I'm designing a full bridge rectifier that takes 110vAC to 4-16vDC for input into a 5v switching regulator.

I opened a few 5v power supplies and they place the bridge on the high voltage side. Why not use the low voltage size so the diodes are smaller/cheaper?

Also, I notice they use an IC on the high voltage side, what is the purpose of that IC and is it required?

Most circuits I see online are basic like this one which should give me full wave recification:

enter image description here

This is the transformer I'm using: http://catalog.triadmagnetics.com/Asset/FS10-600-C2.pdf

  • 2
    \$\begingroup\$ You are comparing switched mode circuits, which are optimised for the size of the transformer, to linear supplies and asking why they are not optimised for the cost of the diodes. \$\endgroup\$
    – Martin
    Sep 1, 2016 at 12:57
  • 1
    \$\begingroup\$ Your probably looking at a switching regulator and not the circuit you have shown. \$\endgroup\$ Sep 1, 2016 at 12:57
  • 1
    \$\begingroup\$ Have you opened a AC mains to 5V converter? I don't think you have if they have the diodes on the high side of the transformer. The diode bridge converts the AC to DC, yet you need AC for the transformer to work. \$\endgroup\$
    – Puffafish
    Sep 1, 2016 at 12:59
  • \$\begingroup\$ If you actually put a diode bridge on the input of a 120VAC 60Hz transformer rather dramatic and unpleasant things would happen were it to be powered up from the mains. \$\endgroup\$ Sep 1, 2016 at 13:59
  • \$\begingroup\$ I found a great resource. It even has the layout and BOM: ac-dc.power.com/sites/default/files/PDFFiles/rdr313.pdf \$\endgroup\$
    – physiii
    Sep 5, 2016 at 15:03

3 Answers 3


You're confusing the "oldfashioned" linear mains supply like this:

enter image description here

Reasons to use this design are:

  • It is a simple design
  • As long as you don't touch the primary side of the transformer, it's pretty safe

With the more modern switched mode power supply:

enter image description here

As you can see the switched mode supply is much more complex !

Reasons to use this design are:

  • the transformer is used a a much higher frequency than the 50 or 60 Hz mains frequency meaning that it can be much smaller and more efficient
  • the smoothing capacitors after the transformer can be smaller
  • being a switched supply means that it is much more efficient
  • it can be made more compact (because of reason one)
  • it can be made cheaper (no expensive transformer)
  • lighter in weight, easier to carry.

For Bonus points:

Here's a "circuit level" example of a simple switched mode mains supply, note that it has 2 outputs, 5 V and 12 V. You could leave out the 12 V output to make it even more simple.

enter image description here

Now compare that to the first schematic !

  • \$\begingroup\$ You're perfectly right. I want to just add one comment. "Switched mode power supply" is more of a consumer name than the engineering name. This is actually a DC/DC voltage converter with AC/DC converter at it's input (diode bridge + capacitor). \$\endgroup\$
    – BJovke
    Sep 4, 2016 at 18:44
  • \$\begingroup\$ Great, I will use this version of the schematic above (fairchildsemi.com/reference-designs/RD-330.pdf). Can anyone recommend a digikey/mouser replacement for the difficult to find EER3016 transformer? \$\endgroup\$
    – physiii
    Sep 5, 2016 at 11:13
  • \$\begingroup\$ Such questions (asking for a product reccomendations) are off-topic. Also, if you do not have much experience then I would strongly recommend against building such a switching power supply. These are really only for experienced engineers ! If this is your first supply design ever, go for the oldfashioned type with the mains transformer. You can combine that with a switching DCDC converter (like LM2596 based design). That is really much safer. \$\endgroup\$ Sep 5, 2016 at 11:43
  • \$\begingroup\$ Thank you, but I don't have a power engineer on the team. I'm an electromagnetics/photonics engineer - if this is the most difficult part of the system I will be happy. \$\endgroup\$
    – physiii
    Sep 5, 2016 at 12:05
  • \$\begingroup\$ You're not even an EE ! Then why not just buy a ready build supply ? If you do not aspire to be an EE, why waste your time on this in the first place ? There are many ready build power supplies available for little money, be safe, get one of those instead of messing around yourself. That is safer and allows you to spend your time on the things that matter to you like the magnetics and the photons. \$\endgroup\$ Sep 5, 2016 at 12:21

The 5 V power supplies you opened use a different topology to the one you are trying to build. Since there are various ways to make a power supply, the real question is why you would expect a random mass-produced one to be the same as yours.

Commercial low-voltage DC power supplies usually rectify the incoming AC to make high voltage DC. This is chopped at much higher frequency than the incoming AC, then run thru a transformer. This allows the transformer to be smaller, lighter, and cheaper than your big, klunky, and expensive 50 or 60 Hz transformer. The chopping circuit also modulates the output, usually with feedback via a opto-isolator.

Big fat power transformers that run at the power line frequency haven't been mainstream for a couple of decades or so.


The circuit shown is correct. Diodes are on the secondary.

IC on the primary side is most likely power factor correction. That in itself is a switching power supply ( AC/AC convertor ) that keeps the voltage and current in phase as far as the primary is concerned. I don't think you need that. but you will need output capacitance to smooth the rectified AC.


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