Electronic – Synchronous DC-DC converter and fast diodes

dc/dc converterdiodesmosfetpowerswitch-mode-power-supply

I'm a bit at a loss when trying to optimize a DC-DC converter (Half Bridge) and selecting my MOSFETs.
Up to 200V I get good results (MOSFET switching times are fast ~ 20ns and the anti-parallel diode is really fast as well – Qrr is less than 1 uC). For higher voltages the MOSFET may still switch really fast ~ 20ns, even though the anti-parallel diode is bad: Qrr in the order of 20uC or even more. This causes very high switching losses at say 20kHz (well – at any frequency – but at around that frequency they become important when compared to conduction losses).

Now for the question: at high(er) voltage(s) the anti-parallel diode is bad in switching (Qrr high), even though its forward voltage drop is usually quite good (less than 1V sometimes). I don't really care since I want to use a synchronous converter (using one MOSFET in reverse conduction mode).
Should I put a fast (with low Qrr ~ 0.3 nC, but with a higher forward voltage drop) diode in parallel to the MOSFET, what would happen? Would the integrated body diode still switch on because of lower V_F (even though this would happen after the fast diode switched on) or could I take advantage of the additional diode's switching speed and short-circuit it with the MOSFET just a few ns after before the body diode could turn on?

Basically: would the slow (but with "good" conducting properties: V_F low) body diode of the MOSFET still turn on even though a fast (but with "bad" conducting properties: V_F higher for the same current) diode is in parallel to it?

I hope my question was clear. Feel free to ask other details if you need them.
Please link / cite any useful documentation if possible.

Additional thoughts: since the body diode may start switching (anyway) when the external diode is already conducting (having therefore a voltage drop lower than say 2V), the body diode would cause very low energy losses because the blocking voltage for it is now 2V, not the (say) 200V when the external diode was blocked. The additional switching losses caused by the unwanted turn-on of the body diode may be nothing worth to be considered. Right?

Best Answer

After discussing this at university here's the answer: no, it's not possible.

Because if the external diode has a higher V_F value, it will be short-circuited by the low-V_F high-Q_rr internal body diode of the MOSFET. Therefore at turn of the internal body diode will have the same switching losses, since the external diode practically never goes in conduction mode.

The only "solution" seems to be to get a hand of a bit higher (~10-20%) R_DS,on MOSFET which has a better diode (10-20 times lower Q_rr) and eventually put some more MOSFETs in parallel in order to lower conduction losses.