0
\$\begingroup\$

I am making a transformer-less AC to Dc power supply and the best design that I have got till now is this:- enter image description here

For better understanding you can download this PDF:- www.nxp.com/documents/user_manual/UM10522.pdf

This circuit uses a TEA1721 ic to convert AC Mains voltage to DC voltage. This design claims to have a 77% efficiency without a transformer and is non-isolated, both of which are fine by me.

The output of this converter is 12V 200mA and I desire 5V 450mA (or greater amperage). Please suggest any changes that I can make in this circuit to get a 5V 450mA output.

If there is a better circuit (without any transformer) which has a better efficiency and an output of 5V 500mA (or greater amperage), then please do suggest me that as well.

\$\endgroup\$
  • \$\begingroup\$ The obvious changes is reduce R3 to get 2.5V at FB and scale L by 5/12th and increase it's current rating by 12/5 at least. However other changes may be necessary and simpler answers offered. dont forget X&Y cap filters \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 Dec 6 '16 at 17:41
7
\$\begingroup\$

You would do well to go to the TI web site and utilize their WebBench design evaluation tool to enter your requirements and see an optimized circuit (or selection of circuits) pop up. For sure you can achieve 80% efficiency or better with todays components.

Hacking an existing design that:

  1. You do not seem to understand how it works
  2. You have not read the data sheet inside out
  3. You fail to understand the functions of R2 and R3

...is not a way of going about designing switching power supplies.

Also make damn sure you understand the implications of usage and safety when designing non-isolated off line power supplies.

\$\endgroup\$
  • \$\begingroup\$ 4. To say nothing of layout requirements. \$\endgroup\$ – Peter Smith Dec 6 '16 at 15:58
2
\$\begingroup\$

Convert 12V 200mA to 5V 450mA

If you want 450 mA, that is a power of 2.25 watts and a power efficiency of greater than 93.75% so, a synchronous buck regulator is the likely choice like this one: -

enter image description here enter image description here

This device is borderline usable given the power efficiency requirements so you may need to find a device that uses an external MOSFET and gives a percent or more greater efficiency.

Of course, it's probably going to be better if you scrap the whole circuit and started again AND used a full bridge rectifier, switching circuit and transformer. You would certainly improve efficiency based on this method and, because a transformer is used you won't mess-up the bridge rectifier by connecting the output to the incoming AC power rail (as you apparently have in the circuit in your question).

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.