p-channel MOSFET switch

Question

I want to use a MOSFET as a switch driven by my microcomputer.

The original circuit using N-channel MOSFET is on the left side. Honestly, I do not understand the choice of the IRLZ44. The circuit is designed for Arduino, which has 5V logic. Which means that for GPIO=True=5V, MOSFET opens and lets the current into the load.

However I have two problems:

• I am using Raspberry Pi, which has 3.3V logic. According to available information 3.3V is not enough to fully open MOSFET.
• I want my load to be connected to the ground (I had to do some voltage measurements).

I know enough electronics to assume that using P-channel MOSFET, as shown on the right side, might solve both of my problems at one stroke. For GPIO=False=0V MOSFET will be fully open, while GPIO=True=3.3V puts -1.7V on the MOSFET gate and practically close it down. If that does not suffice, I could also put GPIO into the listening mode and therefore pull the MOSFET gate to 5V.

Could you please tell me if the idea will work? And what IRLZ44-equivalent P-channel MOSFET should I use?

Answer 1

First off, the rules of the site do state not to ask for recommendations of products, so I will skip that bit. Just read the datasheets as everything will be explained in there. If there is something on a datasheet you do not understand, please post a separate question about it.

Now, on to your problem. From what I think you are trying to do, you may find you might not be able to toggle the PMOSFET fully, or you may have some difficulty unless you understand the datasheets properly. What may be an easier idea, is to use a MOSFET pair, where you toggle an N-channel MOSFET to pull the gate of the P-channel to 0V, like so:

^{simulate this circuit – Schematic created using CircuitLab}

I have used this circuit a few times with no issues. However, as always, make sure to read the datasheets to make sure your components are able to do what you want. You don't always have to use the same components as shown in example circuits. Base your components on your own needs. Example circuits are great for learning how things work, but are not always the most practical. When it comes to designing your own circuit based off an example, you should always consider your own needs, and base your component choice off of that, rather than just use whatever the example has.

Answer 2

The trouble with using a high side P channel MOSFET driven from a signal that doesn't get close (less than 0.5 volts) to the high side voltage is that there is a decent probability that it will appear to be still active when you believe you have it turned off.

However, with some care you can put a zener diode in series with your 3.3 volt GPIO drive voltage to make this work better: -

^{simulate this circuit – Schematic created using CircuitLab}

Now the gate will be switched off and be also capable of being pulled down to 2.7 volts above ground meaning that there will be 3.3 volts between gate and source and hopefully you'll carefully choose a MOSFET that will work. I reckon you could choose a zener of 2.4 volts but you are starting to get to the point where leakage currents through the zener might still activate the MOSFET. Keep R2 low (1 k ish) to avoid this happening.

Alternatively use this two transistor circuit: -

If the high side supply exceeds 15 volts some extra care is needed to prevent gate-source breakdown voltages.

Answer 3

Here is another approach that uses an N-MOSFET in a level shifter configuration that does not invert the control signal polarity.

^{simulate this circuit – Schematic created using CircuitLab}

You need to chose MOSFETs with under 1V gate thresholds, M1 needs to be low Rds-ON. M2 can be a small signal device.

Answer 4

First of all, this wanted to be more like a comment, since it only deals with one of the aspects. For the solution, my preferred one is Andy aka's solution with NPN (nMOS have to be carefully chosen to have a Vgs-th less than 1.5V, while all NPNs work).

As for your question on the "input mode", keep in mind that the rPI pins are NOT 5V tolerant. This is because there is a parasite diode between the input and the power supply. You could theoretically ignore this if you force a very small current inside like

^{simulate this circuit – Schematic created using CircuitLab}

This way you are forcing through the parasite diode a current of (5V-3.3V-0.3V)/100kOhm = 14uA, which should be ok for the rPI. Please be aware, though, that the voltage on the gate will be slightly more than 3.3V in this case (let's say 3.6V).

Consequenly it is better if you use another transistor (I suggest an NPN like Andy proposed) to drive it.

Answer 5

In case you are not driving power for some device, but trying to send some data to 5V device, you can make it a bit easier.

^{simulate this circuit – Schematic created using CircuitLab}

Of course this mean, that "0" level will be 0.6V and "1" level 3.3+0.6=3.9V which is enough, for many applications.

Moreover on level "1" some current will be drained from 5V source to 3.3V, via parasite diode, as described by "frarugi87", but that will be only (5-3.3-0.6)/1000=1.1mA and since rPi consumes a lot more than 1.1mA you will never notice that. Maybe if you use some deep sleep modes and rPI consumes less than 1.1mA, then, you should increase R1 to 10k or 100k.