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I am designing a circuit with 2 48V power supplies to feed a load. My goal is to give priority to Supply 1 when available since I need the load to be fed by only one supply at a time. This is what I've done so far enter image description here

My question is: Is there a way to stop supply 2 from delivering power when supply 1 is on?

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2 Answers 2

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I think we can design a circuit to use MON1 & MON2 as CTRL for priority function. 1st idea is to use an inverter from MON1 to MON2 but how to power inverter (2 diodes ORing from IN1/IN2 to a zener diode linear regulator ?). Another idea will be to use a LTC IC priority circuit to have MON1 = 1 & MON2 = 0 when IN1 & IN2 are available and the opposite when IN1 fail. I have to do a priority circuit with LTC4355 but the design is planned next years only. I will do some test on table with many ideas we can have.

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You need an ideal diode controller with a prioritizer or power path logic. This will also require the power inputs to have two pass elements to block reverse current.

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  • \$\begingroup\$ The ideal diode controller is the LTC4355 in my design, however it does not provide the exact priority I need \$\endgroup\$
    – Roy
    Commented Aug 25, 2023 at 6:52
  • \$\begingroup\$ Are you not able to change the 4355? This is really a part substitution problem. \$\endgroup\$
    – vir
    Commented Aug 25, 2023 at 6:54
  • \$\begingroup\$ I could change the 4355. I found the LTC4418 but it doesn't meet the voltage requirements. It would be better if I can edit my circuit to meet the requirement \$\endgroup\$
    – Roy
    Commented Aug 25, 2023 at 6:59
  • \$\begingroup\$ Aside from permanently changing the lower priority input to be a lower voltage than the other, it doesn't look like it. Even if you e.g. rigged the high priority input to spoof an under voltage condition on the low priority input it looks like the 4355 would only flag an error but keep the lower priority input pass element gate drive unchanged. You need reverse current through the pass element to cut the gate drive. \$\endgroup\$
    – vir
    Commented Aug 25, 2023 at 7:31

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