest A) right at the metallurgical junction B) at the depletion region edge on the n-side C) at the depletion region edge on the p-side D) depends on bias (3 pts) 5. To make an ohmic contact to a p-type semiconductor, one could A) Select a metal with a workfunction smaler than that of the semiconductor B) Select a metal with a workfunction larger than that of the semiconductor C) Use a lightly doped semiconductor D) Insert a thin insulating layer under the metal

Explanation / Answer

Ans 1 b is a correct option i.e limited supply of charge carrier,

In reverse biased ,all the majority carriers will store at the corners of p and n type region and they are not able to cross the junction due to potential barrier due to which very few charge carriers reach on the otherside their fore less current flows

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