AN4149 datasheet(22/43 Pages) STMICROELECTRONICS

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Designing a CCM FOT-controlled PFC

AN4149

22/43

DocID023523 Rev 2

2.3.6

Boost diode selection

Following criteria similar to those used for the calculation of the MOSFET losses, the output

rectifier can be properly selected. A minimum breakdown voltage of 1.2·(Vout) and a

minimum current rating higher than 5·Iout (equation 26) can be considered for an initial

rough selection of the rectifier. The correct selection is then confirmed by the thermal

calculation, as the diode junction temperature must be below 125 °C.

Since this circuit operates in the continuous current mode, reverse recovery is experienced

by the diode, and the MOSFET at turn-on has to carry also the boost diode minority carrier

charge. Then, to minimize the recovery losses, an ultra-fast diode with low trr (reverse

recovery time, the time required to deplete the stored charge) and Qrr (reverse recovery

charges, the charge that must be dissipated on the MOSFET) or a SiC rectifier has to be

selected.

In this 350 W application the STTH8S06 (600 V, 8 A) has been selected and shows very fast

reverse recovery time, 12 ns typical (measured for IF =1 A). The rectifier AVG (equation 26)

and RMS (equation 33) current values, the Vth (rectifier threshold voltage) and Rd (dynamic

resistance) given in the datasheet allow calculating the rectifier losses.

From the STTH8S06 datasheet, Vth is 1.2 V, and Rd is 0.087 Ω, the conduction losses are

equal to:

Equation 63

Since the converter is working in continuous conduction mode the losses in the MOSFETs

due to the recovery of the boost diode have to be taken into account. The energy loss due to

the reverse recovery effect of the diode is:

Equation 64

Where VR is the reverse voltage across the output diode, when it stops conducting, that is

400 V, and Qrr the reverse recovery charges, the charge that must be dissipated through the

MOSFET. On the datasheet the graph of Qrr vs. dIF/dt is represented. Following the

0.5 xIF(AV) curve at Tj = 125 °C a value of 80 nC is found assuming the typical case of

200 A/us.

Then the recovery energy is:

Equation 65

And the reverse recovery losses are:

Equation 66

rms

out

diode

⋅

()

diode

087

⋅

400

⋅

kH z

⋅

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