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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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the gods from engineering
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2 Answers2

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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: -

enter image description here

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.

Andy aka
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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

Oleg GB
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    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. – Spehro Pefhany Dec 20 '17 at 01:19