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In Marty Brown's Power Supply Cookbook, I found the following variation of the floating linear regulator. I'm aware this topic has been discussed in at least one other question; the two-Zener wasn't mentioned in that discussion though. My question for this schematic is why use those particular voltage and power values for the "lower" 5.6V, 500mW Zener. There isn't really an explanation in the text in that regard.

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  • \$\begingroup\$ There are purpose made regulators that are intended for HV applications, their datasheets may give clues as to why other circuits add protection features. \$\endgroup\$
    – KalleMP
    Sep 4, 2015 at 20:19
  • \$\begingroup\$ After reading a lengthy negative Amazon review of this book... I agree with it, that this is a rather confusing (or at least non-illuminating) book in a number of areas (linear regulators in particular). I guess this question is [going to remain] one of those obscure points of the book. \$\endgroup\$ Oct 2, 2015 at 18:24
  • \$\begingroup\$ There is a nice explanation of high voltage regulation with LM317 here: High voltage regulator \$\endgroup\$ Aug 11, 2018 at 15:51

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As per the LM317 data sheet, the max input-output voltage differential is 40V. This is why a 39V zener is placed across the device because, when 100V is first applied, the output is zero volts and therefore the zener rapidly charges the 100uF but maintains 39 volts across the device as a limit. Having said that I think 39 volts is a poor choice and I question its efficacy however, this is what the author of the circuit did intend I believe.

As for the 5.6 v zener, well the normal running voltage between output and adjust pins is 1.25 volts so clearly a 5.6 v zener is not going to be a problem to normal operation so, I would also say that this diode is here to limit voltage across these pins although the data sheet doesn't appear to state a maximum rating for this voltage. However, it's not uncommon to apply normal diodes to these positions to prevent another problem. See this: -

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The diodes above are to protect the device from input voltage shorts when the output capacitor is fully charged. See also figure 24 and figure 37 of the data sheet for this information.

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In the scheme there is an error - the upper resistor (4,7 k) should be 470 ohms Vout=Vref(1 + 27 kOhm/470 Ohm)+(Iadj x 27 kOhm) ≅ 75 V The equation for Vout is taken in datasheet LM317

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    \$\begingroup\$ Good catch. +1 Even 470 ohm is insufficient to draw the minimum load current under typical conditions let alone worst case (typical 3.5mA, maximum 10mA). The output will rise out of regulation if the load is too light. \$\endgroup\$ Dec 20, 2017 at 1:19

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