On every page of my blog, you might notice a chat window. If I’m not busy, we can chat in real-time. If not, the messages come to me by email. Here’s one I got from Matt the other day:
Let’s talk a bit about how (and why) you would use a P-Channel MOSFET. Matt, and he’s not the only one, is probably asking this question based on the “myth” that P-Channel MOSFETs require “negative voltage” supplies.
Keep reading for a how to use only positive voltage in this p-channel MOSFET tutorial.
N-Channel and P-Channel MOSFET Tutorial Video
First, for some background, check out this AddOhms video tutorial about MOSFETs, I already made.
What’s different?
MOSFET Symbols from Wikipedia
The actual construction of the two types of MOSFETs differ. There’s plenty of explanations out there on the layers and how they are made. Instead, let’s focus on how you use them in a circuit.
VGS Threshold
The first thing we need to know about using MOSFETs is one of their critical properties: VGSth. Which stands for Voltage Threshold from Gate to Source. As the voltage difference between those two pins changes, so will the resistance from the DRAIN to SOURCE pins. This is how a MOSFET turns ON and OFF.
How that resistance changes, depends on if it is a N-Channel or P-Channel MOSFET.
P-Channel MOSFET Tutorial and Explanation
Look at the VGSth for a P-Channel MOSFET. You might notice that the VGSth is a negative value. We can use the data sheet from an IRF5305 as an example.
Its VGSth is specified as range: -2.0V to -4.0V. So, how could you possibly use this MOSFET with an Arduino, LaunchPad, Raspberry Pi or any other microcontroller? Do you really have to generate negative voltages?
It’s about the difference
This is where the myth of the “negative voltage” comes in: Since the data sheet says negative, clearly, you need negative voltage to work. And data sheets, never lie (except when they do…).
Let’s literally read what the specification is saying. “Voltage from Gate to Source of negative four volts.” Using different words, you might read it as: subtract the GATE voltage value from the SOURCE voltage value.
Look at the voltages in this “high-side switch” configuration:
The GATE is now at 5 volts. The SOURCE is also at 5 volts. That means the VGS is 5V – 5V = 0V. So the VGS in this case is 0 volts. This means the MOSFET is turned off or an open.
Now this is the same circuit but the GATE is connected to ground instead of 5 volts.
Let’s look at SOURCE and GATE again. The SOURCE is still at 5 volts. However, now the GATE is at ground which means it is 0V. If you take the GATE voltage and subtract the SOURCE, you get: 0V – 5V = -5V. This will turn the MOSFET on.
See what just happened there? We got a “negative” voltage using only positive voltage supplies…
Why use N-Channel over P-Channel
We would need to dedicate a tutorial on when to use an n-channel and p-channel MOSFET. A great use for P-Channel is in a circuit where your load’s voltage is the same as your logic’s voltage levels. For example, if you’re trying to turn on a 5-volt relay with an Arduino. The current necessary for the relay coil is too high for an I/O pin, but the coil needs 5V to work. In this case, use a P-Channel MOSFET to turn the relay on from the Arduino’s I/O pin.
If your load voltage is higher, like 12 or 24V, then you might want to use an N-Channel MOSFET in a “low side” configuration.
Conclusion
Using a P-Channel with positive voltages is easy, when connected to the circuit correctly. We just have to get over the myth of they work on “negative voltages.”
Question: What’s the most creative MOSFET circuit have you seen? You can leave a comment by clicking here.
Hi, Very nice addohms video as usual. I have a logic level n-channel mosfet(FQP30N06L) and switchin on-off by appliying 0 and 3.3v Vgs from a gpio pin. Drain side I have a load operates at +5V. After the pc turns off, PC pulls up +5V all GPIO pins(Vgs). As a result after pc turns off, it turns on the load again. How can I prevent the gate voltage if more than 5V at the logic level mosfet? Best.
I don’t understand your question. You may need to use a P-Channel FET as a high-side switch. Or you may need to add a pull-down resistor to the N-Channel’s gate.
Thanks for the reply. Let me explain again. We simply control a load by using a gpio pin on the gate side of mosfet. From the industrial pc software, it gives 3.3 and 0 volts for high and low to the gate. But when we shut down the industrial pc, it pulls up all gpio pin to 5V(we dont know why) I would like to prevent the load to turn on if the gate voltage is higher than 3.3volts(it can be 4.5V also according to vgs threshold). Hope this is clear. Best.
If your driving signal tends to float high, I would suggest replacing the circuit with a p-channel high-side transistor. On the N-Channel, you could also try using an NPN to invert the driving voltage.
In this circuit when the input to the NPN base is high, it will turn on, which puts 0 volts on the gate of the MOSFET. When the NPN’s base is low, it will turn-off, putting 5 volts on the gate of the FET.
To be honest I dont think the diagram above should work high logic signal at the gate since it floats between 3.3V and 5V. It is ok to control with 3.3V and 0V but not ok with 5V. We dont want to current flow to the load(5V, 2A) at 5Vgs on mosfet(FQP30N06L). As you said (also in my opinion) a P-channel transistor might be placed. According to diagram we get 5V high from the plc after shut down which activate npn either way. Also you can think the load is the 1k resistor at the right hand side. Briefly we need a transistor doesnt turn on above 3.3v at the gate. Many thanks.
Thanks a lot for this article. One point has me confused. Early on, you mention:
“Let’s literally read what the specification is saying. ‘Voltage from Gate to Source of negative four volts.’ Using different words, you might read it as: subtract the GATE voltage value from the SOURCE voltage value.”
But subtracting gate voltage from source voltage is Vs – Vg stated mathematically, and it is the opposite of what you do in your examples.
Funny enough, the way you stated it verbally makes more sense than Vg – Vs. When I think “from point A to point B”, I tend to think B – A, not A – B. For example, if A is 3V and B is 5V, to get from A to B, I need to add 2V, so Vab = 2. I’m just so extremely confused with notation because this isn’t the first place where I’ve read about MOSFETs and saw that Vgs = Vg – Vs. It just makes no sense to my mathematical mind why it seems to be reversed.
Can you help illuminate?
I’ve given up trying to explain it in words. I’ve changed it 3 times in 3 years and people still say it is confusing. So far everyone’s “clarification” confuses someone else. I think it comes down to how you process math in your mind. So, Vgs = Vg – Vs.
I don’t have the project with me any more. It was a gift for a friend who was going away, but I do have all the parts and one of the FETs mounted on an SMD to DIL PCB. I will make up the circuit again on a breadboard and play around with it. If I come up with something, I’ll post it here.
When I did the original project I had limited time so as long as the theory was solid (based, actually, on your explanation, thanks) and the completed project worked for a few hours without any heating I was happy.
Thsnks for this most informative blog. It helped resolve a problem I had with the P-channel MOSFET AO3401. I connected it in exactly the same configuration as in your last circuit with an additional 10K resistor between gate and source. The input went to a CMOS IC and the load was parallel LEDs totaling 120mA. It always remained ON irrespective of the CMOS output being high or low.
If I disconnected the 1K from the CMOS output, connected it to a transistor collector and the transistor base (via a resistor) to the CMOS input, emitter to ground, the LEDs switched on and off with the CMOS output.
I had limited space. I didn’t want to use a transistor.
I ended up connecting the circuit exactly like yours, but with the load between the 5V supply and the source of the AO3401, drain to ground. It worked!
My logic is that since it is a series circuit the MOSFET should handle the current whether the load was in the source or in the drain. My question is: am i right and will there any problems with the MOSFET connected in this way. The reason I ask is I googled a lot of P-Channel MOSFET schematics, but I never saw one connected this way.
That’s creating a “low-side” switch. Usually, a P-Channel will be configured as a “high-side” switch as I show here. However, it just comes down to the voltage difference between gate and source.
The current isn’t an issue in either a low-side or high-side circuit. It comes down to the voltage difference between gate and source. Without a schematic, it is hard to evaluate if your circuit should work as expected.
Thanks for taking the time to reply to my enquiry. Sorry about the delay in my reply. I was looking for a good way to make schematics on my phone. I found a very nice app. I’ve uploaded my schematics to dropbox so that you can have a look. The link is here: https://www.dropbox.com/sh/v139ctjkrcm6d45/AACG_kwgV25Zk4QblPQCmJTya?dl=0.
The TTP223 I/O terminal can only handle 8mA so you need a driver to switch LEDs. I wanted to mount the driver on the 11x15mm sensor board. There was just enough space for one SMD MOSFET and resistor on the touch plate side.
In Fig1, the LEDs remain permanently ON, no matter what the I/O voltage is. In Figs 2 & 3 they switch ON/OFF with the change in I/O voltage.
I measured the I/O voltage and in all cases the high/low swing was 0V to 5V. You can set the jumpers on the sensor board for high on or low on, momentary on or latch on. The I/O terminal is specified as CMOS [probably similar to the push-pull kind of arrangement available in CMOS gates] as opposed to open Drain available in the 16-pin version of the IC.
I was puzzled with the result from Fig1 because I expected the FET to conduct only when the gate voltage came down to near ground voltage, but in this case it was conducting all the time.
I have already completed my USB touch activated night light using Fig3, but I’m still not sure why Fig1 didn’t work. Thanks again for your help.
I’m not sure why it is staying on, I would expect Fig 1 to work. However, I would have implemented with figure 2. I would not have expected figure 3 to work at all.
Great article! For a battery powered ATtiny project I have been using a low-side configured N-channel enhancement mode MOSFET to turn on and off a circuit so that the whole product draws just a few microamps when in sleep. A new requirement is that the circuit is turned on and off on the high-side. A single P-channel MOSFET configured to the high-side seems like it is on the right track but requires that the pin be set high to turn off the circuit so that won’t work to keep the current usage to near zero when in deep sleep. I need to keep the cost and parts count low and would really appreciate any ideas you have on this.
Thank you profusely sir, for explaining so simply yet to the point
What happens if we provide negative supply to the p mosfet? How to turn it on? Do we need a level shifter?
The voltage from gate to source needs to be negative. It does not need to come from a negative supply.
Suppose we want to run it with a negative supply then how should we level the voltage of gate?
I don’t understand what you mean. I don’t understand “run with negative supply.” A transistor is a simple switch.
I mean if we provide say -12 volts to the source and ground the drain. In this case, how to turn the p mosfet on?
Thank you.
Please how do I identify n channel and p channel mosfet on laptop motherboard. Reply asap
If you cannot identify the part from its markings and you do not have a schematic for the circuit, you will probably need to remove it. From some basic testing, you can determine if it is N or P, but you probably can’t determine an appropriate replacement.
I was searching for more information on how the P channel was different than the N channel but nowhere did I find such a great explanation as this page.
Yes, I was confused by this as well.
what happens is your gate value is bigger than the source? does still operating?
Nope. In that case Vgs would be a positive voltage, so the p-channel would not work.
Hello sir we are trying to make H bridge inverter using 4 N channel MOSFET. But when we triggered the gate pulses simultaneously only negative voltage output waveform is fine. But in positive voltage the voltage is too short than Negative voltage. Is there any need to change the positive side N channel MOSFETS to P channel?
Thank you.
Hello,
I have an TP0610K P Channel Mosfet –> VGS – 1 to – 3
And i get on a laptop motherboard (black probe on gnd)
Source -> 19V
Drain -> 19V
Gate -> 3.3V
The gate shouldn’t have 19v- 3v ?
Sorry for a newbie question 🙂
The threshold value given, VGS, is the minimum needed to activate the FET. In the absolute maximum section of the datasheet, you’ll see the maximum value of Gate – Source. 3V – 19V = -16V.
Nice article, but the phrase below seems backwards to me:
“The SOURCE is still at 5 volts. However, now the GATE is at ground which means it is 0V. If you subtract the GATE from the SOURCE you get: 0V – 5V = -5V. ”
If I subtract GATE (0V) FROM the SOURCE (5V), shouldn’t that be read as “5V-0V”, not the other way around?
Thanks
I think the source of confusion is the earlier sentence “Using different words, you might read it as: subtract the GATE voltage value from the SOURCE voltage value.” If it read “… subtract the Source voltage value from the Gate voltage value.” then it would be consistent with math examples that follow.
Does this device just swing from hi to low, or will the drain voltage follow a varying gate voltage?
That behavior isn’t defined by N or P channel types. A JFET’s drain voltage will vary with the gate voltage, in the ohmic (or linear) region of the transistor. Enhancement mode FETs will as well, but their range is extremely limited. They’re really designed for operation in the saturation region. Here’s a good app note on the subject: https://www.vishay.com/docs/70598/70598.pdf
12v or 6v on board and 5v in Proteus simulation
If I have a Source =12V (P mosfet) Drain – Load – Negative power so Vgs just 8 to 10v (because 10-12=-2 or 8-12=-4v ) I will be not connected the Gate to 0v or Ground, Do I do correctly? thank so much
I have a problem with Mosfet P channel when I assembly load on Drain and connect Source to Vcc make 0v at Gate :result fail . when I simulate a mosfet P channel (IRF9540N) with a Load on drain and run project : Result fail. Why? please tell me how to use Load in case P channel and bias of Vgs ( in N channel case is very easy) thank a lot
What is the Gate Voltage you are using?
12v or 6v on board and 5v in proteus simulation , how ever I try bias the Gate by 2 Resistors 0ne is 2k and other is 10k to fix voltage at G is 2V but proteus still get note: error and fail. I have seen more twice your video but when I simulate P channel mosfet it is fault. could you make a video about P channel mosfet and how to assembly Load (as a PNP bjt) and calculating parameters with Load : battery, Cell,(heavy load). finally thank you so much!
I don’t know why it is not working for you.
What if im make H bridge with all N Channel connected to arduino ? it is possible ?
I don’t have a lot of experience designing a discrete H-Bridge. Except for school, I’ve always used an IC.
Hello everyone!
I would like to make a switch to use a screen. My aim is to use a screen with Raspberry Pi. When the Raspberry Pi is booting, the screen is turned off. After that it will work. I need a mosfet to use it. As you know GPIO ports on Raspberry Pi have 3.3V. When 3.3V exists on the GATE, the screen will be turned off. When 3.3V doesn’t exist, the screen will work. How can I do this?
P.S The screen needs 5V and 0.8A.
Please help me. Thank you in advance.
Hi, i try to use pchannel mosfet on highside with 28v rail. And if i use npn with the base connected to my arduino. When logic is high, npn will drain everything to ground and the misfet vgs will be -28 but the problem is misfet maximum vgs is +-20v
What should i use then? If u can give a drawing will be better to understand. Thank you
Sounds like you need a voltage divider.
Very helpful. i got a PA102FDG from a PC motherboard and i was wondering how to use it for high side switching. Thank you
Just like I showed in this tutorial.
I am still confused about when to use N or P. Every example I see on the net is using N-channel by switching the 0V, which leads to load. Is it safe to say that N-channel is always switching ground, can we bring positive voltage to be switched?
In one of the comments you say that we should use N-channel MOSFET if our switching voltage is larger than the command voltage. I have similar scenario like that, I have a command ouput which when 1 (5V Arduino) should allow 12V to the load. The load in question is another input on another device, which needs 10+VDC to become high. I have no control of the load’s ground (0V).
I did not realize that Vgs threshold is actually the difference between gate and source voltage, I though it is the minimum voltage that needs to be brought to gate (disregarding the negative sign). For this reason I have bought a lot of these MOSFETs (http://pdf1.alldatasheet.com/datasheet-pdf/view/123973/TSC/TSM3401CX.html), but as it appears, this transistor keeps source-drain switch always closed, no matter if high or low signal is brought to gate. Due to your explanation it is logical to behave like this, since the voltage difference is always larger than the threshold.
So, what kind of solution do I need to make this work? When high signal (5VDC) on Gate, switch should allow 12VDC to load, when low, switch opens? Since my load is another input, required current is not large (20-30mA max).
Thanks,
Goran
If your current is that low, you might consider a NPN transistor instead. If you stick with a MOSFET, you need a “logic-level” MOSFET like the FQP30N06L.
Thanks – this really helped me use a p channel mosfet in my circuit protection design.
What I don’t understand is that you show the MOSFET is turned on with either 5 volts or 0 volts on input.
it turns on with 0V and turns off with 5V
Using a AO3401 with VGSth -0.9 gate to 3.3 volt Arduino 550Ω to gate 5 volt to source, will there be smoke?
The specified value is the minimum necessary to turn it on. The absolute maximum for the part is 12V