Out of curiosity...when I see a power supply of "9V", I assume that the circuit is intended to run off of a small, rectangular PP3 9V battery. Is this the case for your converter?
Actually , the supply is a 13.5V battery but it will go down to 9V during transient conditions.
-Regulations state we have to design for this voltage (9V) too which is a nuisance but we have to have the capability to give full output power at 9V input.
eem2am,
The switching losses and I squared R losses of the IPD079N should be lower since the rise and fall times and Rds(on) are all lower. In addition, the dissipation in the driver chip should also be lower, since the Total gate charge of the IPD079N is lower.
Regards,
Kral
Is this switching loss higher than the output load current? If no then the relative proportion becomes the point of concern to decide what issue to address first.
Raoof