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What is the maximum input value (Vi) that can be applied without zener diode gets damaged knowing that the maximum power rating of the zener diode is \$ P_m = 500mW \$,the breakdown voltage is \$ V_Z = 10V \$ and the ripple in the output voltage is +-3% ? Any help is appreciated! Thanks in advance!

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    \$\begingroup\$ Do not use Zener diodes as voltage regulators. Use them as voltage references, where you draw next to no current at all and R can be chosen far, far larger. Then again, don't use Zener diodes as references, either, as they are not thermally compensated, but simply use voltage reference ICs that aren't as temperature dependent. And while you're doing that, don't do that, either, but just use a voltage regulator IC: no-one in their right mind would build a voltage source like this. \$\endgroup\$
    – Marcus Müller
    May 2, 2018 at 16:21
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    \$\begingroup\$ I'm voting to close this question as off-topic because it seems like homework and/or not shown what is tried. \$\endgroup\$
    – Michel Keijzers
    May 2, 2018 at 16:21
  • \$\begingroup\$ 40 yrs ago when Bandgap references were becoming popular in LDO's people still used Zeners which are poorman's regulators due to wide Vi variations that the Zener must absorb to maintain regulation ( Izt= knee current) You must design according to Vi min to supply Izt and Pz to dissipate max power at Vi max. \$\endgroup\$
    – Tony Stewart EE75
    May 2, 2018 at 16:24

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I feel like doing some analysis. This is a textbook type homework stuff, but if the professor went over this and you slept in class or they didn't go through this on the chalk board, I'll step you though this.

Ok, first lets look at our expected voltages, and currents then we find the current draw of each parallel branch to get the total current flow. Then we find the voltage drop across the series resistor. then add the expected voltage with the series voltage drop to get the total voltage. Now this is the maximum voltage, and this will answer your question on something like a quiz, but if you are really building this, you only apply 80% of this voltage (25.6 instead of 32V) so that there is a power safety margin.

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  • \$\begingroup\$ Thanks for the easy explanation. I was a bit confused but now i got it. Just one question. Doesn't the ripple in the output voltage play any role at all? (edited my question since i had writen "ripple in the supply voltage" by mistake) \$\endgroup\$
    – MJ13
    May 2, 2018 at 17:09
  • \$\begingroup\$ the ripple voltage not part of the circuit's calculation because it is a side effect of using a power supply instead of a battery. So the end ripple amount would depend on the quality of the power supply. They probably threw that in there to throw some people off if they didn't know its not useful. \$\endgroup\$
    – drtechno
    May 2, 2018 at 17:33

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