What is the exact difference between a series LC resonant converter and an LLC converter. From a topological perspective, they seem to be exactly the same. I believe that for an LLC, also the magnetizing inductance participates in the resonanance phenomenon but I am not certain. Does this only occur when the magnetizing inductance is small compared to the leakage inductance of the converter? Like a factor 10 instead of a factor 100?

Besides, does this difference impact the control of both converters?

Thanks in advance


Small magnetizing inductance + extra inductor = LC;

Large magnetizing inductance + extra inductor = LLC;

Large magnetizing inductance (air gap) + large leakage inductance (coils are separated) + no extra inductor = LLC

The best option is to use a transformer that is deliberately built as bad transformer, with windings separated a part, and with air gap. Then you get two extra inductances L_leak + Lm.

enter image description here A quality built transformer for SMPS has low leakage, this is done by doing stacked partial windinigs: primary, secodary, primary, secondary,... What you get is low magenetizing, low leakage tranasformer. Adding an extra tank inductor Lr you get LC topology, adding an air gap due to saturation you get LLC.

enter image description here

  • \$\begingroup\$ If you add an air gap, doesn't that reduce saturation? \$\endgroup\$ – joe electro Nov 17 '18 at 1:12
  • \$\begingroup\$ Thanks, but I'm not entirely satisfied with your answer. What exactly do you mean by large or small magnetizing inductance? By small I suppose you mean a regular transformer without airgap and by large a transformer with an airgap such that the leakage and magnetizing inductance can be tuned? However, including an airgap should lead to lower inductance (for the same amount of turns) as the reluctance increases and L = N^2/R. Could you put more exact number or percentages to Lm/Lsigma as when it would be considered an LLC? \$\endgroup\$ – Simon R Nov 20 '18 at 8:12
  • \$\begingroup\$ Also, you did not reply on my second question about the impact on the control algorithm.. \$\endgroup\$ – Simon R Nov 20 '18 at 8:12

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