Question

For the given circuit, R = 125 Ω, R_L = 470 Ω, V_z = 9V, and \(I_z^{max}\)= 65 mA. The minimum and maximum values of the input voltage (\(V_i^{min}\ and\ V_i^{max}\)) for which the Zener diode will be in the 'ON' state are

• A. V_i^min = 9.0 V and V_i^max = 11.4 V
• B. V_i^min = 9.0 V and V_i^max = 19.5 V
• C. V_i^min = 11.4 V and V_i^max = 15.5 V
• D. V_i^min = 11.4 V and V_i^max = 19.5 V

Answer 1

The Correct Option is D

To determine the minimum and maximum values of the input voltage \(V_i\) for which the Zener diode will be in the 'ON' state, we need to ensure that the Zener diode maintains its breakdown voltage \(V_z = 9V\) and that the current through the Zener diode is within its maximum allowable current.

1. **Minimum Input Voltage (\(V_i^{min}\))**

• For the Zener diode to be ‘ON’, the voltage across it should be at least \(V_z = 9V\).
• The load resistor \(R_L = 470 \ \Omega\) is connected in parallel to the Zener diode.
• The formula for the minimum input voltage is given by:
  \(V_i^{min} = V_z + I_{L} \cdot R\)
• The load current \(I_{L}\) can be computed when \(V_L = V_z\) as:
  \(I_{L} = \frac{V_z}{R_L} = \frac{9}{470} \approx 0.0191 \ \text{A}\)
• Substitute into the equation:
  \(V_i^{min} = 9 + 0.0191 \times 125 = 11.4V\)

2. **Maximum Input Voltage (\(V_i^{max}\))**

• The maximum Zener current \(I_z^{max} = 65 \ \text{mA}\ = 0.065\ \text{A}\).
• Maximum input voltage occurs when Zener current is at its maximum, calculated as:
  \(I = I_L + I_z^{max} = 0.0191 + 0.065 = 0.0841 \ \text{A}\)
• Apply Ohm's Law to find maximum input voltage:
  \(V_i^{max} = V_z + (I) \cdot R = 9 + 0.0841 \times 125 = 19.5V\)

3. **Conclusion**

• Thus, the minimum input voltage for which the Zener diode is 'ON' is \(11.4V\) and the maximum input voltage is \(19.5V\).

Correct Answer:

V_i^min = 11.4 V and V_i^max = 19.5 V