Related Formula
Voltage drop across series resistor:
Vₛ = Vᵢₙ - VZ$$V_s = V_{\text{in}} - V_Z$$
Load current:
IL = (VZ)/(RL)$$I_L = \frac{V_Z}{R_L}$$
Maximum Zener current:
IZmax = (PZ)/(VZ)$$I_{Z\text{max}} = \frac{P_Z}{V_Z}$$
Core Logic
Given values:
Vᵢₙ = 8~V$V_{\text{in}} = 8\mathrm{~V}$, VZ = 5~V$V_Z = 5\mathrm{~V}$, RL = 1~kΩ$R_L = 1\mathrm{~k}\Omega$, PZ = 10~mW$P_Z = 10\mathrm{~mW}$.
Voltage across Rₛ$R_s$:
VRₛ = 8 - 5 = 3~V$$V_{R_s} = 8 - 5 = 3\mathrm{~V}$$
Current through the load resistor:
IL = (5)/(1 × 10³) = 5~mA$$I_L = \frac{5}{1 \times 10^3} = 5\mathrm{~mA}$$
Maximum current allowed through the Zener diode:
IZmax = 10 × 10⁻³5 = 2~mA$$I_{Z\text{max}} = \frac{10 \times 10^{-3}}{5} = 2\mathrm{~mA}$$
Step 1: Determine the Range of Resistance
Total current through the series loop:
Iₛ = IZ + IL$I_s = I_Z + I_L$
For maximum safety configuration (Zener operating at peak current):
Ismax = IZmax + IL = 2~mA + 5~mA = 7~mA$$I_{s\text{max}} = I_{Z\text{max}} + I_L = 2\mathrm{~mA} + 5\mathrm{~mA} = 7\mathrm{~mA}$$
Rsmin = VRₛIsmax = 3~V7~mA = (3)/(7)~kΩ ≈ 428.6~Ω$$R_{s\text{min}} = \frac{V_{R_s}}{I_{s\text{max}}} = \frac{3\mathrm{~V}}{7\mathrm{~mA}} = \frac{3}{7}\mathrm{~k}\Omega \approx 428.6\mathrm{~\Omega}$$
For minimum Zener current requirement (IZ → 0$I_Z \to 0$):
Ismin = 0 + 5~mA = 5~mA$$I_{s\text{min}} = 0 + 5\mathrm{~mA} = 5\mathrm{~mA}$$
Rsmax = VRₛIsmin = 3~V5~mA = (3)/(5)~kΩ = 600~Ω$$R_{s\text{max}} = \frac{V_{R_s}}{I_{s\text{min}}} = \frac{3\mathrm{~V}}{5\mathrm{~mA}} = \frac{3}{5}\mathrm{~k}\Omega = 600\mathrm{~\Omega}$$
Therefore, the required window for regulation is:
(3)/(7)~kΩ < Rₛ < (3)/(5)~kΩ$$\frac{3}{7}\mathrm{~k}\Omega < R_s < \frac{3}{5}\mathrm{~k}\Omega$$
Step 2: Note on Official Key
None of the given multiple-choice options fall strictly within the stable bounds [428.6~Ω, 600~Ω]$[428.6\mathrm{~\Omega}, 600\mathrm{~\Omega}]$. Officially, the answer key evaluates option (3) as correct, though the problem functions fundamentally as a bonus candidate under rigorous design tolerances.
Pattern Recognition
Always solve the current constraints at both boundaries (IZ = 0$I_{Z} = 0$ and IZ = Imax$I_{Z} = I_{\text{max}}$) to bracket the allowable series resistor zone.
Chapter Mix
Class 12 Physics: Semiconductor Electronics
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