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The document "Under the Hood of Flyback SMPS Designs", talks about the leakage inductance effect (on page 6). I have some questions about that.

The below questions are from fig 5.

  1. Why high clamp voltage will have a smaller Dtr than Low clamp voltage?

  2. In the low clamp voltage figure, there is a "lost voltage - seconds", what does this mean?

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  1. In the yellow part, does this mean increasing the Vleak2 voltage will reduce the IP? If yes, I don't very understand it.

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  1. In the yellow part, the magnetic inductance will increase the voltage to suppress the current change, am I correct?

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  1. Why higher clamp voltage may degrade cross-regulation performance?

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2 Answers 2

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Simplify the circuit: if N2/N1 = 1, then the transformer can be removed entirely; LL and Lm are shown, so the transformer element itself is ideal, justifying this step.

Then you simply have two inductors in series, at the moment of switch turn-off.

Inductor charge corresponds to flux (volt-seconds) corresponds to amperes. (Indeed, Ohm's law for an inductor is \$L=\Phi / I\$.) Thus the voltage must spike up by some amount, at the instant of switch opening, to equalize currents in the two inductors (which are now in series).

The article's remarks follow from this.

  • A shorter Dtr corresponds to a higher Vclamp. \$V = L \frac{dI}{dt}\$ so it takes some \$dt\$ and applied \$V\$ to cause current to rise by some \$dI\$. (For square waveforms like this, change \$d\$s to \$\Delta\$s and that gives the total change.)
  • Some flux has to go into charging the leakage, which must be subtracted from the magnetizing inductance's peak flux at turn-off.

Regarding cross regulation, consider the worst case.

Consider the model at all, to begin with; for this case, we have multiple windings, and we have in general a matrix of leakages between any pair of them. The values of which are constrained by each other, for example no value can be negative, the values are symmetric (LL(1-2) = LL(2-1)), and no value can be greater than the sum of leakages in a given path between windings (that is to say, LL(1-2) + LL(2-3) <= LL(1-3), etc.). (And I think some other particulars; honestly, I don't know the exact/concise definition, as it isn't very important outside of writing simulators/models.)

The worst case is exactly the sum case, which is equivalent to using two completely independent transformers with the primaries wired in parallel. Obviously, we have the case that LL(S1-S2) = LL(S1-P) + LL(P-S2).

If we have unequal loads on S1 and S2, and no clamping at all on the primary, then the flux imbalance between secondaries equals the difference in currents times the sum leakages.

If we have clamping on the primary, then we use up some fraction of output power in the process (there's always a downside!), but each secondary is constrained by its own leakage to the primary leakage. In the limiting case with primary clamping exactly equal to (*), the primary voltage is held constant regardless of secondary load(s) and thus only each individual leakage is in play (cross regulation equals self regulation).

*Well, they don't give it a symbol, but it's the "Current Circulates in Secondary Winding(s)" arrow.

Leakage can be better than the parallel-transformers case, by having better coupling between other pairs of windings. Consider a windup with a full section P, then S1 and S2 together in a single section, say as twisted pair. They will have relatively little leakage between each other, and equal amounts to the primary (LL(P-S1) ~= LL(P-S2)).

As the diagram suggests, physical arrangement/construction completely determines leakage.

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  • \$\begingroup\$ Hi Tim, Thanks for your sharing, A shorter Dtr corresponds to a higher Vclamp: you mean I, L, and V are constant, so higher will have smaller dt, the dt is Dtr? but why I is constant? Regarding cross-regulation, do you have a figure could share with me, it is hard for me to udnerstand your explanation. \$\endgroup\$
    – Jitter456
    May 24, 2022 at 14:17
  • \$\begingroup\$ @Jitter456 I is the current at turn-off, and dt is any relevant change in time, in this case, Dtr, exactly. \$\endgroup\$ May 24, 2022 at 16:42
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1.) Why high clamp voltage will have a smaller Dtr than Low clamp voltage?

When the BEMF voltage is high, dI/dt=BEMF/L , dt is inversely low as the Change in current dI and L are constant. The rectifier or clamp limits that voltage, BEMF. A diode to V+ on the primary side is the slowest release time.a

2.) in the low clamp voltage figure, there is a "lost voltage - seconds", what does this mean.

At switch off, Vmag reverses polarity and the Vleak rises if Vout is low. Vleak=Vmag-Vf.diode-Vout

3.)In the yellow part, does this mean increasing the Vleak2 voltage will reduce the Ip?

No it means the secondary clamp also clamps the primary voltage and thus more leakage loss with the same current decaying as expected.

I think you should be able to answer this if you understand V = L x something

  1. Why higher clamp voltage may degrade cross-regulation performance

Impedance is raised by high dV/dI=Z(t) thus weaker coupling

I think you used up your quota of questions for a week ;)

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  • \$\begingroup\$ Hi Tony, 1.) You mean, I, L, and V are constant, so higher will have smaller dt, the dt is Dtr? but why I is constant? I think it should change. 2.) I still don't understand what's the relationship between "lost voltage - seconds" and Vleak=Vmag-Vf.diode-Vout 3.) What it will increase the leakage loss? 4.) V=L di/dt 5.) Why high impedance cause weak coupling? \$\endgroup\$
    – Jitter456
    May 24, 2022 at 13:54
  • \$\begingroup\$ 1) No Vmag and leak increases with clamp V 2) Lmag lossses absorb V-s with more A-s from lower clamp as in 3), 4)yes V increase with L 5) Why? That should be obvious from Thevenin cct. With ratiometric outputs. \$\endgroup\$ May 24, 2022 at 15:58

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