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Series Resistor for a Zener Regulator

A Zener diode regulates at $V_Z = 10\ \text{V}$ from an unregulated $20\ \text{V}$ supply.

With no load connected, the Zener should carry $I_Z = 20\ \text{mA}$ .

Find the required series resistance.

Given

Vin	20 V	Supply voltage
Vz	10 V	Zener voltage
Iz	20 mA	Zener current

Hint 1

The resistor sees the difference between the supply and the regulated voltage.

Hint 2

With no load, the Zener carries all the current.

Hint 3

R = (V_in − V_Z)/I_Z, with current in amps.

Worked solution — try the problem first

The series resistor drops the difference between supply and regulated voltage:

$V_R = V_{in} - V_Z = 20 - 10 = 10\ \text{V}$

With no load, all current flows through the Zener:

$R_s = \frac{V_R}{I_Z} = \frac{10}{0.020} = 500\ \Omega$

$\boxed{R_s = 500\ \Omega}$

Power check. $P_Z = V_ZI_Z = 10 \times 0.02 = 0.2$ W, so a 0.5 W Zener is adequate. $P_R = 10 \times 0.02 = 0.2$ W likewise.

With a load attached, the current splits: $I_R = I_Z + I_L$ . The resistor must be sized so the Zener still carries enough current to stay in breakdown at maximum load and does not exceed its power rating at minimum load.

Concepts:Zener regulatorVoltage regulationOhm's law

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