Solving Transistor in Cutoff: Vtp=-0.4V

[carnot cycle]

Homework Statement

The threshold voltage (Vtp) of this p-channel transistor is -0.4V. Determine the region of operation of this transistor.

Homework Equations

Vsg + Vtp = Vsd(sat)

The Attempt at a Solution

Since the gate-source voltage is 0V, and the threshold voltage is -0.4V, the saturation voltage should be -0.4V. The source is at 2.2V, and the drain terminal is connected to ground potential. Because this Vsd is greater than the saturation voltage that I found using the above equation, I thought this transistor was in saturation mode. However, the book's answer says that the transistor is in cutoff because the gate-source voltage is zero.

I thought that if the gate-source voltage (0V) is greater than the threshold voltage (-0.4V), the transistor could not be in cutoff.

Any help would be greatly appreciated. This seems like a very easy problem, but I have just started studying transistors so I am still trying to conceptually grasp them.

 

[rude man]

Define Vsg as Vs - Vg. Then a p channel MOSFET is always in cutoff when Vsg < Vt where I define Vt to always be positive. Look again at your equations for cutoff, saturation and linear modes.