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Alex,
This attachment is nearly correct. The Crss has some change in value due to the vari-cap effect but my views are almost a straight line ( during the plateau period) with a very small curve when the Vds is almost fully turned on.
Yar
Alex,
The rise time of the on or off of the fet gate drive is a function of the crss only. the Crss and ciss add up to be the total gate charge. The proff can be seen on the fets data sheet of the Vgs vs Qg. You will see a plateau in this curve. There is no current in the Ciss. All of the gate...
Hi,
I understand you have 3.3 volts and the 10 to 35 volt power supply but some how you need 12volts to drive the power fet. A three terminal regulator is a good way to get the 12 volts from the 35v supply but this is too close to the maximum limit of the regulators without adding a zener diode...
Sorry Alex,
I missunderstood. The correct calcs are using the load voltage needed to switch such as 30 volts and using the Crss of the fet as "C" then the needed gate drive current I is ( I= Crss ( 30 volts/ 200ns )) this gives the exact current needed to switch during the plateau period seen...
Alex,
The 12v is needed if a larger low cost avaliable mosfet is required because of the overheating mentioned earlier. Most of these are driven with 12v. Otherwise you are correct concerning the risetime and current required. The buffer is still needed to covert the processor output current to...
Yes you can. Locate a power buffer ( IC such as a DS0026 or better yet a newer low level mos fet gate drivers like the IR4426 ) capable of 50ns rise and fall times. If lower rise and fall times are used the returns will not be as accurate. A low inductance power supply with very short leads ( 1...
Your solution is not very simple. First a buffer is needed to amplify the output of the processor to drive the gate of the fet. The current needed to drive the gate of the fet at 100ns equals ( I = C x ( 12/100ns )). The C is the total gate charge. Also the load, if it is inductive, will need to...
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