Question

In the given circuit the current through Zener Diode is close to:

• A. 0.0 mA
• B. 6.7 mA
• C. 4.0 mA
• D. 6.0 mA

Answer 1

Answer: (A)

0.0 mA

Answer 2

1. Analyze the circuit: The circuit consists of a 12V source, resistor R1 (500 $\Omega$), resistor R2 (1500 $\Omega$), and a Zener diode with a Zener voltage ($V_Z$) of 10V. The Zener diode is connected in reverse bias.

2. Assume Zener diode is NOT in breakdown: If the Zener diode is not in breakdown, it acts as an open circuit in reverse bias when the voltage across it is less than $V_Z$. In this case, R1 and R2 are in series.

   • The total resistance is $R_{total} = R1 + R2 = 500 \Omega + 1500 \Omega = 2000 \Omega$.
   • The total current from the source is $I_{total} = V_{source} / R_{total} = 12V / 2000 \Omega = 0.006 A = 6 mA$.
   • This current flows through both R1 and R2. So, $I_{R1} = 6 mA$ and $I_{R2} = 6 mA$.
   • The voltage at junction P (between R2 and the Zener diode) is $V_P = I_{R2} \times R2 = 6 mA \times 1500 \Omega = (6/1000 A) \times 1500 \Omega = 9V$.
   • The voltage across the Zener diode is $V_Z = V_P = 9V$ (since its cathode is at ground).

3. Conclusion: Since the voltage across the Zener diode (9V) is less than its Zener voltage (10V), the Zener diode is indeed not in breakdown. In this state, it acts as an open circuit, and the current through it is negligible, approximately 0.0 mA.