I'm trying to understand ways of isolating circuits and general concepts of isolation.

  • I understand how high frequency signals can pass over a capacitor.
  • I understand also how optocouplers work.
  • I understand how transformers work.

Many places talk about using transformers to isolate,but as far as I know this is only for AC sources. They talk about using transformers for isolation but don't take into account how to isolate DC power sources.

How I can isolate a 48 V DC rail that feeds a DC-DC buck regulator with 12 V DC output? I can think of no way to transfer a large amount of DC energy without direct contact.

I understand induction, electromagnetic waves, capacitance, optocoupling, saw waves but I don't know why all this is related to galvanic isolation. Why galvanic? If for example I use 220 V AC isolation transformer, where is the galvanic current, or the galvanic "phenomena"?

  • 4
    \$\begingroup\$ The trick is to chop your DC input into "AC" in order to pass a transformer and be rectified back to DC on the other side. Have you looked into how a forward converter functions? It will provide you with galvanic isolation. \$\endgroup\$
    – winny
    Commented Feb 28, 2023 at 15:57
  • 1
    \$\begingroup\$ Use a flyback converter. The flyback "transformer" (coil) provides isolation between input and output. There are also other topologies which provide isolation, such as "forward" converter, push-pull, etc. \$\endgroup\$ Commented Feb 28, 2023 at 15:59
  • 2
    \$\begingroup\$ Throw your DC-DC buck regulator in the junk drawer and buy another one that's labelled as isolated. \$\endgroup\$ Commented Feb 28, 2023 at 16:03
  • \$\begingroup\$ galvanic isolation just means electrical isolation - no way for current to flow from one side to the other \$\endgroup\$ Commented Feb 28, 2023 at 16:03
  • 1
    \$\begingroup\$ DC-DC converters with galvanic isolation just convert DC to AC, transformer, and AC-DC conversion. Any switched PS convert AC to DC, so it standard solution. \$\endgroup\$
    – user263983
    Commented Feb 28, 2023 at 16:19

2 Answers 2


In short, to generate isolated DC output from DC input, you still use a transformers, but the DC is converted into AC first to pass it through the transformer and converted back to DC.

So to convert 48V to 12V and have isolation means you just need to use an isolated buck converter.

The 48V DC will anyway be pulsed into an inductor in a buck converter, so it can almost just as easily be pulsed into a transformer primary and then the transformer secondary provides the output, that can be rectified back to 12V DC, and optoisolators can provide isolated feedback. Basically, that's exactly how mains switch mode power supplies work, they just rectify the mains AC first to DC.

The "galvanic" part in the word galvanic isolation does not mean anything that special, it just means the parts are isolated, or insulated from each other, so that there is no DC current path between the parts, which formerly seems to be called "galvanic current".

  • \$\begingroup\$ Downvoter; please specify what is wrong with the answer so I can improve it. \$\endgroup\$
    – Justme
    Commented Feb 28, 2023 at 16:53
  • \$\begingroup\$ I think the answer is fine +1 \$\endgroup\$
    – Voltage Spike
    Commented Feb 28, 2023 at 19:37

How I can isolate a 48 V DC rail that feeds a DC-DC buck regulator with 12 V DC output? I can think of no way to transfer a large amount of DC energy without direct contact.

You could use a buck regulator that has isolation built in, there are off the shelf converters and or designs that can provide isolation with a buck circuit.

Galvanic means no DC current path, which means you can eliminate magnetic current loops from devices and also protect users or equipment from fault currents and voltage spikes (if tranisent suppression is used).


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