I get confused with the circuit bellow.

Here is the link to the lecture analyzing the circuit:http://www.youtube.com/watch?v=ZMEpc2o9_gU&feature=youtu.be
When Vin = 0V => Vgs (NMOS) = 0 < Vt => NMOS is in CUT OFF.
And therefore Ids = 0 A.
For PMOS, Vsg = Vs - Vg = 5V > |Vt| => PMOS is NOT in CUT OFF.
PMOS is in TRIODE or SATURATION.
In the lecture:
PMOS has Vsg > |Vt| and null current (Ids = 0) => He concluded that PMOS is in TRIODE.
He doesn't explain the reason why it is in triode. Therefore here is what I guess:
With PMOS in triode region:
Ids = -kn(Vsg -|Vt| - Vds/2)Vds (1)
With PMOS in saturation:
Ids = -1/2kn (Vsg - |Vt|)^2 (2)
As for the case in the video, Vsg =5V > Vt => PMOS is in triode or cut off.
Now because Vsg - Vt≠ 0 => Ids ≠ 0 => This case don't happen.
=> PMOS is in triode

But here is what is confusing me.
With Vin = 0V and NMOS is in CUT OFF therefore Ids = 0.
Can I now consider that D and S of the MOS is not connected and means that it acts as a open switch?
If it can be considered as an open switch then drain of PMOS D is not connected to ground and it is also in CUT OFF.
I think in real life these transistors is not ideal and they has resistance and capacitance but now let consider that they are all ideal.

 

For Vin = 0V NMOS is OFF so you can remove it from the circuit.
PMOS in "ON" because Vgs > Vt. But no current is flow through PMOS so there is no voltage drop across Vsd. And since Vsd < (Vsg - Vt) the PMOS is in triode mode.
Only if Vsd > (Vsg - Vt) the PMOS is in saturation.

 

For Vin = 0V NMOS is OFF so you can remove it from the circuit.
PMOS in "ON" because Vgs > Vt. But no current is flow through PMOS so there is no voltage drop across Vsd. And since Vsd < (Vsg - Vt) the PMOS is in triode mode.
Only if Vsd > (Vsg - Vt) the PMOS is in saturation.

NMOS is OFF and you said that I can remove it? Do you mean that I can remove NMOS and left D of PMOS open?
Can you tell me how to calculate Vsd in PMOS?

 

NMOS is OFF and you said that I can remove it? Do you mean that I can remove NMOS and left D of PMOS open?

Yes, for NMOS OFF and without any external load the PMOS drain is left open.

Can you tell me how to calculate Vsd in PMOS?

In witch situation?
Without any external load Vds = 0V and Vout = 5V.

 

In witch situation?
Without any external load Vds = 0V and Vout = 5V.

I meant that Vds when NMOS is left open.
I think you computed Vds by Ohm's law? Vds = Id* Rds = 0 because Id = 0?

 

I think you computed Vds by Ohm's law? Vds = Id* Rds = 0 because Id = 0?

Yes, you are right.

 

I think I need your help again. Is it correct to apply Ohm's law with MOSFET because it is not a linear device?
And with the assumption that PMOS is in triode then I can apply the equation:
Id = kn(Vsg - |Vt| - 1/2Vsd)Vsd
With Id = 0 then Vsd = 2(Vsg - |Vt|)= 2*(5 - |Vt|)≠ 0???

 

But MOS in triode mode act just like a resistor.
Rsd(on)≈ 1/kp*(Vgs - Vt)

 

But MOS in triode mode act just like a resistor.
Rsd(on)≈ 1/kp*(Vgs - Vt)

Yes, but before that we don't know if the transistor is in triode or saturation?
How you come to know that PMOS is in triode?

 

Hello Jony,
Can you reprove what you did in post #2?
Then we still don't know that PMOS is in triode region and therefore how can you compute Vds?
I think Ohm's law is only applied when we know that it is in triode.

 

For me it is obvious that PMOS is in the triode region. How can PMOS be in saturation region if no current is flow through MOS ?

 

For me it is obvious that PMOS is in the triode region. How can PMOS be in saturation region if no current is flow through MOS ?

Yes. I have just figured it out. My calculation is OK, I know that with Vsd = 2*(Vsg - Vth) then Id = 0 but when I look at PMOS characteristic bellow I see no point that appropriate with above.

Now I have draw Id - Vsd for PMOS with Id is the current flow from S to D.
The characteristic is similar to the one bellow and it is appropriate with the equation.

 

My calculation is OK, I know that with Vsd = 2*(Vsg - Vth) then Id = 0

How can this be true?

 

How can this be true?

Oh, it is my mistake!
With the original circuit we have Id = 0. Therefore, I wrote equation Id, Vsd, Vsg for PMOS both in triode and saturation.
Then from Id = 0, I solve these equation and see if it is happen. Then I see that there is only PMOS is in triode is appropriate.
When Id= 0 => Vsd = 2(Vsg -Vt) ≠ 0
With the first characteristics that I posted in #12, I can't determine the point in the picture.
However with the second chacteristics, it is obvious that there is a point in the triode that has Id = 0 and Vds = 2(Vsg -Vt) ≠ 0. Therefore it make me more confident to say that PMOS is in triode.