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Eurorack Synthesizer Bi-polar Power Supply Design Advise


Apr 12, 2021
Apr 12, 2021
I'm working on designing and building a Eurorack compatible (audio synthesizer) bipolar power supply for my own use, to use with both commercially available synth modules and modules that I will build myself.

My target specifications are:

Clean +12v DC rail, delivering up to 1 amp.
Clean -12v DC rail, delivering up to 1 amp.
Ground output

All reasonable safety features to protect the connected devices, and to protect against fire, etc.

I'm new to electronics, and I'm looking for advice on suitability of parts and for any general feedback/improvements on the design.

I did a first draft, and ordered some parts, and I have a basic version working on a breadboard.

Schematic for first design:


It works, but there's a lot of room for improvement. I've been doing research and here's the things I'd like to improve:

1. Add an inrush current limiter because of the large caps.
2. Add fuses on input and output
3. Improve the design to make the output more stable.
4. Revisit the transformer. I'm currently using a wallwart AC to AC adapter and have questions.
5. Revisit the rectifier.
6. Over/under voltage protection? And what stops connected devices from receiving under voltage power while the capacitors are warming up after first being plugged in?

Here's an oscilloscope reading under load:


The blue is the +12v rail at the output. The yellow is the 12VAC input. There's very little draw on the -12v rail (purple).

I'm not sure why the AC is clipping. But the +12v DC rail is definitely noisy when loaded with ~400mA.

This is a sketch of the improvements I'd like to make:


And I'd like to ask a few questions and get some feedback.

1. For the ICL, I think these are what I want:

It looks like the right thing from the data sheet. Is this a good choice?

2. For D7 and D8, which is intended to prevent voltage from going backwards through the regulators, is this what I want?

I was told I needed a SCHOTTKY diode instead of the 1N4007's used elsewhere.

This one (linked) is rated for 20v and 1 amp. I think it would be running at it's limit, since the circuit is designed to produce 1 amp per rail, and even though the power supply and the output are 12 volts, I measure 20volts on a meter where the caps are, before the voltage regulator, which I assume to be a side effect of the capacitor smoothing. Do I need a diode with a higher voltage /current rating?

What about:

Which is 60v, 5A. Is it better to have a higher voltage and amperage rating so there is overhead?

3. To smooth the output, I've found something called a "capacitance multiplier". I've drawn one in, but need to research this more. Specifically how you can make a negative version of one, and how to size the parts. But, is this a good design pattern for stabilizing audio power supplies?

4a. Transformers. There are two main options: A wallwart going into a "DC" power jack (which is actually AC) or a transformer mounted on the board with a mains connection.

I found this:

Is it suitable?

The reason I'm not happy with the current walwart solution is that it gets very hot. It's the only part that seems to be generating heat. There is also a very limited selection of AC to AC wallwarts and they are hard to find (most are DC output). I really don't like how hot my current wallwart gets.

The one I'm using now is this:

Now I also have a really dumb question. Since the goal is to have 1 amp of +12vDC and 1 amp of -12vDC, do I still only want a 1 amp input transformer, since it's two amps going out (one per rail)?

Will oversizing the transformer help with heat? ie. can I use a wallwart rated for 2.5 amps or something, and fuse it to 1 amp on the input and will it run much cooler since it's running at 30% capacity instead of 100%?

A bigger one:

The goal is to reduce the temperature on the wallwart casing.

4b. Or use a toroidal transformer on the board. What are the advantages to that? Do they run cooler?

5. The rectifier. What I have now, I believe is called a half-wave rectifier, and there's two of them. I found another pattern for a bridge rectifier (four diodes in a diamond pattern) and tried it, but what I get from that is +12 and +6 volts, depending on where it is tapped. Getting +6 volts might be useful, but the bridge rectifier doesn't seem to output the -12 volts which I need. I don't suppose I can have two bridge rectifiers, where one of them is installed with every diode backwards from what it's "supposed" to be , to get a negative bridge rectifier?

The only other pattern I saw like this required a special transformer with a center tap. Which comes back to the transformer thing.

Is the dual half-wave the best bipolar pattern that can be used with a wallwart ?

6. Fuses:

I was going to use these on the outputs:

Do we also need to fuse the ground since it's actually at a higher potential than the "negative" rail?

Is there anything wrong with automotive fuses? (The idea is that I can buy one locally at an auto store to replace).

For the input:ürth-elektronik/696106003002/7244558

That should work with both a wallwart and before a toroidal transformer (if we did it that way), since it is rated for 250 volts.

7. Caps. Do we need less smoothing capacitors (the 4700uf, labeled C1-C6) if you use a capacitance multiplier? Or do they serve a different function?


Harald Kapp

Nov 17, 2011
Nov 17, 2011
Welcome NIck.

A few hints:
  1. Your way of placing links is awkward. It doesn't tell us what you are linking to until we follow that link. A much clearer way is by using the hyperlink button upload_2021-4-16_6-13-55.pngin the menu like this: EPCOS - TDK Electronics B59213J0130A020
    Please consider updating your post.
  2. The design of the -12 V rail has a major flaw: D6 and D8 are backwards and the transistor shoulkd be PNP, no NPN. You'll have to account for the reverse polarity of that output.
  3. D1+ D2 and D3+ D4: Why 2 diodes in series? These 1N4007 are good for up to 1000 V reverse blocking voltage, much more than you'll ever see in that position. A single diode will suffice for each output.
  4. Put the inrush current limiter on the primary side of the transformer. In that position it will also suppress the inrush current of the transformer in addition to the inrush current of the capacitors.
  5. While I'm at the transformer: you are right, the transformer needs to be rated 2 A at the output, 1 A per polarity. A toroidal transformer will also have a higher efficiency than a conventional one.
    Since you are doing linear regulation anyway have you considered using a switch mode power supply to create a raw 15 V to 16 V Dc supply which is the filtered an post regulated to 12 V with the linear ICs?
  6. A wall wart provides more safety as it is an enclosed housing compared to a transformer sitting on the board. But with the transformer on the board you have more options like e.g. the primary side inrush current limiter. The drawback is that you'll have to take care of all the necessary safety measures yourself.
  7. Having 3 × 4700 µF capacitors per output in parallel is possible, but imho overkill. A rule of thumb is 1000 µF / 1 A, so a single 4700 µF capacitor should be enough, maybe two as you are using half-wave rectification.
  8. The 7812 and 7912 each should have small ceramic or foil type capacitors right next to the input and output (connect with really short traces!). 0.33 µF at the input and 0.1 µF (or more) are recommended, see e.g. this datasheet.
  9. Speaking of the voltage regulators: Why the additional transistor in front of the ICs? All these transistors will do is to drop ~ 0.1 V from collector to emitter. If you intended a sort of pre-regulation, you should have used a zener diode based circuit. Imho this is not necessary here and components and cost can be spared. This pre-regulator will not improve the performance of the regulated 12 V outputs noticeably.
    Addendum: I now see that this is meant to be the "capacitance multiplier" you're speaking of. It is imho unnecessary, see top 6 above. It will not hurt either, so if you feel confident, leave it in the circuit.
  10. Also with regard to the regulators: A transformer rated at 12 V means 12 V RMS. That is approx. 16 V peak (accounting for only 1 diode voltage drop across the rectifier, 15.4 V-ish for 2 Diodes) at the input of the regulator IC. This means a voltage drop of 4 V across the ICs and accordingly 4 W of power dissipation at 1 A current. You'll need heatsinks to prevent thermal shutdown or blow out of the ICs.
  11. No need for Schottky diodes instead of 1N4007. Schottkys have a lower pass voltage, but since you have 2 1N4007 in series pass voltage seems not to be a concern.
  12. Never put fuses into ground. If that fuses triggers, the +- 12 V outputs are floating and depending on what is connected and how it is connected a load may suddenly see 24 V instead of 12 V and break down. Also fuses do have a resistance, albeit a very small one. This resistance will create a voltage drop due to the return currents and can thus create cross talk between the positive and negative rail. Not productive when striving for very clean supply voltages

All for now