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I want to use a LM226677 Voltage reagulator to achieve a step-down converter from 12V to 10V. However, in a Schottky diode is used reversely polarized at the output of the LM22677, and the datasheet claims "A Schottky type power diode is required for all LM22677 applications. [...] The reverse breakdown rating of the diode should be selected for the maximum V_IN, plus some safety margin. A good rule of thumb is to select a diode with a reverse voltage rating of 1.3 times the maximum input voltage." datasheet

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Not being able to find/get hold of Schottky diodes with a reverse breakdown voltage of 12V~15.6V (lowest I can find is 20V), I was wondering if it is possible to use 20V reverse breakdown Schottky diode or replace it with a zener diode?

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The calculated reverse breakdown value is a minimum value; you can use any Schottky diode that has a value that's greater than the calculated value.

A zener would be inappropriate because its forward characteristics would be far inferior.

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  • \$\begingroup\$ But why exactly does forward behaviour matter? Isn't it used in reverse? \$\endgroup\$
    – joaocandre
    Commented Apr 11, 2014 at 15:31
  • \$\begingroup\$ Nope. When the switch (SW) is off, D1 is conducting the freewheeling inductor current in the forward direction. The diode is chosen to specifically not break down due to reverse voltage, per Dave's answer. \$\endgroup\$
    – Adam Lawrence
    Commented Apr 11, 2014 at 15:40
  • \$\begingroup\$ Also, it's generally not recommended to accept an answer quickly (even if it's right) since there's little incentive to add to the conversation once the points are awarded. 24 hours is a typical 'cooling off' period before accepting. \$\endgroup\$
    – Adam Lawrence
    Commented Apr 11, 2014 at 15:43
  • \$\begingroup\$ The problem with a zener is it's forward bias drop, it can be quite high, while a schottky is built specifically to keep it very low, it's 200mV against 800mV or more. When the diode is conducting, and it does, all the output current is flowing through it and it disspates power. The lower the voltage drop, the lower the power. Just try to understand how switching regulators work... \$\endgroup\$
    – Vladimir Cravero
    Commented Apr 11, 2014 at 15:58

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