# This circuit uses a transformer on a rectified DC voltage, what purpose does this serve?

I am trying to design a USB charger from scratch, and I have come across this schematic in the data sheet of a AC-DC converter that I would like to use (NCP1011):

Unfortunately the circuit does not have any values, but it looks like there is a transformer (with reversed polarity on the output) on the rectified DC output. Page 17 of the datasheet reveals that if you wanted a 12V output with a 120V rectified voltage, then you should use a transformer ratio of 1:0.1. So the transformer ratio is the same as if it were AC. Since I want 5V output, and I am using 240V (Australian mains power), the rectified voltage will be around 240Vdc, so I will use a transformer with a ratio of 1:0.02.

Also, why is the polarity reversed on the transformer? If the transformer is only converting the ripple, then why would you want that to be negative?

EDIT: Just realized a miscalculation. Australian mains will be closer to 339V (240 * sqrt(2)) after it is rectified, so my transformer would be something 1:0.014 (if such a thing exists)

• This is a fairly common switching topolgy used in SMPS today. The controller IC switches the rectified DC at much higher frequency than the line frequency. This effectively means your taking your 50Hz input, converting it to a much higher frequency AC to drive a transformer. The beauty of this topology is you can get away with much smaller transformers. – Adil Malik Jan 23 '17 at 19:26
• The transformer is actually on an AC signal. – user_1818839 Jan 23 '17 at 19:30