Location of diode bridge on transformer circuit?

Question

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:

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

Answer 1

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

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:

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.

Now compare that to the first schematic !

Answer 2

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.

Answer 3

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.