We feed DC 5V into VCC_EXT_5V and then it outputs 3.3V to the 3V3 in the normal condition. The following is a part of the power circuit.

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My question is followed:

How does it work if I connect the other 3.3V power source to the 3V3 leaving the VCC_EXT_5V open?

As far as I know, the internal circuit may be broken if Vout is higher than Vin. But the manual says there is some protection circuit.

How do you think about it?

Thank you.

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  • 1
    \$\begingroup\$ I don't know that I'd call the '1117 an LDO. \$\endgroup\$
    – Hearth
    Jul 8, 2021 at 2:32
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    \$\begingroup\$ Your question is not very clear. What do " it outputs 3.3V to the 3V3 in the normal condition" and " if I connect the other 3.3V power source to the 3V3" mean? \$\endgroup\$ Jul 8, 2021 at 4:59
  • \$\begingroup\$ @PrathikPrashanth Sorry for my poor English. Normally we feed 5V into input (VCC_EXT_5V) and the LDO outputs 3.3V. The 3.3V is used for the main power for the board. The question is .... Is it OK If I feed DC 3.3V to the 3V3 line(the LDO's output port) without feeding DC 5V into the VCC_EXT_5V? \$\endgroup\$
    – Blue Cloud
    Jul 8, 2021 at 5:04
  • \$\begingroup\$ Im very confused now. How do you expect to get 3.3V at the output of the 3.3V regulator without a 5V input? \$\endgroup\$ Jul 8, 2021 at 5:07
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    \$\begingroup\$ @PrathikPrashanth I think he/she asks about any potential problems if he/she applies +3.3V from a "separate source" while the circuit is unpowered (i.e. no 5V applied). \$\endgroup\$ Jul 8, 2021 at 5:09

1 Answer 1


According to the paragraph you supplied, it won’t cause excess current to flow except perhaps at power-up if the 3.3V source had extremely low source impedance. Say much less than 0.1 \$\Omega\$.

About 2.7V will appear at the 5V input. If that is shorted to ground with 3.3V applied then probably the regulator and/or the EMI filter will be destroyed.


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