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Zener voltage regulator working

Short answer

A Zener regulator uses a reverse-biased Zener diode in parallel with the load and a series resistor. In breakdown, VZ stays constant; changes in input voltage or load current are absorbed by the Zener current, keeping output voltage fixed. Design ensures IZ stays between IZmin and IZmax.

Draw the Zener voltage regulator circuit and explain clearly the working principle of this circuit to produce a regulated dc output.

A Zener voltage regulator keeps the dc output voltage almost constant when the input voltage or the load current changes. It uses a Zener diode operated in reverse breakdown.

Circuit

         Rs
 Vin o--/\/\/--+----------+--o Vo = VZ
  (unregulated) |          |
                Z         RL
          (reverse biased) |
                |          |
 GND o----------+----------+--o

The unregulated dc Vin (from a rectifier and filter) is applied through a series resistor Rs. The Zener is connected in reverse across the load RL.

Working principle

In breakdown, the Zener voltage stays almost constant at VZ while its current can vary from IZmin to IZmax. So Vo = VZ.

  • Series current: IR = (Vin − VZ)/Rs
  • Load current: IL = VZ/RL
  • Zener current: IZ = IR − IL

Input voltage change (line regulation): if Vin rises, IR rises. The extra current flows through the Zener (IZ rises), the voltage drop across Rs increases by the same amount as the rise in Vin, and Vo stays at VZ. If Vin falls, IZ falls in the same way.

Load change (load regulation): if RL decreases, IL increases; the Zener current decreases by the same amount, so IR (and therefore Vo) stays constant. If RL increases, IL falls and IZ rises.

Design conditions

  • Vin(min) must exceed VZ + (IZmin + ILmax)·Rs so that the Zener stays in breakdown.
  • Rs = (Vin − VZ)/(IZmin + ILmax)
  • IZ must not exceed IZmax = PZ/VZ (power rating).
  • IZ must not fall below IZmin (the knee current); otherwise regulation is lost.

Example: Vin = 12 V, VZ = 6.8 V, Rs = 100 Ω → IR = 52 mA. With a 340 Ω load, IL = 20 mA and IZ = 32 mA, and the output stays at 6.8 V.

✓ Answer checked by Claude Opus 5.5 (high effort)

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