I'd like to know if it's possible to find \$V_\mathit{DSsat}\$ knowing \$k_n'\frac{W}{L} = 0.75\,\$m and \$V_\mathit{Th} = 1\,\$V? Also, \$I_D = 1\,\$A.

  • \$\begingroup\$ What is W and what is L? You have an "m" after "0.75" and that's what's confusing me. \$\endgroup\$
    – Andy aka
    Commented Oct 1, 2013 at 8:04
  • \$\begingroup\$ The 'm' is meant to be milli. W and L are the length and width of the substrate. \$\endgroup\$
    – user968243
    Commented Oct 1, 2013 at 8:28
  • \$\begingroup\$ W divided by L is "milli" what then? \$\endgroup\$
    – Andy aka
    Commented Oct 1, 2013 at 8:44
  • \$\begingroup\$ I think I wrote it poorly: I'm trying to say that k times W divided by L is equal to 0.00075. \$\endgroup\$
    – user968243
    Commented Oct 1, 2013 at 8:48

1 Answer 1


No, it is impossible to find \$V_{DS_{sat}}\$ based on the parameters you've provided.


The most basic model for representing NMOS's current is this:

enter image description here

Due to the fact that both \$C_{i}\$ and \$\mu\$ are parameters of a particular technology and are constant across all the NMOSs in a given technology, it is common to replace two constants with a single one: \$k'=C_{i}\mu\$.

If you plot the above equation, you'll find something strange - it predicts a maximum of \$I_{D}(V_{DS})\$. It means that there is some \$V_{DS_{sat}}\$, and for \$V_{DS}>V_{DS_{sat}}\$ the current is decreasing! There might be two explanations to this phenomenon:

  1. There is some very unusual effect takes place.
  2. The above equation has limited validity (in terms of \$V_{DS}\$).

The second bullet is the correct one - this equation is valid up to \$V_{DS}=V_{DS_{sat}}\$. When this threshold is reached, the conducting channel underneath transistor's Gate "pinches-off" and the current does not increase anymore with increasing \$V_{DS}\$:

enter image description here

The names of the regions of operation appear on the graph. Linear region is sometimes referred to as "triode region".

So, how one finds an expression for \$V_{DS_{sat}}\$? Very simple: differentiate the above equation with respect to \$V_{DS}\$ and find when the derivative equals to zero. You'll get the following value:

$$V_{DS_{sat}} = V_{GS}-V_T>0$$

The last inequality represents the fact that the transistor is not in cut-off region.

Substituting this value back to current's equation, you'll find:


Now, back to your question:

As you can see, substituting all the given parameters into the last equation allows you to calculate \$V_{GS}\$, and, since you also know \$V_T\$, you can calculate \$V_{DS_{sat}}\$. However, this requires one additional assumption which you did not state in the question: the transistor should be known to operate in saturation region.

Otherwise, if the transistor is in linear region, you need to know also at which \$V_{DS}\$ it operates in order to be able to calculate \$V_{GS}\$ from the first equation.

  • \$\begingroup\$ @Vasily it's a shame you left the answer like that - it sounds like something I could learn - maybe you could add what it takes to calculate \$V_{DS(SAT)}\$ \$\endgroup\$
    – Andy aka
    Commented Oct 1, 2013 at 20:26
  • \$\begingroup\$ @Andyaka, I indeed have a record of exceptionally long answers, but in this case, the question is very straight and clear. I think it is a bit off-topic to deep dive into explanations and models while OP did not ask for any guidance or explanation at all. \$\endgroup\$
    – Vasiliy
    Commented Oct 1, 2013 at 20:36
  • \$\begingroup\$ @Andyaka, on second thought, I don't see a reason not to do what you're asking for. \$\endgroup\$
    – Vasiliy
    Commented Oct 1, 2013 at 20:47
  • \$\begingroup\$ cool looking forward to it \$\endgroup\$
    – Andy aka
    Commented Oct 1, 2013 at 20:52
  • \$\begingroup\$ @Andyaka, done. \$\endgroup\$
    – Vasiliy
    Commented Oct 1, 2013 at 21:30

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.