Ge and Si diodes conduct at 0.3,V and 0.7,V r | vedclass

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(B) In the given circuit,the $Ge$ and $Si$ diodes are connected in parallel. Since they are in parallel,the diode with the lower threshold voltage wil

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$0.4$

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(B) In the given circuit,the $Ge$ and $Si$ diodes are connected in parallel. Since they are in parallel,the diode with the lower threshold voltage will conduct first.Case $1$: Initially,both diodes are connected such that they are forward-biased. The $Ge$ diode has a threshold voltage of $0.3\,V$ and the $Si$ diode has $0.7\,V$. The circuit will be dominated by the $Ge$ diode,so the voltage drop across the parallel combination is $V_{drop} = 0.3\,V$.The output voltage $V_0$ is the voltage across the $5\,k\Omega$ resistor:$V_0 = V_{source} - V_{drop} = 12\,V - 0.3\,V = 11.7\,V$.Case $2$: If the $Ge$ diode connection is reversed,it becomes reverse-biased and does not conduct. Now,only the $Si$ diode is forward-biased.The voltage drop across the circuit is now determined by the $Si$ diode,which is $V_{drop} = 0.7\,V$.The new output voltage $V_0'$ is:$V_0' = V_{source} - V_{drop} = 12\,V - 0.7\,V = 11.3\,V$.The change in the value of $V_0$ is:$\Delta V_0 = |V_0 - V_0'| = |11.7\,V - 11.3\,V| = 0.4\,V$.

$Ge$ and $Si$ diodes conduct 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......$V$.

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