How to conclude region of operation for a series combination of MOSFET

Question

I understand that M1 is on and for M2 to be on the intermediate voltage(drain on M1 and source of M2) should be less than 1.5V. Also if M2 is ON then M2 is in saturation since Vds>(Vgs-Vt) for M2. Now coming down to M1 and applying the same condition I understand that if the intermediate voltage is less than 1V then M1 is in linear region and if the intermediate voltage is between 1V and 1.5 V then M1 is in saturation.However, I'm stuck at this point. How do I conclude the region of operation for both the Mosfets? Is this problem to be solved using intuition assuming regions of operation and then the condition that same current flows,if then please explain.

Answer 1

This is a bit of a trick question.

INITIAL ASSUMPTIONS:

1. The 3V rail is an infinite current source.
2. As stated the MOSFETS are identical and have the same \$R_{DS_{ON}}\$ when \$V_{GS} > V_T\$.

OBSERVATION:

Since both MOSFETS must be "on", and there is no resistance in this line other than the \$R_{DS_{ON}}\$ for the MOSFETS, the voltage between the MOSFETS must be the simple resistor divider voltage of near half the rail.

^{simulate this circuit – Schematic created using CircuitLab}

MATH:

As you can see, \$V_{GS}\$ Of M1 is 2V so it is definitely on. and \$V_{GS} - V_T = 1V\$, which is less than the half rail 1.5V. It is clear then this MOSFET is saturated.

\$V_{GS}\$ of M2 is then only 1V, it is therefore only at the verge of turning on. As such \$V_{GS} - V_T = 0V\$, so again, if it is on, it must be in saturation.

Now is M2 actually on?

If it is not on, the source voltage will fall and \$V_{GS}\$ will turn on saturated. As such, M2 must be on, all be it, only just.

REALITY BALANCE ACT:

In reality, since \$V_{GS}\$ of M1 is greater than \$V_{GS}\$ of M2, \$R_{DS_{ON}}\$ for M1 will be less than that of M2. As such the midpoint voltage will be lower than 1.5V. As that value lowers, \$V_{GS}\$ of M2 increases and M2 turns on more, so the midpoint will balance somewhere less than 1.5V.

In order to figure out where you need to look at the On_Region Characteristics of the MOSFET. Below is a typical example for an AO8408.

At the balance point \$V_{GS}\$ of M2 must be such that \$I_{M1} = I_{M2}\$. As you can see, for this particular device it ends up balancing at approximately the voltages shown below.

^{simulate this circuit}

CONCLUSION:

According to the Wikipedia formula for MOSFET saturation, both MOSFETS must be in saturation. For real MOSFETS, or at least the one I chose here, by that formula, the top one is saturated and the lower one is not.

Arguably, I'd say in this example, since it is balancing itself, neither are really "saturated" OR acting linearly.

Note: There is some confusion over the term "saturation" when applied to MOSFETS.

Answer 2

For transistor to be in saturation, $$v_{ds} \ge v_{gs} -V_t = V_{gt}$$

For M1, \$V_{gt} = 2 - 1 = 1\$, thus the drain of M1 should be at least at 1V for it to be in saturation. Thus, for M2 \$V_{gt} = 2.5 - 1 -1 = 0.5\$. As current is proportional to \$V_{gt}^2\$, if we assume both the transistors are in saturation then lower transistor (M1) will draw more current compared to upper one (M2). This would cause potential at the drain to go down. This is because the current that discharges the node is more than the one that charges it.
As \$v_{ds}\$ goes below \$v_{gt}\$, lower one will go into linear mode and upper one will go into saturation. So, answer should be C.