MOSFET Speed and power tradeoff

Hi I was reading this document and came across some confusion.
https://www.uio.no/studier/emner/matnat/ifi/INF4420/v11/diverse/Extracting_uCox.pdf

https://www.google.com/search?q=cap...UIECgB&biw=1360&bih=625#imgrc=x0OqxKbzpedvkM:

1.As per the above curve, MOS capacitance is small in depletion mode (when 0<Vgs<Vt) and weak inversion (I don't know exact range for this). Bandwidth represents the speed of the device and BW is inversely proportional to the capactiance. Hence, the speed of the MOSFET should be more in depletion mode and weak inversion mode

2.When Vgs>Vt, the MOSFET enters into strong inversion mode (Saturation region). Then based on chose Vgs value Id is driven, which decides the Power consumed by the device. The more is the chosen Vgs, the more will be power consumption. In this the MOS capacitance is higher than the above case, so less speed.

But in the document it is said that
Always put your transistor in moderate-inversion region (V OV ≈ 150-250 mV) to make a compromise between speed and power consumption. If you need more speed → Strong inversion region. If you want low power → weak inversion

Can someone tell me
1.what are exact range for weak, moderate and strong inversion modes.
2.As per my argument, we get more speed in depletion or weak inversion mode but as per the document it is strong inversion region. Which is correct and why?

Dear circuitking,
the range of the MOS operating region can be determined by gm over Id ratio (gm/ID=2/(VGS-VT)). Hence, if gm is in the range 5-10 S/A the MOS operates in strong inversion, the range 10-15 S/A is the moderate inversion and 15-25 S/A is the weak inversion. The MOS exhibits the maximum speed in strong inversion because it depends by the total gate capacitance that is minimum in this region. I suggest to read about the gm/id methodology in which the mos performance are arranged in lookup tables as function of the gm/ID ratio.
I hope that it is useful

Regards

For a long channel MOSFET, the \[f_T = \frac{1.5\mu}{2\pi L^2}(V_{gs}-V_{th}}\]
If you increase the overdrive, then your transit frequency goes up. But because transconductance is given by \[\frac{2I_D}{V_{gs}-V_{th}}\], increasing overdrive decreases the gm(of course make sure you don't change the current).

Just to clarify something, strong inversion is not the same as saturation. Strong inversion is the point where you have significantly overcome the surface potential and have created an inversion channel between the drain/source. The gate voltage applied is what creates this channel—that is why SI is definite by a gate voltage. Saturation is given by the drain to source voltage, vov=vgs-vt.


If you do not have significant vds, you will be in strong inversion, triode. If you surpass the saturation voltage then you will be in strong inversion, saturation.

vivekroy said:

For a long channel MOSFET, the \[f_T = \frac{1.5\mu}{2\pi L^2}(V_{gs}-V_{th}}\]
If you increase the overdrive, then your transit frequency goes up. But because transconductance is given by \[\frac{2I_D}{V_{gs}-V_{th}}\], increasing overdrive decreases the gm(of course make sure you don't change the current).

Click to expand...

I don't know why the equations came as elektroda.pl

But here is the answer with the expressions:

For a long channel MOSFET, the f_T = \frac{1.5\mu}{2\pi L^2}(V_{gs}-V_{th}}
If you increase the overdrive, then your transit frequency goes up. But because transconductance is given by \frac{2I_D}{V_{gs}-V_{th}}, increasing overdrive decreases the gm(of course make sure you don't change the current).

vivekroy said:

I don't know why the equations came as elektroda.pl

Click to expand...

Your equations were set off by a 'tex' html tag. Apparently the system didn't know how to interpret this, so it defaulted to a logo from our image server as a way of saying 'technical difficulties'.