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Could someone please explain why are there two Schottky diodes in this schematic? And also why is the VOUT connected through one of them to VIN. 3.3V LDO

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The two diodes serve different purposes. The first one prevents C1 from draining if power is temporarily removed from JP1. The second one prevents the condition where Vout > Vin (by more than one diode drop). This may be required if there is a lot of capacitance on Vout. LDO's are normally designed for the condition where Vout < Vin. If this is reversed, there could be large currents that flow back into the LDO and they could cause some damage. Did you read the datasheet for the LDO?

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  • \$\begingroup\$ As Dave Tweed points out, the one on the left also provides reverse voltage protection. \$\endgroup\$ – mkeith Oct 10 '15 at 16:24
  • \$\begingroup\$ I read the datasheet,but it still isn't clear.When the STBY pin is shorted to VIN they say"if STBY is shorted to VIN,the IC will switch to standby mode and disable the output discharge circuit,causing a temporary voltage to remain on the output pin. If the IC is switched on again while this voltage is present overshoot may occur on the output.Therefore,in applications where these pins are shorted,the output should always be completely discharged before turning the IC on.Does that mean that the second diode provides this discharge?In my scheme there is some capacitance on VOUT(decoupling caps) \$\endgroup\$ – Herr Krum Oct 10 '15 at 16:44
  • \$\begingroup\$ Well, the diode would cause Vout to discharge through Vin, but only in the case that Vin goes away. That is not the scenario being discussed in your quote from the datasheet. The scenario they are discussing is when STBY transitions from low to high while Vout is still high. \$\endgroup\$ – mkeith Oct 10 '15 at 20:02
  • \$\begingroup\$ Great. So I made this scheme (added LiPo charger). I think the diodes would work good here: link \$\endgroup\$ – Herr Krum Oct 10 '15 at 21:29
  • \$\begingroup\$ Start a new question with that circuit. There are two main problems. First, if your power source is a single-cell lithium battery, the voltage drop caused by the diode will prevent you from using the full capacity of the battery. Second, if you connect a load and a charger directly to a LiPo battery, it may interfere with the charger's ability to detect end-of-charge. This can be explained more fully if you start another question. \$\endgroup\$ – mkeith Oct 11 '15 at 15:47
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It's a two-diode package for convenience, but the diodes serve two different functions:

  • The one on the left is providing reverse-voltage protection for the regulator. It keeps current from flowing if the connections to JP1 are accidentally swapped.

  • The one on the right protects the regulator from excess positive VOUT to VIN differential voltage. Many regulators can be damaged if VOUT exceeds VIN by any significant margin, and the diode makes sure that VIN is no more than one diode drop below VOUT, even if the input power is removed.

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