I need to isolate the dc input from the dc current inside the circuit (both 9v), in the same way as a 1:1 transformer isolates the AC current inside the circuit from the input AC. I already found out that there are also 1:1 DC-DC converters, but do those converters also really isolate the input from the output? And is this also the best (or only) solution for this? I need this to prevent ground-loops.


edit: Thanks for your comments so far, I've corrected some things in my original question, and below I'll try to explain my problem as well as possible below:

I'm building an effects pedal for my guitar. Effects are usually placed in 2 different locations in the signal chain (before the amp, and between the pre and power amp(effects loop)). If you use the same power supply for pedals in front of the amp and in the effects loop, you can create ground-loops (effect1->audiocable->pre-amp->audiocable->effect2->powercord->effect1), so usually you use 2 different power supplies for this. However, for some reason I need an effect which has parts of it in front of the pre amp, and parts of it between the pre and the power amp*. As a result, I need to find a solution to break the ground-loop inside the effects pedal.

*In theory I could use multiple pedals for this, but then I need to do some crazy tap-dancing on stage to switch multiple pedals at the same time. So this is not a possible solution.

edit2: I forgot to mention, but the only thing still physically connecting both parts of the effect inside the pedal is the power supply. Of course I could use 2 power supplies for one effects pedal, and that would solve my problem, but I think that's an ugly solution.

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    \$\begingroup\$ You need to tell us WHAT you are wanting to achieve. NOT how you think you need to achieve it. We have no real idea of what you want. You MAY be trying to transfer power (DC DC converter). You MAY Be trying to transmit just DC level (various means - not a DC DC converter function) etc. /// What are you trying to achieve? Forget HOW for now. That's our job :-). \$\endgroup\$ – Russell McMahon Jul 26 '11 at 23:28
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    \$\begingroup\$ You'd be better of using a transformer to isolate the signals, not the power supplies. \$\endgroup\$ – endolith Jul 27 '11 at 5:23
  • \$\begingroup\$ And if all you need to isolate is signal, not power, you can use a linear optocoupler. I use LOC110 to good effect. digikey.com/product-detail/en/LOC110/CLA113-ND/95983 \$\endgroup\$ – Stephen Collings Sep 18 '12 at 13:11

First, there are no such things as DC transformers. Transformers inherently work on changing magnetic field, which doesn't change if you put DC thru it.

What you are asking for is called a "DC to DC converter". There may be modules you can get that do 1:1 with isolation. You'd have to look around.

However, this smells like the wrong question. It's a bad idea to ask about a particular solution you've decided on for whatever reason. If you're asking, you probably aren't aware of the various possible solutions. Explain what you are really trying to solve.

It sounds like you have a ground loop problem. There are various ways of attacking them, with a DC to DC converter not usually the best answer. If you've got two separate things to power, why not just use two power supplies to begin with? What connections go between the two units? How far are they? What are the nature of the signals? Can these signals possibly be transformer coupled, or opto-isolated? There may be other ways too.

Step back and explain the real problem.

  • \$\begingroup\$ Of course I meant DC to DC converter, my bad. I've corrected this in my original post and I also added a description of the problem I'm trying to solve. \$\endgroup\$ – Tiddo Jul 27 '11 at 0:14
  • \$\begingroup\$ The magnetic field does change if you put DC through it when there wasn't DC before... \$\endgroup\$ – endolith Jul 27 '11 at 5:25
  • \$\begingroup\$ @endolith, yeah but only once. If you want to keep getting any output, you need to keep changing the field. \$\endgroup\$ – Martin Jul 27 '11 at 10:00
  • \$\begingroup\$ @Martin: Not with an ideal transformer. :) falstad.com/circuit/… \$\endgroup\$ – endolith Jul 27 '11 at 23:06
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    \$\begingroup\$ @endolith: Not only do you need a ideal transformer, you need a ideal dead short load. And then you only get current out, not power. In other words, there is no useful output, even with a ideal transformer. Please let's not confuse innocent bystanders that don't understand all the nuances. For any practical purpose, it remains true that there is no such thing as a DC transformer. \$\endgroup\$ – Olin Lathrop Jul 27 '11 at 23:20

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