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

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(B) In the given circuit,the $Ge$ and $Si$ diodes are connected in parallel. Since the $Ge$ diode has a lower threshold voltage $(0.3 \, V)$ compared

Vedclass · Accepted Answer

$0.4$

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

$Ge$ 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 how many volts?

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