Using TP0610 FET as a power switch

Hi, I'm trying to use a TP0610 P-channel enhancement mode FET to control the switching on/off of a device's Vcc with an MCU's I/O pin. The problem here is that the device's Vcc is 5.0V while my MCU supplies

3.3V only. I'm wondering if a Vgs of 3.3V will be sufficient to turn on the FET? According to the datasheet, Vt(min) = -1.0V and Vt(max) =

-2.4V(condition:Vgs = Vds, Id = -1mA). Does that mean that it takes anywhere from 2.6V to 4.0V at the gate to turn the FET on? If so, 3.3V isn't very safe is it?

Also, how do I calculate how much current I should supply from my MCU to my gate(hence calculating that series resistor value)? My device will have a max current of ~30mA and my MCU supplies a max of 2mA.

Thanks!

I am not sure how you are trying to implement this. I use this device as a power switch for low current devices (Rds ~ 10 ohm).

My typical usage is

TP0610 Source -> Vcc Gate -> 100k -> Source and -> control Drain -> device Vcc net

if the drive for the gate can not reach source potential (as seems the case here), then you'll need something to assist:

I usually use a 2N7002 (or equivalent) as follows:

Drain -> TP0610 gate (control, see above) Source -> ground Gate -> control from processor (or whatever). I usually add a pulldown on the gate (100k or so is usually fine) and sometimes a series resistor if it's a high speed line or otherwise should be damped.

Now if you take the gate of the 2N7002 high, the power switch (TP0610) will turn on.

Hope that helps

Cheers

PeteS

Thanks. What does the 100K G-S resistor do? I'm not sure how the 2N7002 helps either.

I tried wiring up the TP0610 with the following config Source -> 5V Gate -> 50K pullup to 3.3/5V -> jumper to GND Drain -> device Vcc to a dummy load(an oscillator that draws max 30mA)

I nocied that with a 3.3V gate voltage, the "off" Vcc is something like

4.6V, which doesn't seem very "off" to me. With a 5V gate voltage though, the "off" Vcc is lower at ~1.2V, but that still isn't completely off. Am I doing something wrong here?

When I jumper the gate to GND, device Vcc becomes 5V which is good.

OK, so I guess the 100K G-S resistor is just a pullup so that it's by default 5V(off state)? How do you arrive at the resistor value, and does it really matter?

Also, I tried putting a 200K pulldown resistor at device Vcc and now I'm getting an off Vcc of close to 0V. I'm really trying to connect/disconnect the device Vcc though, so it should be either 5V or floating. I'm just not sure why the floating Vcc is so high when I don't connect the pulldown.