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Zener network: VL, IL, Iz, IR

Short answer

For RL=180 Ω, the Zener is off: VL=9 V, IL=50 mA, IZ=0, IR=50 mA. For RL=470 Ω, the Zener is on: VL=10 V, IL=21.3 mA, IZ=24.2 mA, IR=45.5 mA.

Determine VL, IL, IZ and IR for the network shown in figure for the following condition. a) If RL = 180 Ω b) If RL = 470 Ω. [Figure: Vin = 20 V, series resistor Rs = 220 Ω, Zener diode (VZ = 10 V, PZmax = 400 mW) in parallel with RL; currents IR, IZ, IL marked.]

Given: Vin = 20 V, Rs = 220 Ω, VZ = 10 V, PZmax = 400 mW (so IZmax = 400 mW/10 V = 40 mA).

For each RL, first find the voltage across RL if the Zener is removed (V = Vin·RL/(Rs + RL)). If V ≥ VZ the Zener is ON; if V < VZ it is OFF.

a) RL = 180 Ω

V = 20 × 180/(220 + 180) = 9 V, which is less than 10 V, so the Zener is OFF (no breakdown).

  • VL = 9 V
  • IR = (20 − 9)/220 = 50 mA
  • IL = 9/180 = 50 mA
  • IZ = 0

Answer (a): VL = 9 V, IL = 50 mA, IZ = 0, IR = 50 mA

b) RL = 470 Ω

V = 20 × 470/(220 + 470) = 13.62 V, which is above 10 V, so the Zener is ON and VL = 10 V.

  • IR = (20 − 10)/220 = 45.45 mA
  • IL = 10/470 = 21.28 mA
  • IZ = IR − IL = 45.45 − 21.28 = 24.18 mA

Power in the Zener = 10 × 24.18 mA = 242 mW, which is below 400 mW, so the Zener is safe.

Answer (b): VL = 10 V, IL = 21.3 mA, IZ = 24.2 mA, IR = 45.5 mA

✓ Answer checked by Claude Opus 5.5 (high effort)

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