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JEE Mains Physics · Semiconductor Electronics

Zener Diode as a Voltage Regulator

Once a reverse-biased Zener breaks down, the load voltage is fixed at the Zener voltage; the series resistor takes up the rest of the supply, and the Zener carries whatever current the load does not.

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Why this matters

Seventeen PYQs, three from 2026, and nine of them ask for a number: the most numerical page in the chapter. Nine find a current or a power in a drawn regulator circuit: the series current, the load current, the Zener current or the Zener's power. Eight are design questions: the safe series resistor from a power rating, or the range of load resistance the circuit can hold. Every one uses the same three lines of working, once you have checked that the Zener has broken down.

Concept 1 of 2: Currents in a Zener regulator

A Zener in parallel with the load holds the load voltage at V_Z, as long as it is in breakdown. The series resistor then has the rest of the supply across it, which fixes the total current. The load takes V_Z ÷ R_L of that, and the Zener takes what is left. If the supply is too small to reach breakdown, the Zener is simply an open circuit.

Definition

  • Check breakdown first. Remove the Zener and find the load voltage from the divider: VL=VinRLRs+RLV_L = V_{in}\dfrac{R_L}{R_s + R_L}. If this exceeds VZV_Z, the Zener breaks down; if not, it carries no current.
  • In breakdown, the load voltage is VZV_Z.
  • Series current: Is=Vin−VZRsI_s = \dfrac{V_{in} - V_Z}{R_s}. Load current: IL=VZRLI_L = \dfrac{V_Z}{R_L}. Zener current: IZ=Is−ILI_Z = I_s - I_L.
  • Power in the Zener: PZ=VZIZP_Z = V_Z I_Z.
  • The Zener current is largest at the largest input voltage, and when the load draws least.
  • Forward biased, a Zener behaves as an ordinary diode.

Regulator currents

Is=Vin−VZRs,IL=VZRL,IZ=Is−IL,PZ=VZIZI_s = \frac{V_{in} - V_Z}{R_s}, \quad I_L = \frac{V_Z}{R_L}, \quad I_Z = I_s - I_L, \quad P_Z = V_Z I_Z

Worked example

A 14 V supply feeds a 6 V Zener through a 400 Ω series resistor. The load is 1.5 kΩ. Find the Zener current and the power in the Zener.
Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2022 · 26 July 2022 · Q30Moderate

Example 1 · Semiconductor Electronics · Zener Diode as a Voltage Regulator

In the circuit shown below, maximum zener diode current will be ____\_\_\_\_ mAmA

Check breakdown before using V_Z

If the divider voltage across the load is below V_Z, the Zener is off and the load voltage is not V_Z. Assuming breakdown without checking gives a negative or wrong Zener current.

The Zener current is not the series current

The current through the series resistor splits between the load and the Zener. The Zener carries I_s − I_L; it carries all of I_s only when the load is removed.

Load current comes from V_Z, not the supply

In breakdown the load has V_Z across it, so I_L = V_Z ÷ R_L. Dividing the supply voltage by R_L overstates it.
Drill 8 more on currents in a zener regulator

Concept 2 of 2: Choosing the series resistor for a Zener

A Zener can dissipate only so much power, so its current has a ceiling, P_max ÷ V_Z. The series resistor must keep the current below that ceiling in the worst case: the highest input voltage with no load connected, when every milliampere goes through the Zener. With a load, the same circuit regulates only over a range of load resistance.

Definition

  • Largest safe Zener current: IZ,max⁡=Pmax⁡VZI_{Z,\max} = \dfrac{P_{\max}}{V_Z}.
  • Smallest safe series resistor (highest input, load removed): Rs,min⁡=Vin,max⁡−VZIZ,max⁡R_{s,\min} = \dfrac{V_{in,\max} - V_Z}{I_{Z,\max}}.
  • For a fixed RsR_s, the series current is Is=(Vin−VZ)/RsI_s = (V_{in} - V_Z)/R_s, and IL=Is−IZI_L = I_s - I_Z.
  • Smallest load resistance: when the Zener current falls to zero, RL,min⁡=VZ/IsR_{L,\min} = V_Z/I_s.
  • Largest load resistance: when the Zener current reaches its maximum, RL,max⁡=VZ/(Is−IZ,max⁡)R_{L,\max} = V_Z/(I_s - I_{Z,\max}).
  • If the input can fall below VZV_Z, the circuit cannot regulate there; safety is still set by the highest input.

Safe series resistor

IZ,max⁡=Pmax⁡VZ,Rs,min⁡=Vin,max⁡−VZIZ,max⁡I_{Z,\max} = \frac{P_{\max}}{V_Z}, \qquad R_{s,\min} = \frac{V_{in,\max} - V_Z}{I_{Z,\max}}

Worked example

A Zener of breakdown voltage 6 V is rated 0.3 W. It is to be used on a 16 V supply. Find the smallest series resistor that keeps it safe.
Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2026 · 24 Jan 2026 Shift 1 · Q23Moderate

Example 2 · Semiconductor Electronics · Zener Diode as a Voltage Regulator

A voltage regulating circuit consisting of Zener diode, having break-down voltage of 10 V and maximum power dissipation of 0.4 W , is operated at 15 V . The approximate value of protective resistance in this circuit is ____\_\_\_\_ Ω\Omega.

Divide the power by V_Z

The Zener's own voltage is V_Z, so its largest current is P ÷ V_Z. Dividing the power by the supply voltage gives a current that is too small and a resistor that is too large.

Design for the highest input

The current, and so the heating, is largest at the highest input. Using the lowest input voltage gives a resistor that lets the Zener burn out when the supply rises.

No load is the worst case

With the load disconnected, the whole series current goes through the Zener. The safe resistor is found for that case unless the question fixes the load.
Drill 7 more on choosing the series resistor for a zener

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (2)

  • Currents in a Zener regulator

    Regulator currents

    Is=Vin−VZRs,IL=VZRL,IZ=Is−IL,PZ=VZIZI_s = \frac{V_{in} - V_Z}{R_s}, \quad I_L = \frac{V_Z}{R_L}, \quad I_Z = I_s - I_L, \quad P_Z = V_Z I_Z
  • Choosing the series resistor for a Zener

    Safe series resistor

    IZ,max⁡=Pmax⁡VZ,Rs,min⁡=Vin,max⁡−VZIZ,max⁡I_{Z,\max} = \frac{P_{\max}}{V_Z}, \qquad R_{s,\min} = \frac{V_{in,\max} - V_Z}{I_{Z,\max}}

Watch out for (6)

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