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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.

enter image description here

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.

enter image description here enter image description here

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    \$\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

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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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