I want a board I'm designing to accept either 5V or 3.3V power as input. If 5V power is supplied I want to generate 3.3V regulated power on board. The rest of the board all runs off of the 3.3V power. If I call my input power VDDIO, and VDDIO = 5V, then I know I can stick an LDO in there to get my 3.3V for the board. My question then is how do I allow for the case of regulated 3.3V input power.

One solution I thought of was to put in a normally-open solder jumper between Vin and Vout of the LDO, to let someone manually bypass the LDO. A more accessible direct analogy to this would be to include a bypass switch. But I'm interested in whether there may be a more elegant way to achieve the same effect "automatically", or whether the LDO will "handle" this for me already (i.e. if regulated 3.3V is fed into it, maybe it will just pass through to the output).

  • \$\begingroup\$ Will your board function OK with 3.3V - (LDO dropout voltage)? In that case, you can just use the LDO regardless. \$\endgroup\$
    – markrages
    Jan 9, 2016 at 20:15
  • \$\begingroup\$ @markrages I think I have one part on the board whose minimum operating voltage is 3v so that's not a lot of head room. I'm planning to use LD1117DT33CTR with 1V dropout voltage typical... though I don't anticipate drawing more than 20mA on this board actually \$\endgroup\$
    – vicatcu
    Jan 9, 2016 at 20:20
  • \$\begingroup\$ Perhaps a weird buck converter could do that. If the Vout is less or equal 3.3V the PMOS switch is 100% DCycle on \$\endgroup\$ Jan 9, 2016 at 20:55
  • \$\begingroup\$ Use LD39150DT33 and you'll be ok. \$\endgroup\$
    – markrages
    Jan 9, 2016 at 21:44
  • \$\begingroup\$ I was going to suggest a buck-boost regulator but the LD39150DT33 will only drop 80mV max and will most probably work. \$\endgroup\$ Jan 10, 2016 at 11:37

2 Answers 2


If you use a proper LDO then yes the 3V3 will go straight through like you say. I think that you could get an IC that has a P channel mosfet as the pass element. Check the specs for On Resistance of the pass device and if it is low enough you will be able to do this. Example: 300 milliohms Rds(on) means 30mV drop if your 3V3 rail draws 100mA. If you are drawing lots more current then using discretes will give much less on resistance than an IC.

  • \$\begingroup\$ I don't anticipate drawing more than 20mA on this board actually \$\endgroup\$
    – vicatcu
    Jan 9, 2016 at 20:22
  • \$\begingroup\$ @vicatcu ... well you can just buy a pmos 3V3 LDO chip and you will be fine . \$\endgroup\$
    – Autistic
    Jan 9, 2016 at 20:33
  • \$\begingroup\$ yea, I should really just test this with an LED and take some measurements, but thought it was general enough of a problem to ask here for posterity, and let the community weigh in \$\endgroup\$
    – vicatcu
    Jan 9, 2016 at 20:37
  • \$\begingroup\$ NMOS-pass LDOs such as the REG102 are also an option... \$\endgroup\$ Jan 10, 2016 at 6:22

To my knowledge, there's no off the shelf part that does this due to the fact that you need headroom for PFET. You could have a 3.5V Zener Diode and a resistor to GND so that you get 0V on the node if it 3.3V supply and 1.5V if you have a 5V supply. You then could pass that to some sort of IC with an enable, but it also depends what the "output" of the LDO looks like.

The other reason that I do not believe this part exists (as of 8 years ago) is that I made a LDO once specifically to solve this problem based on the work of Chen and Hasler, combined with some special start-up circuits. It give 3.3V out at 5V and 3.2V at 3.3V.

  • \$\begingroup\$ interesting... bypassing the LDO is harmless though right (i.e. applying 3.3V to both its input and its output? \$\endgroup\$
    – vicatcu
    Jan 9, 2016 at 20:15
  • 1
    \$\begingroup\$ bypassing the LDO is harmless, but they draw a lot of quiescent current in reality. I tested the bypass on an a LM317, and it pulls 100mA when you bypass it. I think the answer is it just depends on your power budget. \$\endgroup\$
    – b degnan
    Jan 10, 2016 at 12:17

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