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I was thinking what is the best solution to add a reverse current blocking on switching regulator. My first thought was to look for an ideal diode circuit but all of them were too much expensive. The cheapest solution was to put a simple diode but they've got significant amount of voltage drop. I thought to hide the diode's voltage drop inside the switching regulator. Connecting the feedback pin of regulator at the diode's cathode the Vo would be maintained to just at the desired set point.

  • What are the pros and cons of this circuit?
  • Is there a better solution?

enter image description here

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  • \$\begingroup\$ I tried it long ago, didn't work. Had no time to investigate, so i just shorted the diode. The control loop gone completely crazy. \$\endgroup\$ – Gregory Kornblum Jul 12 '17 at 20:32
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    \$\begingroup\$ Pick a regulator and see if the feature is built in to the chip or addressed in the application notes. \$\endgroup\$ – Transistor Jul 12 '17 at 20:33
  • \$\begingroup\$ Unstable regulation is my top concern. And unfortunately my chip hasn't got this feature. \$\endgroup\$ – MrBit Jul 12 '17 at 20:40
  • \$\begingroup\$ Quick and dirty solutions would include setting the regulator for 5.7 V and moving the diode to after the capacitor. \$\endgroup\$ – winny Jul 12 '17 at 20:47
  • \$\begingroup\$ There are still esd diodes on the feedback pin to vin, so you can still get backfeeding. If its a buck converter current is lower on the input so more efficient to put it there. \$\endgroup\$ – sstobbe Jul 12 '17 at 22:08
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Whoops, my mistake. Should be a P-FET, not an N-FET, otherwise the concept is the same. You still need to be able to set the resistor divider ratio properly based on your Vin, Vout, and the FET threshold voltage. The 10k and 20k I show is just an educated guess for something like a 12V input at Vin and assuming the FET has a threshold voltage range of approximately 1.7V-4V.

Also, the zener may not be totally necessary, it's just there to stop the FET from having too much gate-to-source voltage if Vin is too large.

schematic

simulate this circuit – Schematic created using CircuitLab

It would still be better to use a DC-DC converter with built in reverse current output short-circuit protection though.

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Try connecting a simple diode as shown in the circuit below

schematic

simulate this circuit – Schematic created using CircuitLab

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  • \$\begingroup\$ The diode shunts reverse current around the regulator, it does not block reverse current to source Vin \$\endgroup\$ – sstobbe Jul 13 '17 at 4:22
  • \$\begingroup\$ You mean the current flows through the supply? \$\endgroup\$ – karthik Jay Jul 13 '17 at 4:29
  • \$\begingroup\$ The OP suggests they are trying to eliminate/block current flowing from output to input, your diode provides a low-impedance path for reverse current (this protects the regulator as reverse current will flow through the diode versus regulator potentially causing damage to the regulator itself). \$\endgroup\$ – sstobbe Jul 13 '17 at 4:37
  • \$\begingroup\$ @sstobbe, I thought the aim was to protect the regulator from reverse current, am I wrong? \$\endgroup\$ – karthik Jay Jul 13 '17 at 4:50
  • \$\begingroup\$ I agree with your schematic for protecting the regulator, the OP isn't clear if that's their goal, or blocking any current to flow to Vin \$\endgroup\$ – sstobbe Jul 13 '17 at 5:14

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