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

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Question

(B) Case $1$: Initially,both diodes are forward biased. The circuit will conduct through the diode with the lower threshold voltage,which is the $Ge$

Vedclass · Accepted Answer

$0.4$

Vedclass · Answer

(B) Case $1$: Initially,both diodes are forward biased. The circuit will conduct through the diode with the lower threshold voltage,which is the $Ge$ diode $(0.3 \ V)$.The voltage across the resistor $V_0 = 12 \ V - 0.3 \ V = 11.7 \ V$.Case $2$: When the connections are reversed,both diodes become reverse biased. In a standard idealization for such problems,reversing the diodes means they are now connected in a way that they would conduct if the battery polarity were reversed,or effectively,they are now blocking the current in the current configuration. However,interpreting the question as the diodes being swapped or the circuit being reconfigured such that the $Si$ diode now dictates the drop:If the diodes are reversed,the current path is blocked. Assuming the question implies the diodes are replaced by their counterparts or the circuit is modified such that the $Si$ diode $(0.7 \ V)$ is now the one conducting:The new output voltage $V_0' = 12 \ V - 0.7 \ V = 11.3 \ V$.The change in $V_0 = |11.7 \ V - 11.3 \ V| = 0.4 \ V$.

$Ge$ and $Si$ diodes conduct at $0.3 \ V$ and $0.7 \ V$ respectively. In the following figure,if the $Ge$ and $Si$ diode connections are reversed,the value of $V_0$ changes by .....$V$.

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