Semiconductor Electronics Part 3

Concept:In forward bias, a silicon diode has a constant voltage drop of $0.7 V$.Diodes in forward bias conduct current; reverse bias blocks current.Explanation:Diodes $D_1$ and $D_3$ are forward biased because their anodes connect to the positive source terminal.Diode $D_2$ is reverse biased because its cathode connects to positive, so it acts as an open circuit with no current.The forward voltage drop across $D_1$ and $D_3$ is $0.7 V$ each, but since they are in parallel, the voltage drop is same: $0.7 V$ from the junction to ground.Voltage across resistor $R_1 = 0.3 k\Omega$ is source voltage minus diode drop: $V_{R1} = 12 V - 0.7 V = 11.3 V$.Total current through $R_1$: $I_{total} = \frac{11.3 V}{0.3 k\Omega} = 37.67 mA \approx 37.7 mA$.This total current reaches the parallel junction of $D_1$ and $D_3$.Since $D_1$ and $D_3$ are identical, current splits equally: $I_{D1} = \frac{I_{total}}{2} = \frac{37.7 mA}{2} \approx 18.85 mA \approx 18.8 mA$.Answer:Option B: $18.8 mA$.