Short answer: Maybe.
Medium answer: The FET will have a lot more switching losses than it would with ideal drive. You can try it and see how hot it gets.
Tedious answer:
It does look like 3.3V would be acceptable for the gate voltage, although conduction losses would be slightly lower with 5V.
4mA is conservative, but you have to stay below the 50mA short circuit current rating to avoid damage to the 2148. Lets assume you use 100 or 150 ohms between gate and 2148. That gives a peak discharge current of 15 or 20 mA during turn-off. To make the numbers easy, lets say that averages out to around 7mA.
Notice the total gate charge figure for the FDN337: 7nC typical (this depends on voltage). Anyway, using that number, the guess-timate for turn-off time would be
Lets just assume your voltage and current during turn-off are linear. Then your peak power will be halfway through, with i = .5*Ipk, and v = .5 *Vpk. So the peak power is Ppk = .25*Ipk*Vpk.
We could (should?) integrate to get the total energy, but since the estimate is so crude, we'll just assume the average dissipation during turn-off is half the peak. So
This occurs every cycle, so
Code: Select all P(switching) = f*.125*Ipk*Vpk*Toff = .125*Vpk*Ipk*Toff/T
Using Ipk = 1 amp, Vpk = 10V, Toff = 1usec, f = 40KHz, this gives 50mW.
After plugging in your voltage and current, you have to add your conduction losses, and your turn-on losses (if significant) then multiply total power by thermal resistance and estimate the junction temperature. This is a horribly crude estimate and could be off by a factor of 5 or so. Just ballpark figures to consider whether it is remotely feasible.
