How do I drive a 10 V high side PFET with only 3.3 V logic?

Question

I'm trying to use a high side PFET to drive a load (50 Ω below for example, but actual case is a 200 mA motor). I know that I can use a low side NFET but that is active high, and I want to use active low in my 3.3 V TTL (with a single part). For context, I'm trying to use a PCF8574 to switch a load, which is active high at power on.

Edit: If not possible to reduce the part count, perhaps I can reduce the complexity or cost. E.g. Andyaka's suggestion in the comments to use a Zener diode.

I'm able to turn the PFET off by using an NPN to drive the gate up to 10 V, but I'd like to use a single part if possible. Being familiar with NFETs, I tried driving the PFET gate to 3.3 V, but that didn't turn the PFET off (I'm still trying to understand PFETs, hence this question). I see that you have to drive the gate all the way to 10 V to turn it off. I read the datasheet, and \$Vgs(th)\$ is -1 V to -3 V, which I don't understand.

I suppose I could also use an NFET to drive the PFET instead of the NPN below, and in this case maybe that's what a complementary PFET/NFET IC is designed for? If so, that'd reduce it to a single part. Is that the best way to keep my part count to 1 and achieve active low?

^{simulate this circuit – Schematic created using CircuitLab}

Answer 1

You can fudge it with a zener diode like this: -

But you need to be careful in selecting the resistor and, it only works when the supply remains at 10 volts and your logic level is 3.3 volts. But, you might as well use an N-channel device like this: -

So, this time you drive the source of the N-channel device (common gate circuit). This permits you to have the input active low i.e. when the drive input is low, the P-channel MOSFET is activated.

I read the datasheet, and \$V_{GS(TH)}\$ is -1 V to -3 V, which I don't understand.

It means that the gate has to be more negative than the source to start activating a P-channel device. In comparison, for an N-channel device, the gate has to be more positive than the source to activate the device.

I suppose I could also use an NFET to drive the PFET instead of the NPN below, and in this case maybe that's what a complementary PFET/NFET IC is designed for? If so, that'd reduce it to a single part. Is that the best way to keep my part count to 1 and achieve active low?

You will still need a resistor and that makes two added components. You need at least two components added to make this work.