# Explain in layman's terms Vgs and Vgs(th) of MOSFET's

I'm trying to understand $$\V_{GS}\$$ of MOSFET transistor. From what I understand $$\V_{GS}\$$ normally stands for voltage gate to source breakdown, but other than that I lack an understanding. $$\V_{GS(th)}\$$ is the threshold voltage at which the mosfet will turn on, so I have some questions about the threshold voltage;

1. What happens if I go over the max threshold as told by the data sheet?

2. What happens if I'm under it?

• Not quite sure everyone here is on reddit or knows what ELI5 (Explain like I'm 5 years old) means. Feb 25, 2015 at 20:19
• Great analogy! I understood this concept easliy. Thankyou so much :) May 28, 2018 at 8:21

Vgs is just the voltage from gate to source (with the red lead of the multimeter on the gate and the black one on the source). Everything else is from context.

The Absolute Maximum Vgs is the maximum voltage you should ever subject the MOSFET to under any conditions (stay well away). Usually the actual breakdown is quite a bit different (borrowing from this datasheet):

Vgs(th) is the voltage at which the MOSFET will 'turn on' to some degree (usually not very well turned on). For example, it might be 2V minimum and 4V maximum for a drain current of 0.25mA at Tj = 25°C (the die itself is at 25°C).. That means that if you want your 20A MOSFET to really turn on fully (not just conducting 250uA) you need a lot more voltage than 4V to be sure about it, but if your Vgs is well under about 2V you can be pretty sure it's well turned off (at least around room temperature).

Rds(on) is always measured at a specified Vgs. For example, it might be 77m$\Omega$ with Vgs = 10V and Id = 17A and Tj = 25°C. That 10V is the Vgs you need to feed your MOSFET for it to be happily turned on so it looks like a very low resistance.

Vgs also comes up when you want to know the gate leakage. Igss might be +/-100nA at Vgs = +/-20V and Tj = 25°C.

Vgs is the gate to source voltage.

In the datasheet you'll find an absolute term Vgss this is the maximum voltage that can be applied between the gate and the source. Beyond this, you risk damaging the mosfet.

An N channel mosfet is essentially a P type sandwiched between two N type regions.

Party time.

You are hosting a party and inviting all the neighborhood electrons to attend. So you broadcast "PARTY AT MA HOUZE YOLO #SWAG!". I.e. you apply a positive voltage at the gate with respect to the source. Given that your neighbors are some distance away (next door), your broadcast isn't loud enough (You are below Vgs(th)) Once you yell your invite loud enough (i.e. Have Vgs at Vgs(th)) your next door neighbors hear and come party with you.

Vgs(th) is the voltage at which the mosfet channel begins to conduct. At this voltage, a positive voltage, it creates an electric field, which attract electrons (since our applied voltage is positive, so positive charges on gate). These accumulated electrons near the gate, form a bridge between the source and the drain (which are both n type). Now you have a "continuous" n type path from source to drain.

You only attracted your next door neighbors, so your party is kinda lame. How do you get more people? You broadcast louder (Increase the range of your electric field - increase your Vgs).

Now, there's loud, and then there's HOLY CRAP, A PLANE IS ABOUT TO CRASH INTO OUR HOUSE, WHAT THE \$#@! IS THAT NOISE!?! We don't want to freak people out too bad, so we need to know how much louder (the difference between) our call is than the minimum needed for just our immediate neighbors (Vgs(th)). This difference is called by a couple of different names, but the two that I've heard most often are V-on (Von) or V-overdrive (Vov). This quantity (Vgs-Vth) represents how much more potential is between the gate and the source than is needed for the transistor to turn on, and it influences just about every other behavior of the MOSFET: current in triode (how many people are at your party when just your neighbors can hear), current in saturation (how many people are there when it's full of people), and transconductance (how much volume you need per person at the party), just to name a few.

So now you've increased your Vgs to the point that you cannot accept anymore people on your property. You are completely full. You can broadcast your party as much as you want, but there simply isn't enough room to accommodate everyone. Your transistor is now in saturation. [Technically, more people can be at the party: if your immediate neighbors start broadcasting the party (increasing Vds), and people start partying at their houses (the source and the drain), this increases the number of people at the party (increased current). This is called channel-length modulation. However, this can only happen if your house is already full (channel-length modulation only happens when the device is in saturation).]

If you start increasing your Vgs to the point of Vgss, well...cops show up and shut you down. Underage drinking, drug use etc.. You go to jail (Your transistor has been damaged).

• I really did love this explanation though! lmao Feb 26, 2015 at 3:27
• @Yasindu I have rejected your proposed edit. English is a strange language with many different standards. Your edit was about changing this post from one standard to another. Even in the UK they have difficulty deciding what the correct use of English is. For the future avoid doing this where it does not interfere with the readability of the post.
– RoyC
Feb 12, 2018 at 14:28
• What the hell did I just read? :) That's what you get when you ask "Explain in layman's terms".
– Rev
Oct 5, 2018 at 12:06
• Perfect explanation. Yes, I used to run parties Feb 14, 2020 at 15:45