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# LLC resonant converter

Other Parts Discussed in Thread: UCC256404

Hi all,

The diode bridge rectifier efficiency is around 81 %. But in case of a LLC converter the secondary is connected to a diode bridge rectifier. But still the efficiency is more than 95 %. How it is possible?

The efficiency of a diode bridge rectifier depends on the rectified voltage, rms currents, etc. Diode bridge rectifier losses are not the same for every design and therefore the efficiency will vary depending on the design. Which design are you referring to?

Best Regards,

Ben Lough

• Hi Ben,

I am referring to diode bridge rectifier at the secondary side of LLC converter.

The efficiency of a diode bridge is not always 81%. This is an extremely rough approximation. You need to consider the RMS voltage and current in the diodes as well as the output voltage and current. For example, take a look at the UCC25640EVM, a 180W design. The average current through each rectifying diode is 8.252A. As schottky diodes are used, the forward voltage is ~0.5V for both of the recitifying diodes. The conduction loss in the rectification stage would be 2x8.252A*0.5V = 8.252W. For 180W output, this loss is only 4.5% of the total output power, not 19% as the approximation would suggest.

Best Regards,

Ben Lough

• Hi Ben,

As per the data sheet of UCC25640EVM, the output power is 180 W with a DC output voltage of 12 V.

This indicates the average output current would be then (180/12) = 15 A.

Assuming the average current flowing through output filter capacitor is negligible, the average current through each diode will be 7.5 A.

In that case the total power loss would be = 4*7.5*0.5 = 15 W (Min. loss)

Please correct me if I am wrong.

This is not correct. The current through each diode is sinusoidal. You cannot just split the output current in half. In addition, there are only two diodes used for rectification, not 4.

Best Regards,

Ben Lough

• HI Ben,

In case of LLC converter, the output current is also sinusoidal. The average output current from the full wave rectifier is equal to (2*Im/Pi) and the average current through each diode is (Im/Pi).

No, that is not correct. Please refer to the design procedure in the UCC256404 datasheet.

Best Regards,

Ben Lough

• Hi Ben,

As per the design procedure mentioned in the UCC256404 datasheet, the secondary side is connected to a center tapped full bridge rectifier and the diode currents are assumed with an 110 % overload condition.

The average output current is 15 A. Considering an overload factor of 110 %, the current becomes 16.5 A.

So 16.5 A is (2*sqrt(2)*Is/Pi), where Is is the secondary rms current.

Hence the diode current is sqrt(2)*Is/Pi = 8.25 A.

As the configuration is center tapped, so two diodes (with a forward drop of 0.5 V).

Power loss in the present case = 2*0.5*8.25 = 8.25 W.

My calculation is based on assuming the configuration to be a full wave rectifier without a center tap transformer with 4 diodes.

Assuming the same diodes i:e a forward drop voltage of 0.5 V, the power loss will be 4*0.5*8.25 = 16.5 W.