# Negative value for saturation mode voltage of MOSFET

I want confirmation of whether a MOSFET is in saturation or not. I have calculated the condition using the formula $V_{GS}-V_{tn}$, and my answer is a negative value. Does it matter if the value is negative? If yes, what is the value actually showing?

• What does $V_{DS} - V_{tn}$ get you? I don't think I've heard an equation that requires you to subtract the threshold voltage from the drain-source voltage. I've heard, however, of subtracting the threshold voltage from the gate-source voltage, thus getting the drain-source voltage, which can determine the FET's mode of operation. Perhaps you mean $V_{GS} - V_{tn}$ instead? – user103380 Apr 10 '17 at 3:34
• Did you switch channels? from N to Pch_en – Tony Stewart Sunnyskyguy EE75 Apr 10 '17 at 6:18
• yaa, it is actually Vgs not Vds, my mistake. i have edited it. – Haziq Mahmur Apr 11 '17 at 8:33

If $V_{GS}$ is smaller than your than your threshold, $V_{tn}$, then that means your FET is in cutoff region, meaning that your drain current $I_D=0$ and your FET will not conduct. Here's a good table from the Sedra\Smith textbook that gives a good summary of the regions of operations for the NMOS device, where $V_{ov}$ is the overdrive voltage. • tq for the answer and the reference u gave.! – Haziq Mahmur Apr 11 '17 at 8:38
• You're very welcome :) glad it was a typo. It confused me at first. Just remember to add channel length modulation to the formulas if the question asks to include it. This table also neglects the Body Effect because it's a three terminal device implying that the substrate is tied to the source. – user103380 Apr 11 '17 at 15:06

The check for saturation has two conditions:

$V_{GS}>V_{th}$ and $V_{DS}\geq V_{GS}-V_{th}$

To rearrrange the second equation for $V_{GS}$, you would get the following equation: $V_{DS}+V_{th}\geq V_{GS}$. If you somehow ended up with $V_{DS}-V_{th}$, that suggests a likely algrebra error.

If you find that $(V_{GS}-V_{th})<0$, then that means that you do not meet the other requirement, which is $V_{GS}>V_{th}$.

Your transistor is off (cutoff/subthreshold) and will be in neither linear mode or saturation mode.

• yaa it is actually Vgs not Vds. tq for the answer! – Haziq Mahmur Apr 11 '17 at 8:36