The voltage V_0 in the network shown is | vedclass

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Question

(A) In the given circuit, there are two diodes connected in parallel: a Silicon $(Si)$ diode with a threshold voltage of $0.7 \, V$ and a green $LED$

Vedclass · Accepted Answer

$V_0=11.3 \, V$

Vedclass · Answer

(A) In the given circuit, there are two diodes connected in parallel: a Silicon $(Si)$ diode with a threshold voltage of $0.7 \, V$ and a green $LED$ with a threshold voltage of $2.2 \, V$.When a voltage is applied, the diode with the lower threshold voltage will conduct first.Since the threshold voltage of the $Si$ diode $(0.7 \, V)$ is less than that of the $LED$ $(2.2 \, V)$, the $Si$ diode will become forward-biased and conduct, while the $LED$ remains effectively off (open circuit).Therefore, the circuit behaves as if only the $Si$ diode is present in parallel with the $12 \, V$ source.The voltage $V_0$ across the $2.2 \, k\Omega$ resistor is determined by the voltage drop across the $Si$ diode.$V_0 = 12 \, V - 0.7 \, V = 11.3 \, V$.

The voltage $V_0$ in the network shown is

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