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I am in the process of designing a switched mode power supply as I need rather high power(12V 2A). I took a look at the TI tool that throws a basic circuit at you. I understand the danger of such a circuit and I understand the circuit itself almost completely. My main problem is finding the right coupled inductor. I stumbled upon this pdf in another question. The transformer in this pdf is rated for up to 265 VAC. Why is it driven with rectified mains which equals to 375 VDC? Regarding the pwm signal: If I understood correctly the duty cycle does in fact, like in a buck or boost converter, affect the ouput voltage?

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Why is it driven with rectified mains which equals to 375 VDC?

It isn't, the rectified mains is alternately switched onto the primary of the flyback transformer (in order to store energy) then it is disconnected thus allowing most of that stored energy to be transferred to the secondary winding. U1 does that switching: -

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the duty cycle does in fact, like in a buck or boost converter, affect the ouput voltage?

Correct.

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  • \$\begingroup\$ Therefore regarding windings a close enough voltage say from 185-265 vac rms to secondary side 20-30 volts is okay due to regulation of the duty cycle? \$\endgroup\$ – schgab Feb 13 at 15:50
  • \$\begingroup\$ I'm not sure what you are asking me? Duty cycle controls how much energy is given to the secondary and, controlling duty cycle means you turn the basic power regulator into a voltage regulator. \$\endgroup\$ – Andy aka Feb 13 at 15:56
  • \$\begingroup\$ A coupled inductor in this configuration behaves like a normal inductor. We just use it for isolation purposes. But actually it is a transformer with different amount of windings in secondary and primary. My question is: Is the voltage on the secondary affected by the duty cycle and the ratio of windings on primary and secondary side like in a normal transformer application? \$\endgroup\$ – schgab Feb 13 at 16:21
  • \$\begingroup\$ When the primary is driven, the secondary voltage does no work but respects the turns ratio i.e. it acts like any other transformer. When the primary is disconnected, the current that flowed in the primary at the instant of disconnection transfers to the secondary (respecting the inverted turns ratio) but, given that the transformer at that point is not driven any more, the current reduces towards zero as it replenishes the charge in the output capacitor. The rate that the current drops is proportional to the output DC voltage. \$\endgroup\$ – Andy aka Feb 13 at 16:42
  • \$\begingroup\$ Ok just like I meant in my first comment. Thanks for clarification! \$\endgroup\$ – schgab Feb 13 at 17:00

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