Say I have this very simple power supply: enter image description here Variant 1
A = +12V
C = -12V

Disregard any problems with additional filtering, regulation, and the fact that the transformer's primary isn't connected. This is just a simplified example to show what the problem is.

Say I want to change it to this:
A = +24V
B = +12V
Ideally I'd move the capacitor AB to AC: enter image description here Variant 2

But for simpler routing it would be nice if I didn't have to to do that. I.e. if I could leave the capacitor AB where it is: enter image description here Variant 3
Ideally I would like to select the voltages simply by soldering the diodes (and jumper) into different positions while keeping the copper traces the same.
But the placement of the capacitor AB feels wrong.

What are negative (or possibly positive) consequences of doing it this way?
If I understand right, the effective capacitance between A and C in variant 3 is half of what the capacitor AC would provide in variant 2 (assuming the same capacitance for each idividual capacitor). I assume that makes it less effective at smoothing out ripples?

Please consider these details about my situation when writing your answer (if they are relevant), but a general answer is always appreciated:
This is part of a switching power supply at 132kHz (driven by a TNY268).
The diodes are MUR420.
The capacitors are 470µF electrolytic capacitors and you can safely assume their voltage rating is sufficient.
To the right of what's shown in the schematics above, there will be a 100µH inductor in all non-ground rails, and after those another set of the same capacitors plus 100nF ceramic capacitors.

  • \$\begingroup\$ Which tap will it be regulating from? Is Variant 3 still stable with half the capacitance form AC? \$\endgroup\$
    – Aaron
    Oct 14, 2020 at 23:28
  • \$\begingroup\$ Going from Variant 1 to Variant 3, how do you have 100uH on all non-ground rails? There is only one possibility (rail A), unless you are double stuffing those as well. \$\endgroup\$
    – Aaron
    Oct 14, 2020 at 23:31
  • \$\begingroup\$ @Aaron It's less a question of "Can it handle a defined load?" but "Can it handle less load?". I don't have a fixed goal to get to. I'm trying to find a balance between "good" and "simple". \$\endgroup\$
    – Niko O
    Oct 15, 2020 at 4:54
  • \$\begingroup\$ @Aaron The schematic I actually use has inductors in all three lines. Depending on which configuration I need, I'll solder in an actual inductor, or a piece of wire. \$\endgroup\$
    – Niko O
    Oct 15, 2020 at 4:57
  • 1
    \$\begingroup\$ Positive consequence is that you don't need to double the cap's voltage rating. Negative : (a) see calculation of capacitor values in series. (b) +24V is contaminated by noise and ripple on +12V, and vice versa. \$\endgroup\$ Oct 15, 2020 at 8:09

2 Answers 2


FYI, variants 2 and 3 are worse, as the current from the top winding goes through the bottom, requiring thicker wire there (or more strands thereof). Better this way:

Variant 1a

Which you will notice is just variant 1 with a diode flipped. You can put the diodes on either side, of course, as long as the phasing works out right. (Which sometimes helps for EMI purposes, putting it in the +V or GND return side. Depends on transformer windup, and if they're in tabbed packages, heatsinking.)

In any case, I don't get the motivation. What's wrong with the bipolar arrangement, and just defining the negativemost one as GND? Electrons aren't equipped with voltmeters, they won't judge you if you do something slightly out of order. :)

Similarly, after the chokes doesn't matter, up to a small difference in EMI (and, again, depending on transformer windup).

From what I've seen, a tapped arrangement is usually chosen when the lower tap doesn't need to deliver much current, so the windings can be equally sized, and it's literally just taking a tap out of a larger winding. And this is supported by the above argument about wire sizing. If the loads are similar, the bipolar arrangement is better.

Note that leakage inductance between the two secondaries can be saved, if they are common ground (in the AC sense -- as in this variant). A coupling cap placed between the dotted ends of the windings will serve to equalize their peak voltages, improving cross regulation. Works much as a SEPIC stage does, but powered by a separate flyback primary instead. Assuming the voltages are equal, of course! (Which is the case in this example, so it's relevant.)


Variant 2 would be reliable.

A dedicated capacitor for each supply would not only guarantee functioning of either of them, isolated from the loading effects of the other, but also facilitate tailoring of both of them to their respective loads.

enter image description here

The transformer rating would be determined by the combined load and that of the diodes and the capacitors by the individual loads.

  • 2
    \$\begingroup\$ This does not answer my question of whether the capacitor should be between the top and middle terminal, as shown in your schematic, or between the top and bottom terminal. \$\endgroup\$
    – Niko O
    Oct 15, 2020 at 7:05
  • \$\begingroup\$ @Niko O, The top filter capacitor is for the + 12 V supply and the bottom one for the - 12 V supply. The two supplies, in series, form the 24 V supply. There is no need for a separate capacitor from the + 12 V line to the - 12 V line. \$\endgroup\$
    – vu2nan
    Oct 15, 2020 at 9:48
  • \$\begingroup\$ Define "no need". Brian Drummond already commented that noise on +24V will be coupled into +12V and vice versa. It has a non-zero impact. I want to know what and how big those effects are. \$\endgroup\$
    – Niko O
    Oct 15, 2020 at 11:07
  • \$\begingroup\$ @Niko O, My answer has undergone a transformation, further to your comments. \$\endgroup\$
    – vu2nan
    Oct 16, 2020 at 3:02

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