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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?

schematic

simulate this circuit – Schematic created using CircuitLab

Measurements

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    \$\begingroup\$ You can't reduce your part count to 1; think again; be realistic. State what might be acceptable and you might stand a chance. You can fudge a P channel MOSFET with a drive that is in series with a zener diode but, that requires a zener diode and some resistors. Part count doesn't equal 1. \$\endgroup\$
    – Andy aka
    Commented Nov 5, 2022 at 11:41
  • \$\begingroup\$ @Andyaka Thanks for confirming that I can't reduce the part count, that should be the answer. I'll stop barking up that tree! Edit: Perhaps I can reduce cost or complexity instead of reducing the part count? \$\endgroup\$ Commented Nov 5, 2022 at 11:42
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    \$\begingroup\$ But you can reduce the part count to one. Buy a suitable all-in-one component. These are called "power switches" and are available for both high side and low side switching, with active low or active high logic control input. Want that as an answer? \$\endgroup\$
    – Justme
    Commented Nov 5, 2022 at 11:50

1 Answer 1

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

enter image description here

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: -

enter image description here

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.

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  • \$\begingroup\$ Thanks for trying to explain negative Vgs(th), but the penny still hasn't dropped for me. Perhaps there's a chapter in Practical Electronics for Inventors or Art of Electronics that covers this? \$\endgroup\$ Commented Nov 5, 2022 at 12:26
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    \$\begingroup\$ I have no idea about books and what they may contain. Never bought books or read them. I just read data sheets and use the internet. Do you understand for an N-channel device, the gate has to be more positive than the source to activate the device? \$\endgroup\$
    – Andy aka
    Commented Nov 5, 2022 at 12:28
  • \$\begingroup\$ > "I just read data sheets and use the internet." -- Wow, ok I need to rethink a few things. \$\endgroup\$ Commented Nov 5, 2022 at 12:29
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    \$\begingroup\$ Yeah but I've been in this business since 1975 so I've learned a lot and luckily, remembered a lot of it. \$\endgroup\$
    – Andy aka
    Commented Nov 5, 2022 at 12:30
  • \$\begingroup\$ @NickBolton Negative threshold voltage is seen in P-channel enhancement-mode (the standard type) devices and in N-channel depletion-mode devices. Positive threshold voltage is seen in N-channel enhancement-mode devices and P-channel depletion mode devices. All the polarity means is that the threshold is when the gate is at a lower voltage than the source (for negative Vth) or when the gate is at a higher voltage than the source (for positive Vth). You can even get parts with a threshold voltage of zero. \$\endgroup\$
    – Hearth
    Commented Nov 5, 2022 at 18:25

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