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I am struggling to understand how to model a resistance loss of a transformer secondary winding. In a typical flyback converter, if the primary and secondary transformer have a resistance loss, how do we model the the secondary transformer resistor during the second interval (when the mosfet is off and diode conducts). If the turns ratio is 1:n, the current during the second interval would be i/n. So the voltage drop across the secondary winding resistor would be i/n*Rsecondary. As per my understanding, the voltage reflected on the primary winding during the second interval should be -Vload/n - i/n * Rsecondary/n. But this does not seem to be correct. The schematic of a flyback converter is:

enter image description here

Any help/pointer will be appreciated. Thanks

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1 Answer 1

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Any help/pointer will be appreciated

And

how do we model the the secondary transformer resistor during the second interval (when the mosfet is off and diode conducts).

You can model the secondary resistance as a simple resistor in series with the diode.

If the turns ratio is 1:n, the current during the second interval would be i/n

Here's where it falls apart. At the instant the MOSFET disconnects the primary, the primary current can be called \$I_{P(PK)}\$ and this transfers to a secondary current peak of \$I_{P(PK)}/n\$. Thereafter, the current is falling from that peak towards zero like this for example: -

enter image description here

Picture from here showing the \$\color{green}{\text{secondary current}}\$ of a 2:1 step-down flyback transformer operating (in this example) in DCM at a duty of 0.34.

As can be seen, the secondary current ramps down from \$I_{P(PK)}/n\$ towards zero amps.

So the voltage drop across the secondary winding resistor would be i/nRsecondary.*

Correct if assuming an ideal diode and we are talking about \$I_{P(PK)}/n\$ at the instant the 2nd phase of the switching cycle begins. Clearly, as \$I_{S}/n\$ falls, the volt drop gets smaller because the current gets smaller.

*As per my understanding, the voltage reflected on the primary winding during the second interval should be -Vload/n - i/n * Rsecondary/n.*

Given the dot notation in the transformer, the reflected voltage on the primary (at the MOSFET drain) becomes more positive than the supply rail i.e. the minus sign is an error. The flyback on the primary raises the drain above the power (source) voltage by: -

$$\dfrac{V_{LOAD}}{n} + \dfrac{I_{LOAD}\cdot R_{SECONDARY}}{n} + \dfrac{V_{DIODE}}{n}$$

$$ = L_{SECONDARY}\cdot \dfrac{d(I_{LOAD})}{n\cdot dt} + \dfrac{I_{LOAD}\cdot R_{SECONDARY}}{n} + \dfrac{V_{DIODE}}{n}$$

This does not take into account spurious overshoots caused by leakage inductance in the windings.

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