--- title: "Zener network: VL, IL, Iz, IR" subject: "Basic Electronics Engineering" chapter: 2 key: "semester_2/basic_electronics_engineering:zener-regulator-220-ohm" source: https://www.bctengineer.com/q/semester_2/basic_electronics_engineering/zener-regulator-220-ohm site: BCT Engineering --- # Zener network: VL, IL, Iz, IR Basic Electronics Engineering · Chapter 2: Diodes - Asked in: 2070 Bhadra, 3+3 marks - Video: [Solved Problems on the Zener Diode](https://www.youtube.com/watch?v=kYtebfx-yHc) — ALL ABOUT ELECTRONICS - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_2/basic_electronics_engineering/zener-regulator-220-ohm - Whole chapter: https://www.bctengineer.com/md/semester_2/basic_electronics_engineering/chapter-2 ## Question 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.] ## 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. ## Answer **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** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. How do you decide whether the Zener is ON? A. Compare Vin with VZ B. Remove the Zener and find Vin RL/(Rs+RL); if it is at least VZ the Zener is ON C. Check if RL is greater than Rs D. Check the power rating first Answer: B. If the open-circuit load voltage reaches VZ, the Zener breaks down. (Key point: Zener ON/OFF test) 2. For RL = 180 Ω (VZ = 10 V, Vin = 20 V, Rs = 220 Ω), what happens? A. Zener ON, VL = 10 V B. Zener ON, IZ = 40 mA C. Zener OFF, VL = 9 V and IZ = 0 D. Zener OFF, VL = 20 V Answer: C. The divider gives 9 V < 10 V, so the Zener is OFF and VL = 9 V. (Key point: Case a) 3. For RL = 470 Ω, what is IZ? A. 45.45 mA B. 21.28 mA C. 66.7 mA D. 24.18 mA Answer: D. IR = 45.45 mA and IL = 21.28 mA, so IZ = IR - IL = 24.18 mA. (Key point: Case b Zener current) 4. Why is the Zener safe in case b? A. PZ = 242 mW is below the 400 mW rating B. IZ is less than IL C. VL equals Vin D. Because IZmax = 24 mA Answer: A. Power = 10 V x 24.18 mA = 242 mW < 400 mW. 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