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I am trying to find out the 'inductor' responsible for secondary side voltage ringing observed when flyback MOSFET(SW) turns on using LT8300 simulation in LTspice.

enter image description here

Frequency of ringing observed is around 195 MHz. enter image description here

Capacitance of D1 is ringing with inductor and creating oscillations of frequency 195 MHz.

Considering the very high oscillation frequency, its probably the leakage inductance of secondary side and diode capacitance.

Why secondary side inductance of 7.91 uH is not participating in this ringing?

Capacitance across diode sees negative voltage at secondary transformer node and easier to transfer the energy into secondary side winding?

Thanks

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  • \$\begingroup\$ k = 0.998 seems suspiciously high. What happens if you vary it? And what effect does that have on the nonideal transformer equivalent circuit? \$\endgroup\$ Commented Jun 29, 2023 at 7:51
  • \$\begingroup\$ Tim is right. The leakage should be 0.4% of the open-circuit inductance to get that high coupling factor. This is normally what you get with no air gap. Flyback transformer (coupled inductor) has air gap, so the coupling factor can't be that high. The calculated coupling factor for that transformer is around 0.989, so I assume there's a type in your coupling directive. \$\endgroup\$ Commented Jun 29, 2023 at 8:01
  • \$\begingroup\$ @RohatKılıç how did you determine the coupling would be 0.989 for the transformer. I agree that without a gap it would be about 0.99 but that's only from experience. Maybe you have a clever formula? \$\endgroup\$
    – Andy aka
    Commented Jun 29, 2023 at 8:08
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    \$\begingroup\$ @Andyaka I use \$k=\sqrt{1-L_S/L_{OC}}\$ for measured values, where \$L_S\$ is the leakage (inductance when other winding is shorted) and \$L_{OC}\$ is the open-circuit inductance. Datasheet gives 8uH, and 400uH, respectively. \$\endgroup\$ Commented Jun 29, 2023 at 8:10
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    \$\begingroup\$ @Ronnie there you go. The magnetising inductance can be thought like a pulse source (it's not but let's do so for the sake of simplicity). Because it's involved in energy transfer. Without any parasitic components this voltage would be transferred to the rectifier then the output capacitor. But since there are parasitic components (leakage inductance, diode reverse capacitance, etc) the rising/falling edges of the so-called voltage source "excites" the LC tank hence you have an oscillation. The parasitic resistive components (diode, winding etc) bring damping. \$\endgroup\$ Commented Jun 29, 2023 at 15:02

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The (unfeasible) leakage inductance in the secondary circuit is \$(1-0.998)\times 7.91 \text{ }\mu\text{H} = 16\text{ nH}\$. It will form a series resonant circuit with the reverse biased capacitance of the PMEG3010 (D1). It's capacitance is about 40 pF for a 6.5 volts average reverse voltage: -

enter image description here

The 6.5 volts comes from here: -

enter image description here

So, 16 nH and 40 pF resonate at 199 MHz (not far off).

Of course your coupling factor is probably not going to be as good as 0.998.

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  • \$\begingroup\$ Why secondary side inductance of 7.91uH not participating in oscillations ? in DCM primary side inductance and ouput cap of MOSFET oscillates when primary current reaches zero . like that why not secondary and output cap of Diode forming rining here ? is there a specific criteria to understand which inductance (magnetizng or leakge ) will participate during ringing ? \$\endgroup\$
    – Ronnie
    Commented Jun 29, 2023 at 9:05
  • \$\begingroup\$ @Ronnie because 99.8% of the secondary inductance is coupled to the primary and it is forced to a fixed voltage during the period when the MOSFET is activated. Therefore 99.8% is basically clamped leaving just the leakage inductance of 16 nH to form a resonant circuit. \$\endgroup\$
    – Andy aka
    Commented Jun 29, 2023 at 10:13
  • \$\begingroup\$ Put another way: if you took the transformer as a component and short circuited the primary winding and measured the inductance of the secondary winding, you'd measure 16 nH. However, if you measured the secondary inductance with the primary open-circuit, you'd measure 7.91 uH. \$\endgroup\$
    – Andy aka
    Commented Jun 29, 2023 at 10:51
  • \$\begingroup\$ so when primary is clamped to fixed voltage , is it acting like a short circuit because already inductor current (IL) is flowing through it ? \$\endgroup\$
    – Ronnie
    Commented Jun 29, 2023 at 14:22
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    \$\begingroup\$ @Ronnie correct but current doesn't need to flow for it to be regarded as a short circuit; it could be clamped to 0 volt and still be regarded as an AC short just the same as clamping it to 1 volt or 1000 volts; the DC current that may flow in the primary winding doesn't affect how it appears to be when analysing it at AC. \$\endgroup\$
    – Andy aka
    Commented Jun 29, 2023 at 14:26

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