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This question already has an answer here:

I don't understand how a zener diode placed in series with a resistor creates a constant voltage for a load in parallel with the diode. The load voltage drop must be the same as the zener diode voltage drop since they are in parallel, right?

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marked as duplicate by ThreePhaseEel, JRE, Community Dec 25 '16 at 18:48

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    \$\begingroup\$ Uhm you answered your own question there !?! Why not have a look at EEVBlog's Dave excellent tutorial on Zener diodes: youtube.com/watch?v=O0ifJ4oVdG4 It is all explained there. \$\endgroup\$ – Bimpelrekkie Oct 22 '16 at 20:58
  • \$\begingroup\$ Actually if the voltage divider ratio is always slightly greater than the Zener Vz, it will absorb the excess voltage, as long you take into account this Pd=VI to handle the difference in Imax-Imin \$\endgroup\$ – Sunnyskyguy EE75 Nov 22 '16 at 5:05
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That's right. The zener's voltage drop is constant, as long as current is flowing through the zener, so the voltage across the load is also constant, since it's in parallel with the zener.

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    \$\begingroup\$ Fair enough, I just wondered if it was worth spelling out what happens when I=0. I'll delete. \$\endgroup\$ – Brian Drummond Oct 22 '16 at 21:39

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