Ge and Si diodes start conducting at 0.3,V an | vedclass

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(D) In the initial configuration,both diodes are forward-biased. The diode with the lower cut-in voltage will conduct first and determine the output v

Vedclass · Accepted Answer

$0.4$

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(D) In the initial configuration,both diodes are forward-biased. The diode with the lower cut-in voltage will conduct first and determine the output voltage. Since $Ge$ has a cut-in voltage of $0.3\,V$ and $Si$ has $0.7\,V$,the $Ge$ diode conducts,and the output voltage is $V_{0_i} = 12\,V - 0.3\,V = 11.7\,V$.When the $Ge$ diode connection is reversed,it becomes reverse-biased. Since it is assumed to have a large breakdown voltage,it acts as an open circuit. Now,only the $Si$ diode is forward-biased and conducts. The output voltage becomes $V_{0_f} = 12\,V - 0.7\,V = 11.3\,V$.The change in the value of $V_0$ is $|V_{0_f} - V_{0_i}| = |11.3\,V - 11.7\,V| = 0.4\,V$.

$Ge$ and $Si$ diodes start conducting at $0.3\,V$ and $0.7\,V$ respectively. In the following figure,if the $Ge$ diode connection is reversed,the value of $V_0$ changes by: ...... $V$ (assume that the $Ge$ diode has a large breakdown voltage).

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