Semiconductor Devices (8hrs) · Basics of Electrical & Electronics Engineering

Zener diode as voltage regulator

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Zener diode as voltage regulator

1. What a Zener Diode Is for Regulation

Zener diode = heavily doped p-n designed to break down (Zener) at precise VZV_Z (22–200200V) and operate stably in breakdown with current limiting, maintaining near-constant VZV_Z despite current variation — used as shunt voltage regulator (and reference).

2. Circuit — Shunt Regulator

Unregulated DC Vin (e.g., rectified+filter, ~15V) → series resistor RS → node → Zener (reverse breakdown) cathode to node, anode to ground → parallel load RL across Zener
  • RS limits current:
IRS=(Vin−VZ)/RS=IZ+ILI_{RS} = (V_{in} - V_Z)/R_S = I_Z + I_L
  • Zener in reverse breakdown at VZV_Z clamps node to VZV_Z (if IZI_Z within knee (IZKI_{ZK}) to max (IZMI_{ZM})).
  • Load RLR_L sees VZV_Z constant.

3. Working and Regulation

  • Line regulation: if VinV_{in} ↑, IRSI_{RS} ↑, extra current goes to Zener IZI_Z↑ (since VZV_Z constant, IL=VZ/RLI_L = V_Z/R_L constant), so VLV_L stays VZV_Z.
  • Load regulation: if RLR_L ↓ (load current ILI_L ↑), IZI_Z ↓ to keep IRSI_{RS} ≈ constant (since Vin,RSV_{in},R_S fixed) ⇒ VLV_L stays.

Condition for regulation: Zener must stay in breakdown:

IZK<IZ<IZMI_{ZK} < I_Z < I_{ZM}

and

RSR_S

chosen so

IRS>IZK+IL(max)I_{RS} > I_{ZK}+I_{L(max)}

and

IZ⋅VZ<PZMI_Z·V_Z < P_{ZM}

(power).

4. Design Example

Need Vout=5.1V from Vin=10V, load RL=500Ω (IL=10mA), Zener 5.1V, IZK=5mA, choose RS:
Total current needed I_RS = I_Z + I_L. Pick I_Z=10mA (mid) ⇒ I_RS=20mA ⇒ RS=(10-5.1)/0.02≈245Ω ⇒ standard 220Ω ✓
Check line variation Vin 12V: I_RS= (12-5.1)/220=31mA ⇒ I_Z=21mA still <IZM (e.g., 50mA) ⇒ okay.
For load variation RL=250Ω (IL=20mA): I_Z= I_RS -20mA =0mA? If I_RS=20mA, I_Z=0 ⇒ dropout ⇒ need larger RS or Vin.

5. Limitations and Modern Use

  • Low efficiency (RS dissipates), only small currents (mA to 100mA).
  • Modern: replaced by IC regulators (7805 etc., next topic) for higher current/efficiency, but Zener still used as reference.

6. Exam Points

  • Draw shunt regulator circuit (Vin→RS→Zener||RL) — 5-7 marks, very frequent (almost certain to ask Zener regulator circuit).
  • Explain VZV_Z constant and IZI_Z adjusts for line/load — 5 marks, very frequent.
  • Give design RS=(Vin−VZ)/(IZ+IL)R_S=(V_{in}-V_Z)/(I_Z+I_L) and condition IZK<IZ<IZMI_{ZK}<I_Z<I_{ZM} — 5 marks, frequent.
  • Mention PZ=IZVZP_Z=I_Z V_Z limit and efficiency low — 3 marks.
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