i have the following circuit:

can one of you mosfet experts tell me whats the biggest R1 i can use that will allow me to switch the Pfet at like 500kHz for a buck convertor? The DIO pin is supposed to be able to sink 25 mA.
thanx mike

correction: the transistor is upside down in this picture

 

See

 

I am sorry.

 

A back-of-the-napkin calculation says about 10 ohms or less.

Borsdodynov's picture is worth a thousand words and those words are saying that driving that much capacitance at such a high frequency is not practical unless using active drive in both directions.

 

i don't wanna sound dumb but i don't get the implication here. i'm not trying to figure out Vout, it's Vg, the gate of M1, that i wanna know. I wanna gate it on and off without the extra mosfets, if it's feasible. So the gate capacitance is too much to achieve 500kHz. i scrounged around and found a IRF9Z34N pfet. it has a Vgs(th) of -2 to -4v, and an input capacitance of only 620pF. so if Vsource = 5 then it should be willing to turn off at Vgate = 3, right? and i won't need a pullup.

 

What current you are going to to commute?
Iout_mcu=20mA ==> Time_on(Gh-->Gl)=C*V/i=620pF*5/20mA=155ns (on transistor), Time_off> Time_on. F=500kHz ==> T=200ns

 

You've drawn the circuit with the transistor backward.

It looks as if you'll have to do without the resistor and just run the FET gate off a processor pin. I'm not totally comfortable with playing with the minimum gate voltage, but you are quoting the specs correctly; if the processor runs on 3V and Vsd is 5 (or Vds is -5) then the FET should be fully off with Vgs at -2. But then, your maximum Vgs in the ON state is -5. Will the FET be conductive enough at that voltage to give you the current flow you need at a voltage drop you can tolerate? You said Rds (on) = 11m, but that's with a Vgs of -10.

 

What current you are going to to commute?
Iout_mcu=20mA ==> Time_on(Gh-->Gl)=C*V/i=620pF*5/20mA=155ns (on transistor), Time_off> Time_on. F=500kHz ==> T=200ns

the pFET Vsource is connected to the 5V rail of my usb port, and my program continually adjusts the duty cycle so that the whole circuit draws a constant 450mA.
i kinda see what you mean, but i still don't get some of the details. 155ns to turn on. i didn't know about that formula. and >155ns to turn off. but i don't get the last part. i thought that if f=500kHz then it means that the period is 2us, which means that i can still get a fairly broad range of duty cycle. correct?

 

You've drawn the circuit with the transistor backward.
It looks as if you'll have to do without the resistor and just run the FET gate off a processor pin. I'm not totally comfortable with playing with the minimum gate voltage, but you are quoting the specs correctly; if the processor runs on 3V and Vsd is 5 (or Vds is -5) then the FET should be fully off with Vgs at -2. But then, your maximum Vgs in the ON state is -5. Will the FET be conductive enough at that voltage to give you the current flow you need at a voltage drop you can tolerate? You said Rds (on) = 11m, but that's with a Vgs of -10.

oh yeah, i did.
well i may have a solution. i found an old power supply and i'm gonna wire it up to my Discovery developer board http://www.st.com/content/st_com/en...tm32-mcu-discovery-kits/stm32f0discovery.html so now the MCU will have 3.6v. As for the Vgs, i think -5 is enough because i've already tested the whole circuit with a 5v arduino board and it works fine at 500kHz. so i know it's turning on and flowing plenty of current. so, if i can just get it to turn off....
i also scrounged around and found a FQP7P06 pFet https://www.fairchildsemi.com/datasheets/FQ/FQP7P06.pdf. it has a Vgs(th) of -2 to -4v, and an input capacitance of only 225pF. hopefully it'll do the job.
the only Q that remains is the second half of my circuit is a boost convertor that uses an nFET and i hope 3.6v is enough to turn it sufficiently on....
time will tell. THANX