Worked example: a 5 V Zener regulator (1N4733A)

Design a 5 V regulator using the 1N4733A Zener ($V_Z = 5.1\,\text{V}$, $I_{Z,\text{min}} = 5\,\text{mA}$) given$V_{in} = 12\,\text{V}$ and a load $R_L = 1\,\text{k}\Omega$. Pick the series resistor R_S.

1. Topology. V_in drives R_S in series with the parallel combination of the reverse-biased Zener (cathode-up) and the load R_L. The Zener clamps V_out at V_Z.
2. Load current.
   $$I_L = \dfrac{V_Z}{R_L} = \dfrac{5.1}{1000} = 5.1\,\text{mA}$$
3. Series current. KCL at the output node demands $I_S = I_Z + I_L$. To keep the Zener in breakdown, set $I_Z = I_{Z,\text{min}} = 5\,\text{mA}$:
   $$I_S = 5\,\text{mA} + 5.1\,\text{mA} = 10.1\,\text{mA}$$
4. Series resistor. R_S drops V_in − V_Z:
   $$R_S = \dfrac{V_{in} - V_Z}{I_S} = \dfrac{12 - 5.1}{0.0101} \approx 683\,\Omega$$
   Round down to the nearest E12 standard value: $R_S = 680\,\Omega$ (rounding down keeps I_Sslightly higher, ensuring the Zener stays well above I_Z,min).
5. Verify with R_S = 680 Ω.
   $$I_S = \dfrac{12 - 5.1}{680} = 10.15\,\text{mA}$$
   $$I_Z = I_S - I_L = 10.15 - 5.10 = 5.05\,\text{mA}\ \checkmark$$
   I_Z > I_Z,min, so regulation holds.
6. Power dissipation — sanity check.
   $$P_{R_S} = I_S^2\, R_S = (0.01015)^2 \cdot 680 \approx 70\,\text{mW}$$
   $$P_Z = I_Z\, V_Z = 0.00505 \cdot 5.1 \approx 26\,\text{mW}$$
   The 1N4733A is rated 1 W, R_S needs only a 1/8 W resistor.