in Circuit Maker I am using a couple of Power Mosfets (e.g. IRF624S) as simple switches. but I cannot get the drain voltage passed from the mosfet unless I bias it by the same voltage.
I mean for example: Vdrian= 100v DC, R(source-ground)=4k Ohms. I think when I apply 10 or 12v to gate I should have 100v across the Resistor. but I cannot!
can any one help me? I can send the schematic of my circuit to you.
ASCII art 100V ! G !!- D Vgate -----!! !!--------- To load S
If this is what you did, then you should expect that the load will always have less voltage on it than is on the Vgate. The MOSFET is turned on by the voltage difference between its shource and gate.
(2) Use a P MOSFET and put the load in the drain and reference the drive to the plus power. You may have to add a small NPN to the disign to shift the drive level. 100V ! +---------+------- / ! \\ R1 /---/ D1 / ^ ! ! +---------+------- To gate ! !/ --! Q1 !\\ e ! / R2 \\ / ! GND
R1 = R2 = about 10K
D1 = a 20V zener to protect against mistakes
Q1 = a > 100V NPN transistor
(3) Arrange it so that the gate voltage is developed referenced to the source. The easiest to explain is to use a "photo-voltaic opto-isolator"
(4) Gate on and off a AC (RF) voltage to something like this:
C1 D2 AC ----!!---+--->!-----+--------+----- To gate ! ! ! ---D1 / /---/ D3 ^ \\ R1 ^ ! / ! ! ! ! ----------+--------+------ To source
The AC (RF) input needs to be something like 20Vp-p at lets say 10KHz to
10MHz. The frequency needs to be so that at least 10 cycles will happen in the time you want the MOSFET to switch in.
D3 is a zener to prevent overvoltages from happening.
R1 and the gate capacitance of the MOSFET controls the turn off speed.
D1 and D2 are 1N914s
C1 is so that 1/(2*PI*F*C) is about 10 times less than R1.
ElectronDepot website is not affiliated with any of the manufacturers or service providers discussed here.
All logos and trade names are the property of their respective owners.