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I have a circuit which currently runs off of either a 12VDC wall wart adapter, or it falls back to a 9V battery if the wall wart adapter is not connected. My circuit currently uses a chip which requires a clean 8VDC source in order to run.

Right now, I just have an 8V LDO regulator burning off the excess power as heat, but I've just realized that my 9V cells can drop as low as 7.8, and still have 50% capacity (mAh) remaining, and my LDO requires that the supply power be at least 0.2V greater than the output voltage, or it just shuts off.

What would be the best approach to constantly having an 8V low-noise output assuming a potential input voltage swinging between 7V and 12V? My initial guess would be to use a flyback or buck-boost switching transformer, which could handle the input swing being so large and the undesired circumstances where Vi < Vo.

However, I can't seem to find any 8V fixed output buck-boost regulators. Also, one thing that has me confused is that both fixed-output and variable-output regulators both require external components (diode, capacitor, inductor). Why do both require external components?

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  • \$\begingroup\$ Either of space or power issues. \$\endgroup\$
    – Ignacio Vazquez-Abrams
    Commented Sep 26, 2014 at 19:31
  • \$\begingroup\$ @IgnacioVazquez-Abrams I'm afraid I don't follow. I don't have space issues, since it's a prototype circuit. \$\endgroup\$
    – Cloud
    Commented Sep 26, 2014 at 19:33
  • \$\begingroup\$ On the die itself. \$\endgroup\$
    – Ignacio Vazquez-Abrams
    Commented Sep 26, 2014 at 19:33
  • \$\begingroup\$ Perhaps the best approach would be (if possible) to redesign your circuit to run from a slightly lower voltage. Or use a buck-boost (eg. Sepic) but the cost will generally be higher and you'll be adding another noise source. \$\endgroup\$
    – Spehro 'speff' Pefhany
    Commented Sep 26, 2014 at 21:11
  • \$\begingroup\$ At this point, I'll likely be using a single switching power supply, followed by two LDOs, as I'm attempting to decouple a pair of microphones, and the easiest solution (electronics.stackexchange.com/questions/130957/…) seems to be to just use independent power supplies. \$\endgroup\$
    – Cloud
    Commented Sep 26, 2014 at 21:24

2 Answers 2

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What would be the best approach to constantly having an 8V low-noise output assuming a potential input voltage swinging between 7V and 12V? My initial guess would be to use a flyback or buck-boost switching transformer

It's a good approach. A buck-boost regulator can be implemented in a small area with a minimum component count (no isolation).

both fixed-output and variable-output regulators both require external components (diode, capacitor, inductor). Why do both require external components?

The reason is that components such as capacitors and inductors occupy too much volume when they are integrated into an IC. For this reason, they are left as external components.

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[...] both fixed-output and variable-output regulators both require external components (diode, capacitor, inductor). Why do both require external components?

  • inductor - it accumulates energy (ON-state), so during this stage power comes from:
  • capacitor - it gets charged by the inductor (OFF-state)
  • diode - serves for rectification

As @MartinPetrei says these components cannot be integrated into the IC capsule.

This is how a buck-boost converter works:

Buckboost operating.svg
"Buckboost operating". Licensed under CC BY-SA 3.0 via Wikimedia Commons.

You can read more on Wikipedia.

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