question_answer

If in a p-n junction diode, a square input signal of 10 V is applied as shown                                              Then the output signal across \[{{I}_{AC}}={{I}_{EF}}\] will De

A)

B)

C)                

D)                

Correct Answer: D

Solution :

Key Idea For negative potential the diode becomes reverse biased whereas for positive potential diode is forward biased. For            \[=\frac{m{{d}^{2}}}{12}+\frac{m{{d}^{2}}}{4}\left( as\,{{I}_{EF}}=\frac{m{{d}^{2}}}{12} \right)\] the diode is reverse biased and hence offer infinite resistance, so circuit would be like as shown in Fig. (2) and \[{{I}_{AD}}=\frac{m{{d}^{2}}}{3}=4{{I}_{EF}}\] For \[{{V}_{i}}<0\], the diode is forward biased and circuit would be as shown in Fig. (3) and \[{{V}_{0}}=0\] Hence, the optical is correct.