bjtmosfet
I'm seeing the schematic of a Notebook HP Pavilion dv9000 (download). On page 17 in the corner left you can see this part:
Two questions:
What is wrong with your circuit is irrelevant from the point of view of getting a working unit.
There is no need for a MOSFET driver in this case. Use the IRLML2502 MOSFET and drive its gate directly from the digital output of the microcontroller. Check the IRLML2502 datasheet and you will see that it has a maximum guaranteed on resistance of only 45 mΩ with 4.5 V on the gate. You also don't need a resistor in series with the gate:
Yes, it really is that simple.
The voltage spike created by the leakage inductance of your transformer is breaking down your MOSFET each time it switches off. You need to limit this voltage somehow; the usual technique is to put an R-C snubber across the source and drain of the MOSFET, sized so that the capacitor captures the energy of the spike before the voltage rises too high.
The values required depend very much on the characteristics of your specific transformer, so you'll have to do some experimentation to determine them.
One way to start is to find the value of the peak primary current in the transformer; the resistor should be sized so that this current, multiplied by the resistance, gives a peak voltage that's comfortably less than the rating of the MOSFET.
At the moment the MOSFET switches off, the drain voltage initially spikes high because of the leakage inductance of the transformer, but then it settles down to a voltage that's proportional to the secondary voltage (by the turns ratio of the transformer). The capacitor needs to be sized so that it charges to that voltage level in a time that's somewhat longer than the duration of the spike. This depends on both the inductance value and the resistor value. Be conservative at first (i.e., use an over-large value) and then fine-tune it (for better efficiency) once you have the circuit working.
Best Answer
A mosfet requires usually a fair bit more gate voltage than the transistor ~0.7V so that's an immediate advantage. For a crude N-fet switch / emitter follower it also means that you've got that same ~0.7V voltage drop over the transistor instead of the mosfet >2V.
BJT is a current-driven device so the base will drain the line charge by itself even if there is no pull down/up resistor. Transistor base is also not bothered at all by a high voltage drive signal as long as you have a suitable series resitor in base. A mosfet will fizz if you connect e.g. 48V signal to the gate. Beware of the reverse voltage rating thought.
The transistor base current on the other hand is a source of many headaches if you are using a transistor as an inverter. That 60µA current is perfectly fine for lighting up a signal LED for example..