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I want to understand the following statement that describes a flyback converter when the switch is turned off :

" When the PWM controller instructs the power switch to turn off, the voltage across the primary inductor suddenly reverses, in an attempt to keep the ampere-turns constant. The voltage developed across Lp now appears in series with the input voltage, forcing the upper switch terminal voltage (the drain for a MOSFET) to quickly jump to

Vdsoff=Vin+VLp ..........(1) "

Vdsoff: is the voltage of the drain to source of the MOSFET when the switch is off Vin: is the input voltage. VLp: is the primary inductor voltage

Question:

How does the primary inductor voltage reverses in attempt to keep ampere turns constant and at the same time it keeps its polarity positive in the equation (1)?

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  • \$\begingroup\$ Learn Design are we done here? Did you understand the answer? \$\endgroup\$
    – Andy aka
    Commented Mar 26, 2023 at 10:33

1 Answer 1

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How does the primary inductor voltage reverses in attempt to keep ampere turns constant and at the same time it keeps its polarity positive in the equation

We're talking here about the red waveform in the picture below.

  • The area of voltage•seconds above the \$V_{BUS}\$ level equals the area below \$V_{BUS}\$. That ensures that the coupled inductor (aka transformer) behaves correctly.

  • It is said to "reverse" but another way to look at it is to see that the secondary voltage produced when the MOSFET deactivates becomes transformer-coupled back to the open-circuit primary by \$N•V_{OUT}\$.

enter image description here

Image source.

What is meant by the voltage reversing is that the non-dotted end has \$-V_{BUS}\$ applied when the MOSFET is activated and, that voltage reverses to a \$+N•V_{OUT}\$ level when the MOSFET deactivates. To fully understand this you have to regard \$V_{BUS}\$ as the voltage reference point.

If instead of the above circuit we used a high side MOSFET to control the transformer primary (dot to 0 volts) then, during MOSFET activation, the undotted primary would reach \$V_{BUS}\$ and, during MOSFET deactivation it would go to \$-N•V_{OUT}\$. Maybe that's easier to see?

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  • \$\begingroup\$ you have clarified thinks to me, but as far as I understood, when the switch closes, the transformer behaves like a no load transformer, so the energy is stored in the primary inductor (what we refer in transformers as magnetization current),when the switch is opened, what forces the diode to conduct? since the Vout is initially 0 and the primary inductor is initally energized? I think the answer is the statement "To keep the ampere turns constant" but I did not grasp that meaning, is it like a conservation of energy in electromagnetism ?? \$\endgroup\$ Commented Sep 18, 2020 at 15:24
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    \$\begingroup\$ 1st point yes, correct. When the switch opens, the natural back emf of the primary cause the voltage to reverse very rapidly but this is pretty soon clamped by the secondary voltage equally reversing and then forward biasing the diode. Given that the load voltage can be regarded as constant per switching cycle, the secondary output is clamped to that Vout while it soaks up the energy from the magnetization originally setup before primary opened...... \$\endgroup\$
    – Andy aka
    Commented Sep 18, 2020 at 16:02
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    \$\begingroup\$ I think the ampere-turns bit being constant is a big mental leap to take (although true) and is a cause rather than an effect. \$\endgroup\$
    – Andy aka
    Commented Sep 18, 2020 at 16:08

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