Are there any limitations of connecting two non-isolated buck converters in series? Just taking the output from the first converter and connect it as the input of a second converter to produce two DC voltages.

I've seen and tried this before but are there any relations between the two switching frequencies that I should avoid or any other parameters?

  • \$\begingroup\$ I wouldn't know of any special issues apart from this setup having lower overall efficiency. \$\endgroup\$
    – Bart
    Commented Apr 14, 2016 at 7:59
  • \$\begingroup\$ Likely depends on the quality of the them, I have seen cases where one causes the other to fall out of regulation due to drawing currents in just the right frequency. \$\endgroup\$
    – PlasmaHH
    Commented Apr 14, 2016 at 8:06
  • \$\begingroup\$ Thank you both for your answers, really appreciate it. Just started to experiment with DC/DC converters and will soon send my first design to the fab. Just wanted to cross check this before I send it away. \$\endgroup\$
    – Fever
    Commented Apr 14, 2016 at 8:18
  • \$\begingroup\$ Although series connection of linear regulators can be a good idea to spread the power dissipation, power dissipation should not be an issue with a switched converter. So I do not see an advantage to place the converters in series. I would just have them in parallel, without shorting the outputs obviously. \$\endgroup\$ Commented Apr 14, 2016 at 10:22

1 Answer 1


Make sure you have ferrite beads that can filter out the higher harmonics of the 2nd buck switching frequency.


simulate this circuit – Schematic created using CircuitLab

This ferrite is shown by L1. Suppose that U2 switches at 200kHz, the MOSFET/switch, keeps switching at that frequency. There will be an impact on the Vin (P1) of the 2nd buck. That noise might percolate ionto the input rail of the 2nd buck or rather the output rail of the 1st buck. As such you might not have much issues. But are you providing the output of the 1st buck to other loads (sensitive) ? if so,especially in case of RF load, this noise can be a killer.

Filter out this with proper ferrites and capacitor banks.

  • \$\begingroup\$ Using an inductive source to the second buck will normally require a snubber at the input of that second buck (there can be high voltages due to di/dt in the second buck. \$\endgroup\$ Commented Apr 14, 2016 at 11:02
  • \$\begingroup\$ You need a snubber circuit for a ferrite bead ???? \$\endgroup\$
    – Board-Man
    Commented Apr 14, 2016 at 11:06
  • \$\begingroup\$ 1 microhenry can be a significant inductance at the input of a switchmode device. \$\endgroup\$ Commented Apr 14, 2016 at 11:08

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.