2
\$\begingroup\$

I have different designs that can be powered either by the USB port or an external 5v power supply.

I want to avoid blowing up something and think I have solved it using two different strategies:

  1. By using two diodes

enter image description here

  1. By using a MOSFET

enter image description here

Both solutions seem to work and do its job but ... which one is more reliable avoiding any possible problem if both the USB port and the external power supply are connected at the same time?

\$\endgroup\$
3
  • 2
    \$\begingroup\$ Normally you use a MOSFET because there is no voltage headroom from 5V to lose. \$\endgroup\$
    – DKNguyen
    Commented Dec 14, 2021 at 16:34
  • \$\begingroup\$ It does not matter as in both drawings the CP2102N connections are wrong for a self-powered device. \$\endgroup\$
    – Justme
    Commented Dec 14, 2021 at 17:17
  • 3
    \$\begingroup\$ Does this answer your question? Switching between power supplies - USB dominant \$\endgroup\$
    – Marcus Müller
    Commented Dec 14, 2021 at 17:43

1 Answer 1

Reset to default
1
\$\begingroup\$

The APK2112 has a maximum dropout voltage of 1.3V, if you are going to use a diode in either leg, you will need to select another LDO with lower dropout.

5V - 500mV = 4.5V, that gives you 1200mV of maximum dropout, and you don't want to design with no margin as regulation accuracy will suffer right at the minimum dropout margin.

There are single chip solutions for 5V or'ing that have internal mosfets, that is probably your best bet. Two LM66100's will do the job, I vaguely recall that someone makes a single chip solution that has dual ideal diodes for 5V switching but I didn't find it on a quick search.

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.