MOSFETs rated for >250V with Rds(on) < 0.1 Ohm and with a Qg of around 20nC
don't exist yet unfortunately.
However, some improvement can probably be made over the IRF740. The IRF740A
is a lower gate charge version of the IRF740 but is otherwise pretty much
identical. Some performance advantage could likely be had using that
device, however I might suggest the IRFB9N30A (available at Digikey) as an
even better choice still:
http://www.irf.com/product-info/datasheets/data/irfb9n30a.pdf
The device is a 300V 0.45 Ohm 33nC max device.
Be wary. Selecting a gigantic MOSFET rated for >250V and less than 0.1 ohms
on resistance will not likely yield the best efficiency possible. Not only
will such a beast have giant gate charge, but of additional interest it will
also have relatively large Coss (output capacitance). Some of the switching
loss is caused by slow transistions in the MOSFET itself (IE, periods of
blocking relatively high voltage but passing high current still), but the
MOSFET output capacitance can cause non-negligible loss contributions for
high voltage MOSFET run at high frequency.
Your MAX1771 uses a funky pulse frequency modulation technique but it can
easily decide to run at up to 300kHz. That classifies as high frequency,
especially when dealing with such large output voltages.
The output capacitance loss can be estimated if you know the switching
frequency, the maximum voltage the output capacitance gets charged to each
cycle, and the actual output capacitance value at the voltage it will be
seeing.
The formula for estimating the output capacitance switching loss is:
Power Loss = 0.5 * f * c * v^2
Where f is the switching frequency, c is the MOSFET output capacitance in
farads as measured in the datasheet at the drain-source voltage the MOSFET
will be seeing (250V in your application), and v^2 is the drain-source
voltage the MOSFET will be seeing squared. This formula assumes you are
charging the output capacitance from an inductive device. Since you are
using the MAX1771 I'm assuming you are using a plain ordinary boost
converter topology. In this case when the MOSFET switches off, the drain
voltage must rise from 0V (conducting) to 250V (off) in order for the output
diode to turn on. During this period the inductor transfers some of its
stored energy to the output capacitance losslessly. If the boost converter
is operating in continous conduction mode then this energy stored in the
output capacitance is lost when the MOSFET turns on for the next cycle (it
discharges through the MOSFET channel and heats the die). In discontinuous
conduction mode the inductor current drops to zero, but then reverses
direction because the MOSFET output capacitance is at 250V (which is clearly
larger than 12V supply). Then energy will flow back through the inductor
and try to charge up the input capacitors on the 12V side somewhat.
Eventually the output capacitance voltage will drop below 12V, but the
inductor will keep the current flowing for awhile and thus it may pull the
MOSFET output capacitance below ground and activate the MOSFET body diode.
There will be some ringing, damped by the Q of the system and any snubbers
if present, but in the end some of the energy may be returned to the supply,
while other energy may get lost. If the MOSFET was switching a resistor
instead of an inductor the above power loss forumula would not have the 0.5
factor in front, but would be 1 instead since there would be energy loss for
both charging and discharging the output capacitor then.
Anyway lets take a look at your application. Lets assume either you are
operating in continuous conduction mode or almost all of the energy in the
output capacitance is lost so the above formula is fairly accurate. With
the IRFB9N30A the datasheet claims the output capacitance is 52pF when
measured at 240V Vds. 240V Vds is very near your 250V figure, so we will
assume the 52pF is accurate (is practice it would be every so slighly
smaller at
the slighly higher voltage, but the difference is small). Now assume your
MAX1771 is switching the device at the full 300kHz.
Power loss = (0.5)*(300000)*(5.2x10^-11)*(250^2) = 0.49 Watts
Half a watt is not to be sneered at. A bigger MOSFET with lower
on-resistance will almost certainly have a larger output capacitance and
subsequently higher output capacitance switching loss contribution (not to
mention the other switching loss mechanisms) which will most likely more
than offset any gains in lower Rds(on).
What kind of diode are you using?
My suspicion is you could probably eek our better performance by modifying
your inductor (although considering the topology and your input output
voltage requirements your efficiency is already pretty good). A boost
converter with such a large input output voltage range is likely to have
high AC losses in the inductor. These can be minimized by paying close
attention to skin effect, proximity effect, and core eddy and hysteresis
losses. But that is all another big topic in and of itself.
