I am working on a project to power some audio equipment that requires +/- 15V DC. Output current load is about 250mA, maximum. I am wondering if there’s any reason I couldn’t create this using two switching boost converters and a DC wall wart, as shown:enter image description here

Edit: Would this circuit work better? enter image description here

  • 2
    \$\begingroup\$ Theoritically yes, but it would nit be perfectly simetrical (adjustment needed), and also it could be very noisy. \$\endgroup\$
    – fifi_22
    Commented Nov 12, 2020 at 20:44
  • 3
    \$\begingroup\$ That is going to give you a more noisy ground rail then generating a negative voltage using an inverting converter. \$\endgroup\$ Commented Nov 12, 2020 at 20:44
  • 5
    \$\begingroup\$ If the +15V boost convertor has no means to pull down, the load between it and +30 can pull it up to +30V. \$\endgroup\$
    – user16324
    Commented Nov 12, 2020 at 20:45
  • 4
    \$\begingroup\$ Why not use a boost and an inverting buck/boost? As @BrianDrummond said a boost converter can't necessarily sink current without special care, so your scheme may not work. \$\endgroup\$
    – John D
    Commented Nov 12, 2020 at 20:45
  • 4
    \$\begingroup\$ I would try my dearest to have your input ground. = output ground and use a boost to get to +15 V and an inverting boost to get to -15 V. \$\endgroup\$
    – winny
    Commented Nov 12, 2020 at 21:08

1 Answer 1


Thanks to the helpful replies, I’ve learned about inverting buck boost converters. In case anyone reading this wants to know more about them, here’s some papers by TI:




I’m still not sure why the noisy ground and symmetry would be any less of an issue with this setup, in case someone can explain that further.

  • 1
    \$\begingroup\$ You can also just use isolated converters and hook them up together however you want. \$\endgroup\$
    – Hearth
    Commented Nov 13, 2020 at 3:44
  • \$\begingroup\$ In your first circuit, transients on the negative rail move the positive rail's ground. Since the output ground is the same for both the positive and the inverted signals, transients on the negative rail don't move the positive's ground, and vice versa. \$\endgroup\$
    – Dave X
    Commented Apr 22, 2022 at 14:07

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.