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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\$
    – user76844
    Commented Jul 12, 2017 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
    Commented Jul 12, 2017 at 20:33
  • \$\begingroup\$ Unstable regulation is my top concern. And unfortunately my chip hasn't got this feature. \$\endgroup\$
    – MrBit
    Commented Jul 12, 2017 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
    Commented Jul 12, 2017 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
    Commented Jul 12, 2017 at 22:08

3 Answers 3

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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
    Commented Jul 13, 2017 at 4:22
  • \$\begingroup\$ You mean the current flows through the supply? \$\endgroup\$ Commented Jul 13, 2017 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
    Commented Jul 13, 2017 at 4:37
  • \$\begingroup\$ @sstobbe, I thought the aim was to protect the regulator from reverse current, am I wrong? \$\endgroup\$ Commented Jul 13, 2017 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
    Commented Jul 13, 2017 at 5:14
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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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I assume you want to prevent power flow back through the switching or linear regulator from a secondary power source on the output side of the regulator (EG:backup battery)

If the voltage drop on the DC-DC converter or linear regulator is greater than about 2.5 volts then the following circuit using an N-channel mosfet will work.

enter image description here enter image description here

In the diagram the gate is connected to the power input side, drain to the circuit to be powered and the source to the regulator output. Here you can see the mosfet is not allowing power to flow back to the regulator.

The zener diode and gate series resistor are not necessary and may be omitted as long as the input voltage to the circuit is less than the mosfet gate breakdown voltage (typically 20V). The zener voltage should be greater than ~5V to allow the mosfet to fully turn on though it must be less than the mosfet maximum Vgs value. The resistor value is not critical.

The mosfet only needs a Vds rating greater than your output voltage so most logic level mosfets will work fine. Many logic level mosfets even integrate a zener diode for gate protection.

The difference between the input and output voltages is the mosfet gate voltage so this determines whether the mosfet turns on. If there is very little voltage difference between the input and output voltages, the mosfet acts as a diode and gives the usual 0.7V drop.

If you have less than 2V of voltage drop between the regulator input and output the following circuit should work down to around 0.5V of difference.

enter image description here enter image description here

Resistor values are not critical. Reducing the upper resistor value to 10K or 1k with the same 100k gate resistor would improve operation for low voltage differences.

Both the above circuits should have very low reverse leakage current and be suitable for circuits with backup battery supplies. They are possible because the regulator input is accessible and significantly higher then the regulator output voltage. This makes it easy to detect whether there is a power source connected to the input using either a mosfet gate or a BJT as a crude sort of comparator. To switch between two power supplies whose voltages are very close an ideal diode circuit is necessary.

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