# MOSFET: Calculating $V_\mathit{DSsat}$ from $k_n'\frac{W}{L}$ and $V_\mathit{Th}$

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.

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

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

Theory:

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

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}\$$:

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:

$$I_{D_{sat}}=\frac{1}{2}k'\frac{W}{L}(V_{GS}-V_T)^2$$

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.

• @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 $V_{DS(SAT)}$ Commented Oct 1, 2013 at 20:26
• @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. Commented Oct 1, 2013 at 20:36
• @Andyaka, on second thought, I don't see a reason not to do what you're asking for. Commented Oct 1, 2013 at 20:47
• cool looking forward to it Commented Oct 1, 2013 at 20:52
• @Andyaka, done. Commented Oct 1, 2013 at 21:30