I dismantled a cheap Chinese 240VAC to 48v-20A SMPS to try to understand the basic working principle.

There is a CBB22 335J630V polypropylene film capacitor in it. I have drawn it in red on this basic schematic I downloaded from ti.com of a half bridge SMPS:

enter image description here

What is the purpose of this capacitor? The Chinese wouldn't have put it there for no reason at all. Half of the time they don't even add a fuse so it must have some important purpose. Why would it be omitted from a basic schematic?

  • \$\begingroup\$ Maybe they are using it to partially resonate with the primary inductance? Maybe they are using it to prevent massive flux overload if the MOSFET operating frequency was too low? \$\endgroup\$
    – Andy aka
    Mar 15, 2023 at 12:53

2 Answers 2


This is the resonant capacitor of an LLC resonant converter. Your PSU is likely using the stray inductance of the transformer's primary winding as the resonant inductor.

The PSU is probably of higher quality than you'd think - LLC resonant converters are one of the most efficient topologies for off-line switchmode power supplies, which is why you see this topology a lot in high-power (>800W) PSUs. Yours definitely fits that description with its (almost) 1kW of output power.

The capacitor's datasheet also hints at its purpose as a resonant capacitor: It has a dissipation factor of at most 0.0015 at 1kHz, which is equivalent to an ESR of 70 mOhms. This is low enough to handle the typical resonant currents on the primary side of a 1kW LLC resonant converter, which are on the order of 5A, assuming that it's fed with 350V DC from a boost PFC. This results in approximately 1.75W of dissipation in the cap, which is fine.

Regular non-resonant half-bridge or full-bridge buck-derived converters (which you see up to about 500W) don't have this capacitor, as you've already noticed.

  • 2
    \$\begingroup\$ It also blocks DC if your circuit does not do a good enough job of preventing staircase saturation of the transformer. \$\endgroup\$
    – winny
    Mar 15, 2023 at 13:12
  • \$\begingroup\$ many thanks, i will have a read of the LLC resonant converter link. it was part of a backup power feed of my house that has worked for years, i limited it to 18a not wanting to push look letting it run at 20a. most of the time it would never hit 18a anyway, but my wife put the electric car on charge during a power cut, long and short of it, it supplied 18a for about 6 hours before going pop. i have replaced it with a mean well equivalent now im just using it for the learning curve. \$\endgroup\$
    – Jay Dee
    Mar 15, 2023 at 13:25
  • \$\begingroup\$ This is NOT an LLC. An LLC converter normally doesn't need the capacitive divider (C1-C2). And in an LLC converter no one would put the resonant cap between the switch point and the transformer, because the voltage across it is used to control power or to have an idea about the load current (i.e. lossless current sense). Therefore the capacitor would be placed in a way that one end is ground. The converter shown in the question is a classic hard-switched half bridge converter and the cap is a DC blocker to prevent core saturation. Nothing else. \$\endgroup\$ Mar 15, 2023 at 13:35
  • \$\begingroup\$ @RohatKılıç This is not true - see the linked Infineon application note; it shows a typical full-bridge LLC converter with floating resonant capacitor in figure 2.1. Infineon also details that a full-bridge design is recommended for designs with high primary current (= high power). It's also advantageous to use a capacitive divider in a half-bridge variant instead of a single cap to ground, the app note also shows this. (OP's design must be a full-bridge, though, otherwise the cap wouldn't be floating) \$\endgroup\$ Mar 15, 2023 at 17:50
  • \$\begingroup\$ @JonathanS. 1) for a 1 kW design, regardless of being a half- or full-bridge, it is NOT feasible and practical to avoid lossless current sensing. How much would be the loss with resistive current measurement for this power level? 2) for a 1 kW design, the resonance frequency won't be high, so assume 85 kHz, and the resonant choke (or leakage) for Cr=3.3uF would be 1uH. It's known that the magnetising inductance will be 3 to 10 times the resonant inductance, so the magnetising inductance would be 10~15uH. Calculate the primary current for this mag inductance and freq, and think about 1 again. \$\endgroup\$ Mar 15, 2023 at 19:35

The schematic shown is a classic half-bridge converter.

Theoretically, volt-seconds product for the transformer's primary (i.e. magnetisation and de-magnetisation) is equal for each half cycle. But due to the imperfections of the real world components there might be a very small offset which will push or pull the B-H loop up or down. This, eventually, will result in core saturation which is never desired in an half-bridge converter.

To prevent this, a DC blocking cap is placed. Its value should be selected so that the reactance it shows at switching frequency can be negligible.

  • \$\begingroup\$ Since this is a half-bridge, isn't it already impossible for DC to flow in the primary because of C1 and C2? \$\endgroup\$
    – John D
    Mar 15, 2023 at 20:49
  • \$\begingroup\$ @JohnD No. The primary voltage swing is, theoretically, ±V/2 because of the capacitive dividers (hence the term "half bridge". With full bridge the swing would be ±V). Practically, with a capacitive divider, it won't be possible to divide the input voltage exactly in half. So the volt-seconds for primary won't be equal for each half cycle. This causes flux walking, and leads to core saturation, eventually (after a couple or maybe a couple of trillion switching cycles). \$\endgroup\$ Mar 15, 2023 at 21:00
  • \$\begingroup\$ Of course, you're right, I just had a momentary lapse of brain involvement. \$\endgroup\$
    – John D
    Mar 15, 2023 at 22:46

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