High side switch with N-FET

Question

I know nothing about MOSFET, yet, so I'm fumbling in the dark here.

What I want to achieve is a high side switch using an N-FET. All guides I've been reading says "low side switch = NPN or N-FET", "high side = PNP or P-FET". I want to control a very small 3.3V-load (50mA) using a microcontroller (5V), and since a BJT has about 200mV for Vcesat I thought that a FET might be better to get rid of that drop.

I setup a simulation just to test the idea, and I can see 3.3V on the source pin when the gate is high. But obviously, current cannot flow in the direction I want. Is it possible to make this work with an N-FET, or do I need to bite the bullet and buy a P-FET?

EDIT: datasheet for the FET in question. ON-characteristics for Vgs(th) says "max 1.5V". 5-3.3 = 1.7V in my case, shouldn't that be enough?

Answer 1

So if you are just turning an LED on and off with 20mA, I know I would use an NPN transistor (2N3904 or 2N2222 type). But a MOSFET will certainly do the job just as well. The BSS138 as a low side switch will work fine for an LED. I do it all the time as do many other designers.

However, if you are enabling power to other circuitry, low side switching gets very tricky, because once you do low side switching you no longer have a single ground plane. I strongly recommend you avoid low side switching if you are trying to control power to other IC's that you are communicating with.

In your case, since you have a 5V VCC available to switch a 3.3V rail, you actually can use a high-side NMOS without any special additional circuitry. But you need to find one that will fully turn on with Vgs of 1.7V. And remember, 1.7V is the nominal voltage. If your 3.3V rail is running a little high and your 5V rail is running a little low, you can find yourself with a lot less than 1.7V left to turn on the FET.

The Vgs(th) rating of the FET is not what you should be looking at when you are using the FET as a switch. You should note what is the lowest voltage at which Rds(on) is specified. In your case, you want a low Rds(on) to be listed in the datasheet when Vgs is 1.5V or even less if possible.

Purely as an example (shopping questions are off-topic here) I did find one such FET that might work for you. It is the Diodes Inc part number DMN1019USN-13. Rds(on) is specified as 41mOhms max at Vgs=1.2V. So that should work for you.

A word of caution: for MOSFET's with such low switching thresholds, you need to make sure that you have it turned off when you want it off. Any little voltage may actually cause the channel to start conducting a bit. In your case that should be no problem because when you drive the gate to ground, Vgs will be negative. But others who may be reading this may be in a slightly different situation.

Answer 2

Yes, you can use a NFET as a high side switch, with some aditional components.

To make this work, you need a bootstrap circuit. Basically what this circuit does is raises the Vgs of the NFET to meet the required turn-on voltage for the MOSFET. You can either build one, or search for a MOSFET driver that will drive a NFET on the Highside.

That said, if all you're doing is turning on an LED, then it would make more sense to use a PFET. You would only want to drive a NFET on the Highside if you couldn't/didn't want to use a PFET because the Rds value of it was too high for your application (i.e. 75 mOhms vs 12 mOhms).

Answer 3

In order to use an N-FET, you will have to make sure Vgs is well in the saturation region (the voltage between the gate and source). For logic level FETs, this is usually around 5V and for non logic level FETs this is usually around 8-9V.

When the MOSFET is on, the Source voltage is 3.3V (the drop across the mosfet is very close to 0V). This means the gate voltage needs to be 3.3 + 9V for a standard n-fet or 3.3V + 5V for a logic level N-FET.

Answer 4

\$V_{TH}\$ is more or less the gate-source voltage where the thing is just barely turned on. If you want it to be really turned on then you need to hold \$V_{gs}\$ at the recommended voltage for the part in question -- which for that part is 10V (basically, you look at their spec for \$R_{DSon}\$ -- whatever the \$V_{gs}\$ is there, that's what they've designed the part for).

Answer 5

Even with a bootstrap circuit it may not fully get you there with a 3.3volt supply. That BSS138 gives 1ma with 1.5V on the gate. With bootstrapping you can get max of maybe 3 volts through the mosfet, there will probably be some drop since its rdson is pretty high.., not sure how much current that will give you, check the datasheet. It may not reach 50ma at that gate voltage .

Datasheet seems to indiate 3V on the gate can give up to 80ma. So maybe it could work but i dont think it will, i do not think you will reach your desired current, even.

However you arent going to get 3.3volts on the source, the rdson is too high, if your load is voltage sensitive then i think you need a better mosfet choice