enter image description here 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.

  • \$\begingroup\$ There's a trick you can use, assume M2 is off (no current), what would happen? Since M1 is on, the source of M2 is pulled to ...., that means Vgs of M2 is .... Does that confirm that M2 is off or does it prove the opposite (that M2 is on)? Also note applying a Vgs > Vt means an Id can flow. But the reverse is also true, if an Id flows, it means that the MOSFET must have a certain Vgs > Vt, if the Vgs < Vt then that current cannot flow. \$\endgroup\$ Commented Nov 28, 2017 at 6:49
  • \$\begingroup\$ @Bimpelrekkie thank you for the comment, but how does it help me conclude if the region of operation is linear or saturation? \$\endgroup\$
    – Fawaz
    Commented Nov 28, 2017 at 6:56
  • \$\begingroup\$ this looks like a school assignment or a test even .... review your lessons \$\endgroup\$
    – jsotola
    Commented Nov 28, 2017 at 6:58
  • \$\begingroup\$ This was a question asked in a national level competitive exam and I was trying to solve it out of interest. Please contribute if you could help \$\endgroup\$
    – Fawaz
    Commented Nov 28, 2017 at 7:02
  • \$\begingroup\$ Same thing: assume it is in linear region, then check if that is a possible solution. What external condition determines if the MOSFET is in linear region or in saturation? The review your lessons still stands. It is impossible to answer this question correctly (apart from guessing and luck) without having had some education about MOSFETs. If you didn't have that education then you should not expect any help from this site. \$\endgroup\$ Commented Nov 28, 2017 at 7:02

2 Answers 2


This is a bit of a trick question.


  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\$.


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


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.


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.

enter image description here

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


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.

  • 1
    \$\begingroup\$ I wouldn't call that a controversy. It's simply a terminology mismatch between BJTs and MOSFETs. People should simply understand that: saturated BJT "=" MOSFET in linear (aka triode, aka ohmic) region, and BJT in active region "=" saturated MOSFET. I admit that the terminology is confusing for a newbie, especially if one studies MOSFETs after having used BJTs for a while. \$\endgroup\$ Commented Nov 28, 2017 at 14:15
  • \$\begingroup\$ @LorenzoDonati yup. reworded. \$\endgroup\$
    – Trevor_G
    Commented Nov 28, 2017 at 14:19

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.


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