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Continous Mode of Flyback SMPS

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It's quite simply down to the formula for an inductor: -

$$V = L\dfrac{di}{dt}$$

When the fly-back transformer's primary is being "charged" with energy, a DC voltage is switched in parallel and this gives rise to a ramping current. In the 2nd half of the process, the energy is released and the current ramps back towards zero and, the rate of change of this current (\$\dfrac{di}{dt}\$) IS the output voltage produced divided by the turns ratio: -

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The above is for a DCM fly-back circuit but the same is true of CCM. There are some worked examples on this page.

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  • \$\begingroup\$ OK i got it. One more thing that during the transistor off time why does current in primary side is zero.Shouldn't be there linearly decreasing current because inductor opposes sudden changes?(shown in image attached with question) \$\endgroup\$ – Electro_learner Apr 18 at 13:38
  • \$\begingroup\$ Simple version: The primary and secondary inductors are coupled so either can be used to "discharge" the magnetic field. Longer version: primary and secondary are never 100% coupled so there will always be a bit of primary current that needs to be got rid of (using a snubber for instance). \$\endgroup\$ – Andy aka Apr 18 at 14:18
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Voltage does not stay constant. There will be ripple voltage that depends on the output capacitor ESR and capacitance.

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