# Please explain MOSFET output characteristics and load line when used as a switch

My end goal is to switch a 9V-12V LED strip with an STM-microcontroller (3,3V level logic). So, I'm reading about MOSFETS (n-channel enhancement type in particular), but I'm very confused about the output characteristics and load line graphs.

• In what region should MOSFET operate in "on" state? Linear or saturation?
• What is the point of the load line in the output characteristic? We build a load line with 2 points: Vd=0, I=Vsupply/Rload and Vd=Vsupply, Id=0. Then we find an operating point at our Vg (Vg=3,3V in my case). But what is the point of this if maximum current is at Vd=0? And there is no graph part that corresponds to Vd=0. If Vd=0, Id=0.
• How do I find a MOSFET with a low enough threshold voltage to turn fully on at 3,3V? Thank you!
• You need to specify the current taken by the load and how much volt drop is tolerable and what your power supply voltage range will be for powering the load. BTW is your name a reference to Allfather D'Aronique from the preacher? Commented Nov 21, 2020 at 14:05
• The load is 3 pieces of 9-12V LED strip in parallel. Each piece of LED strip is 3 LED in series and a resistor. So, the total load current is approx. 30mA*3=90mA. The supply voltage is 9V battery. (My name is a reference to Odin the Norse God.) Commented Nov 21, 2020 at 14:28

In what region should MOSFET operate in "on" state? Linear or saturation?

For this I've assumed the load current is 30 mA because it matched with a graph of MOSFET characteristic I had close to hand. For your load of 90 mA the green line will cross near the Y axis at 90 mA if you can supply enough $$\V_{GS}\$$ voltage: -

When "on" the MOSFET will be in the linear or ohmic region. When "off" the MOSFET will be in a part of the saturation region called "cut-off" i.e. right along the curve for when $$\V_{GS}\$$ is at 0 volts. Current leaking through will be around 1 μA.

What is the point of the load line in the output characteristic?

For your application there is no point at all - you have two points; one close to the Y axis and one close to the X-axis providing you can supply enough $$\V_{GS}\$$ voltage: -

But your $$\V_{GS}\$$ voltage is only 3.3 volts hence it will look more like this: -

And, for the characteristic picture I chose, both ends of the load line will be in the saturation region.

How do I find a MOSFET with a low enough threshold voltage to turn fully on at 3,3V?

You have to dig a little and possibly use website comparison pages like those produced by Farnell, Mouser and Digikey to name but three. You are probably going to narrow down your search by selecting the maximum voltage ($$\V_{DS}\$$) to be no more than 30 volts and you will probably find that selecting a device with maximum drain current of at least 5 amps is going to put you in the right area for picking a device.

• Thank's for the answer! But I still don't understand the following. If I use "your" transistor and apply 3,3V Vg:1) I will have a voltage drop Vd=~4V? That means that not enough voltage will be supplied to my particular load. 2) Basically what I don't get is why do we plot a point of maximum current Id if this point is unachievable? 3) So ideally I need to find a MOSFET that has an output curve at Vg=3,3V that is as close to Y-axis as possible? So that the voltage drop Vd will be minimum and Id will be as close to maximum as possible? Commented Nov 21, 2020 at 15:29
• 4) Why does an "on" MOSFET have to be in a linear region? Because the curve in that region is as close to Y-axis as possible? Commented Nov 21, 2020 at 15:29
• Comment 1: Yes, using my curve the volt drop will be about 4.5 volts - not great but, it's just a picture. You plot the load line as if perfection were possible then you see what the MOSFET characteristic delivers to make things less than perfect. Of course you don't have to plot a load line. I don't any more; I just look at the characteristic directly and know what I want. Point (3) yes. Commented Nov 21, 2020 at 15:34
• Comment 2: It doesn't have to be (as I showed in my lower picture) but you will get a significant power wastage if you don't AND also if it's not close to the Y axis. Commented Nov 21, 2020 at 15:35
• Thank you! I get it now : ) Commented Nov 21, 2020 at 15:39