# Why does this MOSFET always turn on in saturation?

Consider the following case:

In Razavi's book, it says that this MOSFET will always turn on in saturation mode.(assuming Vgs is just greater than Vth) I'm trying to figure out why. For Saturation: $$V_{DS} > V_{GS} - V_{TH}$$ Simplifies to: $$V_{out}>V_{in} - V_{TH}$$

How does this prove that it will turn on in saturation? I mean yes we could probably prove that with the small-signal model, but there is no small signal here.

• this looks like a JFET Sep 30, 2018 at 0:22
• It's definitely a MOSFET. Supposed to be a common-source stage with resistive load. Sep 30, 2018 at 0:24
• @Sparky256 Are you claiming that an n-channel mosfet cant have a threshold voltage less than or equal to zero? You would be wrong if you are.
– Matt
Sep 30, 2018 at 4:53
• @Matt. If you read the first comments there was confusion as to whether the drawing was a JFET or a primitive MOSFET model. Get sloppy with the arrows and use non-standard symbols and you get a bit of confusion. Btw, you can tie a JFET gate to its source pin and it will still have some current flow.
– user105652
Sep 30, 2018 at 5:09
• It might be a depletion-mode MOSFET. Uncommon, but they (kinda like JFETs) do conduct with 0 Vgs. And the schematic symbol looks almost right for it to be one of those. Sep 30, 2018 at 9:33

## 3 Answers

It seems, from reading the comments that there is some confusion with some members about MOSFET saturation (not to be confused with BJT saturation): -

MOSFET saturation is when the device is operating in its constant current area. It is not when in its "ohmic" area. The ohmic area is when the device can be said to be "turned on".

In Razavi's book, it says that this MOSFET will always turn on in saturation mode.

And here we have some confusion - to say the MOSFET will **turn on" in the saturation region is misleading but, I don't have the referenced book so I can't be sure whether this is a mistake by the OP or a mistake in the book. Or maybe the words are not interpreted well...

Going back to the circuit, the device will initially be in the saturation region as the gate-source voltage begins to rise but, if the gate-source rises high enough it will clearly be in the ohmic (turned-on) region.

So, maybe the author, if the quote was verbatim, meant to say that it will **initially* "activate" in saturation as the gate-source voltage rises past VGS(th).

• according to the graph, as Vgs increases the mosfet goes from ohmic to saturation?or Am I looking at it wrong? Sep 30, 2018 at 9:36
• @Navaro for a given Supply voltage (say 10 volts), VDS will be close to 10 volts as the MOSFET starts to conduct a little bit. This will be when VGS slightly exceeds VGS(th) and this puts it firmly in the saturation region. Of course as VGS rises, VDS drops but it has to drop well below 3 volts before it is fully out of the saturation region. However, if $R_D$ is very high in value, only a slight increase in VGS will push it from saturation to ohmic region but, it will still "begin" in the saturation region as VGS rises. Sep 30, 2018 at 9:39
• How about the depletion-mode MOSFET? With Vgs = 0, you can still get it into saturation. Sep 30, 2018 at 9:42
• @RichardtheSpacecat then you would begin with Vgs at a negative value but, it's not clear what the symbol is in the OP's question because it is the "generalized" symbol for a MOSFET and unclear in that respect. Sep 30, 2018 at 9:46
• @AlfroJang80 you have the graph in my answer so I think you should easily see that if Vdd is too low it begins immediately in the ohmic region hirrespective of what the book might say). Sep 30, 2018 at 14:18

I have seen the video Lectures of the Author in which he posed a problem like" prove that MOSFET will always turn on in saturation region, given VDS > 0 and you start increasing VGS from 0V."

Now this above statement can be easily justified as whenever VGS reach infinitesimal value above VT (VGS - VT< VDS ) will be satisfies for atleast short range values initially which will lead the device directly into the saturation region.

This is how I see it. When Vin = 0 the device is turned off and Vout = VDD (This point is important). So when Vin = small voltage = Vgs; Vout is finally something less than VDD. Now as you mentioned; Vout > Vin - Vth is the condition for to work in saturation. So when you are starting, Vin is low and naturally it can't cross the Vout as it's close to VDD, the supply voltage. That means, Vout > Vin - Vth because the difference between Vin - Vth is small.

This changes when Vin is large as Vdsat or Voverdrive (its nothing but Vin - Vth) becomes large and output is much smaller and the device goes into triode region.