I have a circuit with a couple pairs of p-channel mosfets that control output from a couple inputs (see attached picture). The inputs are 5v on one side which also powers a buck converter, and the other input is the buck converter output of 3v3. The mosfet 'outputs' are tied together to get either 5v or 3v3, or no output. The back-to-back mosfets prevent the 3v3 side from conducting via the body diode when the 5v side is enabled. The extra p-channel on the 5v side is not needed, but these are 2-mosfets in one package so just added it.
A micro controls the output selection via a couple pins in open drain mode. The buck converter enable pin is also controlled by the micro. The micro runs from 5v.
None of this is actually tested, only simulated somewhat in my (parts limited) simulation software.
Now, looking closer at the buck converter datasheet I noticed it has 100% duty cycle capability.
Maybe I misunderstand, but it seems I can eliminate my mosfet switching circuit-
buck converter EN low- internal mosfets off = no Vin->LX = 0V out
buck converter EN high- FB = R1,R2 (to get 3V3 out) = 3V3 out
buck converter EN high- FB = >0.3v (prevent over current protection I think), <0.6v (keep duty at 100%) = Vin->LX = 5V (less pch mosfet loss)
(dac pin could drive fb pin to something less than 0.6v to get 5v/max output, or just use the pin to provide a path to ground for another resistor in fb)
these are my current buck converter choices (just need up to 1A, easy to use, low parts count, and larger than a grain of sand)
Diodes Inc. AP3924A or AP3417C
https://www.mouser.com/datasheet/2/115/AP3429_A-766309.pdfhttps://www.mouser.com/datasheet/2/115/AP3417C-271484.pdfAm I missing something, or can the buck converter take over my switching duties? And am I correct in assuming just keeping FB below the reference voltage of 0.6v gets me 100% duty cycle (in these parts)?
Thanks.