Question

Assuming in forward bias condition there is a voltage drop of \(0.7\) V across a silicon diode, the current through diode \(D_1\) in the circuit shown is ________ mA. (Assume all diodes in the given circuit are identical) 

• A. \(11.7\)
• B. \(17.6\)
• C. \(20.15\)
• D. \(18.8\)

Hint:

For identical diodes connected in parallel, always find the total current first and then divide it equally to obtain current through each diode.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
The problem requires identifying the biasing of diodes. A diode is forward-biased if its anode is at a higher potential than its cathode.
Step 2: Key Formula or Approach:
Kirchhoff's Voltage Law (KVL): \(V_{in} - I \cdot R - V_D = 0\).
Current division in parallel branches: \(I_{branch} = \frac{I_{total}}{n}\).
Step 3: Detailed Explanation:
Analyzing the circuit:
1. Diodes \(D_1\) and \(D_3\) are oriented such that they are forward-biased by the \(12\ \text{V}\) source.
2. Diode \(D_2\) is reverse-biased and acts as an open circuit (no current flows through it).
3. The parallel combination of \(D_1\) and \(D_3\) will have a constant voltage drop of \(0.7\ \text{V}\).
4. The voltage across the resistor \(R_1 = 0.3\ \text{k}\Omega\) is \(V_{R1} = 12\ \text{V} - 0.7\ \text{V} = 11.3\ \text{V}\).
5. The total current in the circuit is \(I_{total} = \frac{V_{R1}}{R_1} = \frac{11.3\ \text{V}}{300\ \Omega} \approx 37.66\ \text{mA}\).
6. Since \(D_1\) and \(D_3\) are identical and in parallel, the total current splits equally between them.
7. Current through \(D_1\) is \(I_{D1} = \frac{37.66\ \text{mA}}{2} \approx 18.83\ \text{mA}\).
Step 4: Final Answer:
The current through diode \(D_1\) is \(18.8\ \text{mA}\).