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# Calculation of FET switch ON time for Boost PFC?

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#### cupoftea

##### Advanced Member level 5
Hi,
What value of CRSS do you take for calculating switch ON time of a Boost PFC, when the inductor current at switch_ON is 6A, and the FET is STW6N95K5? Vout is 390VDC.
Now, Page 9 of Lazlo Balogh……

Laszlo Balogh:
http://www.radio-sensors.se/download/gate-driver2.pdf

….gives the switch_ON time as being the sum of “t2” and “t3”…
Where t3 = CRSS * VDSoff/IG3 (page 9)

What value do you take for CRSS? (As you know, “CRSS” is obviously “CGD”)

Now, Page 4 of Balogh gives: CGD(equiv) = (1 + gfs * RL) * CGD

…..though what is “RL”?

….Presumably its "390V/6A" in this case?

I am taking gfs for this FET as 1.

STW6N95K5 FET datasheet

you are over thinking the whole process - the gate drive should supply enough current and have enough rising dv/dt to turn on the fet completely in 30 - 60nS ( empirical data ) quicker = EMC issues,

slower = increased switching losses

Some ppl switch a lot faster and use clever methods to limit the EMC arising, now with SiC diodes you can get away with a lot faster switching ( turn on ) than you ever could due to lower peak reverse currents in the diode.

Turn off should be maximally fast, i.e. < 30nS fall time on Vgs, to almost eliminate turn off losses in the fet.

Thanks, also, we are trying to calculate how to size the series gate resistor in order to get that 60ns turn on time.
To do this we will be charging up the CGD capacitor.
However, what value do we use for CGD?

Our driver is TC4428 and has vcc of 11.3V (TC4428 has internal resistance of 7 ohms)

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