Everything seems fine, except right at the end. You are equating the average power delivered by the inductor L, with the average power delivered by the power supply (\$V_{in}\$). As you found out, this is only true if \$D=1\$ (duty cycle of 100%).
Simply put, during \$T_{off}\$ the power supply adds its voltage to the voltage that L develops, and share the same current, so the energy delivered to the load HAS to be more than what the inductor delivers (which can't be more than what it accumulated during Ton).
In the general case, during the on-time (\$T_{on}\$) the power supply delivers an energy of \$V_{in}I_{avg}T_{on}\$ exclusively to the inductor L. All of this energy has to be delivered to the load during the off-time (\$T_{off}\$), because this is supposed to be perfectly cyclical. But during \$T_{off}\$, \$V_{in}\$ is still delivering current, with the same average as in \$T_{on}\$ (we're calling it \$I_{avg}\$), and that energy \$V_{in}I_{avg}T_{off}\$ is going to the load as well.
So in the end the load gets, per cycle, an energy of \$V_{in}I_{avg}(T_{on}+T_{off})\$, which means the average power \$P = V_{in}I_{avg}\$. But this is not the same as the average power delivered by the inductor, which is only \$V_{in}I_{avg}(\frac{T_{on}}{T_{on}+T_{off}})\$.
A way to look at it is that, per cycle:
- [Total Energy received by the load] = [Total Energy delivered by power supply] = [Energy delivered by the power supply during Ton] + [Energy delivered by the power supply during Toff].
- [Energy delivered by the power supply during Ton] = [Energy delivered to the inductor during Ton] = [Energy delivered to the load by the inductor during Toff] < [Total energy delivered to the load].
So your last equation is correct:
\$ P_L=ft_{on}V_{in}\hat{I}_{in} \$
But it just talks about the average power delivered by the inductor, and is not all the average power delivered by the power supply:
\$ P_{PS} = V_{in}\hat{I}_{in} \$
Well, only if \$ ft_{on}=1\$ (100% duty cycle), or Iavg=0 (0% duty cycle). But we wouldn't even be in the correct operating mode in those cases.
If the Spice model of the IRFP4668 is at all accurate, it will not be turned on well with only 5V gate drive. You should increase the gate drive to 10V to support higher current. Expect the power modulator to have a resonant frequency of ~ 1200 Hz, with a Q of ~ 4. So, it will take a lot of simulated time to settle, probably at least a second and the output will be ringy.
Peak switch current with a 34W output and 11V input and 64uH inductor will be ~3.4A. The LM2585 switch can limit as low as 3A, so don't expect to get 2A output. It might be able to put out 1.75A with a 3A switch limit. The inductor will need at least a 5A rating to avoid saturation and thermal problems.
Also, the 12V input will need to be able to supply the 3+ amps.
Best Answer
The RMS current is the equivalent DC current, so if you switch 1 Amp of current with a 50% duty cycle, your RMS current would be 707mA,
However this is not the only part that causes power dissipation in mosfets, there is also switching losses, to do with how when the mosfet is switching on or off, there is a breif moment each time when its not fully on or off, but rather acting as a resistor,
Boost converters depending on there topology may also see some heating through the use of its body diode instead of an external diode, but this is rarer these days.