Q. A voltage regulating circuit consisting of Zener diode, having break-down voltage of 10 V and maximum power dissipation of 0.4 W, is operated at 15 V. The approximate value of protective resistance in this circuit is ____ Ω.
Explanation (Complete Step by Step Calculation)
In a Zener voltage regulator, the protective resistance limits the current through the Zener diode so that its power rating is not exceeded.
Given Zener breakdown voltage:
\[
V_Z = 10 \, \text{V}
\]
Maximum power dissipation of Zener diode:
\[
P_{max} = 0.4 \, \text{W}
\]
Maximum Zener current is:
\[
I_{Z(max)} = \frac{P_{max}}{V_Z}
\]
\[
I_{Z(max)} = \frac{0.4}{10} = 0.04 \, \text{A}
\]
Supply voltage:
\[
V_S = 15 \, \text{V}
\]
Voltage across the protective resistance is:
\[
V_R = V_S - V_Z = 15 - 10 = 5 \, \text{V}
\]
Protective resistance is:
\[
R = \frac{V_R}{I_{Z(max)}}
\]
\[
R = \frac{5}{0.04} = 125 \, \Omega
\]
Hence, the approximate value of protective resistance is 125 Ω.