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Regulador shunt com Zener

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Exemplo R must be 156 Ω or smaller — the nearest E24 value is 160 Ω, which is ABOVE the ceiling and would starve the zener (only 4.375 mA left for it at the worst corner) — step down to 150 Ω.

Exemplo With the recommended 150 Ω, the design holds: at 9 V in with the full 20 mA load the zener still keeps 6 mA (you asked for 5 mA). At the other corner (15 V in, 0 A load) the zener swallows 66 mA and burns 336.6 mW — fit a 1 W zener; the resistor burns 653.4 mW — a 2 W resistor. The pair draws 66 mA from the supply whether the load wants it or not.

A zener holds the output at $V_z$ by swallowing whatever current the load leaves behind, so the series resistor is sized by the worst corner and never by the nominal one: $R\le \frac{V_{\mathrm{in,min}}-V_z}{I_{\mathrm{load,max}}+I_{\mathrm{z,min}}}$ — the smallest headroom (input at its minimum) against the largest demand (full load, plus the current the zener itself needs to stay in breakdown).

The zener's own worst corner is the opposite one — input at its maximum with the load unplugged, where it carries the lot: $P_z=V_z\times (\frac{V_{\mathrm{in,max}}-V_z}{R}-I_{\mathrm{load,min}})$. The resistor's worst case is the same maximum input: $P_R=\frac{{(V_{\mathrm{in,max}}-V_z)}^2}{R}$. Size R from the nominal supply and the typical load and the regulation collapses the moment the supply sags, then the zener cooks the moment the load disappears. Leave the last field blank to check the resistor this tool picks, or type in the one you were about to fit and watch what it does at both corners.