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When I use a USB Serial Programmer, the 3.3V line from the programmer gets connected to the 3.3V line from the circuit below.

The problem is that I also get 3.3v on the voltage regulator's VIn, supplying power to my other components hooked to the 5v net (not shown on the schematic). Could I use a Schottky diode in serial either in Vo->3V3 or 5v->Vin in order to prevent this? I think it might be better to use it on the input side of the voltage regulator since the voltage drop (0.3V) is less significant but then the voltage regulator will still be "on"

enter image description here

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    \$\begingroup\$ what would the diode on the input of U2 actually protect? \$\endgroup\$
    – Marcus Müller
    Sep 20 at 15:52
  • \$\begingroup\$ It would prevent powering my 5v line with 3.3v while I connect my USB programmer. I have a LED Matrix panel connected to my 5v line. \$\endgroup\$
    – rasco22862
    Sep 20 at 15:57
  • \$\begingroup\$ You could add a diode, but if this is a one time thing during initial programming and you don't expect another 3.3v supply to ever be connected, I would disconnect the 3.3v from the programmer and run off of USB power. \$\endgroup\$
    – user1850479
    Sep 20 at 16:01
  • \$\begingroup\$ @rasco22862 but if you connect external 5V, then that diode doesn't do anything, because it's forward biased (5 V > 3.3 V). So, it solves no problem while there's external 5V. \$\endgroup\$
    – Marcus Müller
    Sep 20 at 16:05
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    \$\begingroup\$ @MarcusMüller I guess it'd prevent the current from going to the other components I have connected to the 5v net (schematic is not showing them) \$\endgroup\$
    – rasco22862
    Sep 20 at 16:43

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