This a theoretical question, I am happy with my current design, no intentions to try this yet.


Some PCB that has requirements of 5V, 3.3V, and 1.8V. The 5V is the supply from some standard linear dropout regulator, and the 3.3V and 1.8V current requirements are low and also fed from some linear regulator powered from the 5V supply.

Let's also pretend maybe our only goal is to reduce the number of components on a board, and that the components aren't particularly sensitive to noise or the exact voltages required.

Theoritically, would it be possible to only populate the boards with the 5V regulator to act as the supply and the 3.3V to act as a supplementary source for the required components. Then for the specific component that needed 1.8V, connect the VCC to 5V and the VDD to the 3.3V plane/source.

Would this not create a sufficient ~1.7V voltage potential across the chip? We could even design say to have a 5.05V source and a 3.25V secondary to get a closer 1.8V approximate assuming the other chips have good tolerance as well.

What would the greatest penalties be and considerations? Obviously a dirtier 1.8V source with power fluctuations and spikes. Also introducing the tolerances in output voltage for 2 regulators instead of one. But if our acceptable input range was 1.65-1.95V, could this work?

  • 1
    \$\begingroup\$ What happens when you need to interface to 3.3v I/O, but your chip is actually putting out 5v relative to actual ground? \$\endgroup\$
    – Ron Beyer
    Commented Jun 29, 2020 at 3:13
  • \$\begingroup\$ what does this mean? ... connect the VCC to 5v and the VDD to the 3.3v plane/source ... please draw a schematic diagram \$\endgroup\$
    – jsotola
    Commented Jun 29, 2020 at 3:26
  • 4
    \$\begingroup\$ @jsotola they're trying to use the difference between their 5v and 3v3 rails as their 1v8 supply, which is of course not going to work, but additionally the supply rail terminology is all confused. \$\endgroup\$ Commented Jun 29, 2020 at 3:31
  • \$\begingroup\$ @ChrisStratton, that's what i thought, but i was not sure that my understanding was correct \$\endgroup\$
    – jsotola
    Commented Jun 29, 2020 at 3:37

1 Answer 1


No, in virtually all cases the 1.8V rail must share a ground with the 3.3V and 5V supplies so that won't work.

Even if you did have some strange load that didn't care about grounding (say an infrared LED) there's another more subtle reason- that most regulators can't sink current, so your 1.8V load would run the risk of increasing the voltage on the 3.3V supply and damaging everything connected to it.

  • 2
    \$\begingroup\$ The second thing won't happen if the 3.3V load is always greater than the 1.8V load, but you will have to be careful to ensure that. \$\endgroup\$ Commented Jun 29, 2020 at 11:56
  • \$\begingroup\$ Also, the start-up sequence is worth considering. Briefly getting close to 5V across that 1.8V component might be well outside the spec. \$\endgroup\$ Commented Jun 30, 2020 at 4:37
  • \$\begingroup\$ Protection from transient overvoltage can easily be handled by a pair of Zener diodes (say 2.0 V and 3.6 V) between the respective rails. But yes, the main point is that passing any signal between the 1.8 V and 3.3 V worlds without a common ground will require optocouplers, which might not be practical. \$\endgroup\$
    – TooTea
    Commented Jun 30, 2020 at 8:00
  • 1
    \$\begingroup\$ Back before speed-controllable computer fans were common, a popular way to slow 12v fans down was to connect them between the 12v and 5v rails. This worked because 1) fans are one of those strange loads that don't care about grounding, and 2) loads on the 5v rail at the time had plenty of capacity to sink the current from doing this. \$\endgroup\$
    – Mark
    Commented Jun 30, 2020 at 20:37

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