0
\$\begingroup\$

I'm trying to design my own power supply and I decided to use flyback topology. But before chosing a controller for the circuit this question came up: how far can i get with the output power? since I will design the feedback winding for the current measurement why can't I just "fool" the controller by changing this parameter(feedback turns)?

\$\endgroup\$
5
  • \$\begingroup\$ Check out the Texas Instruments WebBench Designer. It can design most of the supply for you in a few minutes. ti.com/lsds/ti/analog/webench/overview.page \$\endgroup\$
    – user4574
    Nov 9 '16 at 17:59
  • \$\begingroup\$ thanks, but i need to understand the principles (more than the power supply itself actually...) \$\endgroup\$
    – iMohaned
    Nov 9 '16 at 18:02
  • 1
    \$\begingroup\$ Most designers (including me :) ) prefer Flyback for up to 100-120W (less than 5A output current). \$\endgroup\$ Nov 9 '16 at 19:00
  • \$\begingroup\$ @RohatKılıç So.. Can i use TL3843 for a 60W (5A) power supply without any problems? \$\endgroup\$
    – iMohaned
    Nov 9 '16 at 19:02
  • 1
    \$\begingroup\$ Of course you can, but PWM controller is not a limiting component; MOSFET, transformer and some other components are. By the way, I don't recommend Flyback topology for 5A output current. Because many problems show themselves. \$\endgroup\$ Nov 9 '16 at 19:06
1
\$\begingroup\$

There is no inherent limit to the power you can get from a flyback topology. It just happens to be more cost-effective for relatively low power applications. Many flyback converters control the peak current by a low-value external sense resistor in the switch source (or emitter) circuit.

I suggest auditing the Coursera courses on switchmode design (from the University of Colorado, I think). They're free to view (but unfortunately not the quizzes).

\$\endgroup\$
0
\$\begingroup\$

I would say the you can calculate power capability by considering the peak current limit on the primary, and the duty cycle. For boundary mode, the duty cycle is known with the constraint the magnetizing inductance must have volt-second balance. I get D = VoutNps/(Vin+VoutNps), not considering the diode forward drop. Then the input current is .5*D*Ipk, on average. Multiply that by Vin and some efficiency derating, that's the answer I think.

\$\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.