Question

Three silicon diodes connected parallel to each other as shown. Forward voltage of diode is 0.7 V. Find current through diode A : 

• A. \(\frac{113}{3}\)mA
• B. \(\frac{113}{6}\)mA
• C. \(\frac{113}{9}\)mA
• D. \(\frac{226}{3}\)mA

Hint:

Always check the orientation of diodes first. A reverse-biased diode (like B here) is effectively invisible to the current, simplifying the parallel branches.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
In a DC circuit, a silicon diode conducts current only when it is forward-biased.
Once forward-biased, it acts like a voltage source of magnitude equal to its barrier potential (0.7 V for Silicon).
Diodes that are reverse-biased act as open circuits and do not conduct current.
Step 2: Key Formula or Approach:
1. Identify biasing: Diodes A and C are forward-biased, while diode B is reverse-biased.
2. Apply Kirchhoff's Voltage Law (KVL): \(V_{source} = I_{total}R + V_{diode}\).
3. Divide current: Since diodes A and C are in parallel and identical, \(I_A = \frac{I_{total}}{2}\).
Step 3: Detailed Explanation:
The supply voltage is \(V = 12 \text{ V}\) and the series resistance is \(R = 0.3 \text{ k}\Omega = 300 \Omega\).
The voltage across the parallel combination of forward-biased diodes (A and C) is fixed at the forward voltage drop: \(V_f = 0.7 \text{ V}\).
The voltage drop across the resistor is:
\[ V_R = V - V_f = 12 - 0.7 = 11.3 \text{ V} \]
The total current flowing out of the resistor is:
\[ I_{total} = \frac{V_R}{R} = \frac{11.3 \text{ V}}{300 \Omega} = \frac{11.3}{0.3} \text{ mA} = \frac{113}{3} \text{ mA} \]
This total current splits equally between the two parallel forward-biased diodes A and C:
\[ I_A = \frac{I_{total}}{2} = \frac{113/3}{2} = \frac{113}{6} \text{ mA} \]
Step 4: Final Answer:
The current flowing through diode A is \(\frac{113}{6}\) mA.