# P-Channel MOSFET Tutorial with only Positive Voltages

### From the mailbag (or chat… bag?)

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.

## What’s different?

MOSFET Symbols from Wikipedia

The actual construction of the two types of MOSFETs differ. There are plenty contraction explanations out there on the layers. 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 threshold is how a MOSFET turns ON and OFF.

How that resistance changes, depends on if it is an 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 VGSth is a negative value. We can use the data sheet from an IRF5305 as an example.

Its VGSth is specified as a 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?

This spec 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 “GATE voltage value minus 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 voltage 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 the 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 minus 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. An excellent 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 that they work with “negative voltages.”

Question: What’s the most creative MOSFET circuit have you seen? You can leave a comment by clicking here.

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## 87 thoughts on “P-Channel MOSFET Tutorial with only Positive Voltages”

1. Terence Pinto says:

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.

• James Lewis says:

ut with the load between the 5V supply and the source of the AO3401, drain to ground.

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.

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

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.

• Terence Pinto says:

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.

• Glenn Bever says:

I don’t see how your last circuit could work as drawn. Since the Vgs in a P-channel MOSFET must be negative to turn on, and the source pin is held at ground, the gate can never go negative with respect to it since it only provides +5v or ground. That is why P-channel MOSFETS have the source pin tied to the high-side load–so that Vgs can be negative. Were the drawn circuit using an N-channel MOSFET (with the source tied to ground), or the P-channel MOSFET reversed with drain to ground, then I could see it working.

• Glenn Bever says:

For that matter, the first circuit drawings that describe the negative Vgs (using only a positive supply) for a P-channel MOSFET are drawn using N-channel MOSFETs. Based on the circuit, it looks like it is a ‘typo’ where the arrow is pointing the wrong way (for a P-channel MOSFET).

• Every circuit drawn in this post uses a P-Channel symbol.

• Glenn Bever says:

For some reason, the way that the ‘MOSFET enh (no bulk)’ symbol is standardly drawn is the opposite of every other FET symbol: the arrow on the source points into the P-chan device instead of out. Most confusing. (Reference the Wikipedia drawing.) Other standards (e.g. JEDEC and IEEE) aren’t real consistent either. The one you chose to use is an ‘outlier’ to the most commonly used symbols that I have encountered—in terms of arrow direction. Because of the potential for confusion, I think the best approach is to explicitly state what the source of symbology is (which you did) and explicitly state whether your symbol is P-channel or N-channel (on the drawing). Including a note to the unwary in a tutorial that the arrow direction is ambiguous would be a good idea, too. When designing with actual FETs, use the symbol used on the manufacturer’s datasheet for the part you are using.

• Thanks for your opinion. I have no plans to change this post or its graphics. So far, only 1 person has had an issue understanding the schematics (out of hundreds of thousands.)

• The source is tied to 5 volts in every schematic in this post.

• Glenn Bever says:

The diagram with the NPN transistor buffer does show the source tied to ground, not 5v—the source being the lead with the arrow on it.

• If you’re talking about the comment from Nov-21, it is an N-Channel.

2. 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.

3. Jaikumar Balgobind says:

Thank you profusely sir, for explaining so simply yet to the point

4. Lehar Kewalramani says:

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.

• Lehar Kewalramani says:

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.

• Lehar Kewalramani says:

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.

• Emmanuel says:

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.

5. 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.

6. Yes, I was confused by this as well.

7. tiago antonio says:

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.

8. Harsha_Reddy says:

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.

9. Rui Almeida says:

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.

10. Jerry H. says:

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?

• Don Mundinger says:

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.

• RICHARD ROESNER says:

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