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 in parallel. The diode with the lower threshold voltage will conduct first. Since $Ge$ has a thr

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$0.4$

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(B) In the given circuit,the $Ge$ and $Si$ diodes are in parallel. The diode with the lower threshold voltage will conduct first. Since $Ge$ has a threshold voltage of $0.3\,V$ and $Si$ has $0.7\,V$,the $Ge$ diode conducts.Case $1$: Initial connection.The voltage drop across the circuit is determined by the $Ge$ diode,$V_{drop} = 0.3\,V$.The current $i = \frac{12\,V - 0.3\,V}{5\,k\Omega} = \frac{11.7\,V}{5\,k\Omega} = 2.34\,mA$.The output voltage $V_{o1} = i 	imes R = 2.34\,mA 	imes 5\,k\Omega = 11.7\,V$.Case $2$: $Ge$ diode connection is reversed.Now,the $Ge$ diode is reverse-biased and does not conduct. The $Si$ diode is forward-biased and conducts with a threshold voltage of $0.7\,V$.The current $i = \frac{12\,V - 0.7\,V}{5\,k\Omega} = \frac{11.3\,V}{5\,k\Omega} = 2.26\,mA$.The output voltage $V_{o2} = i 	imes R = 2.26\,mA 	imes 5\,k\Omega = 11.3\,V$.The change in the value of $V_0$ is $\Delta V_0 = |V_{o1} - V_{o2}| = |11.7\,V - 11.3\,V| = 0.4\,V$.

$Ge$ 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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