### MOS Calculations Example

Below is an example of an NMOS with the given parameters:

μn = 300 sq. cm / v.sec
tox = 200 x 10-10 m
Ɛo . Ɛr = 3.9 x 8.85 x 10-14 F/cm
W = 10μm, L = 1μm
VGS = 1V, VT = 0.5V
λ ~ 0 (device is long channel, so VA and ro are high)
C0v ~ 0
VDD = 1.8
RL = 10K

Let us calculate:

ID = Bias current or DC current = 1/2μn Cox . (VGS - VT)2 (1 + λ VDS) = 1/2μn Cox . (VGS - VT)2

Since lambda value is 0.

Calculating: Cox = Ɛo . Ɛr / tox

=> tox = 200 x 10-10 m = 200 x 10-8 cm

Hence, Cox = Ɛo . Ɛr / tox = 3.9 x 8.85 x 10-14 / 200 x 10-8

=> Cox = 0.172 x 10-6 F/cm2

Now, μn = 300 cm2  / V.sec

=> ID = 1/2 . 300 . 0.17 x 10-6 . (10/1) . (1 - 0.52)
= 63.75 μ
= 63.75 x 10-6 A

Now, let us calculate gm = 2ID / (VGS - VT) = 2 x 63.75μ / (1/2) = 255μs

Our transistor is working in the saturation region. To show that, VDS > VGS - VT

To find VDSVDS = VDD - (voltage drop over this transistor which is RL x ID)

Since, VDD = 1.8V

VDS = 1.2 v which is higher than VGS - VT i.e. > (1 - 0.5)

Now we are sure that the transistor is working in the saturation region.

Let's calculate the gain (G or Av) now.

G = gm (ro || RL) = gm. RL = 255μ x 10k ~ 2.5 appx.

CGS = 2/3 WL Cox + WCov  (Here, WCOV is zero)

CGS = 10 x 10-15 F = 10fF

At last, we will calculate f

fT  gm / 2πCGS = 4GHz for this transistor.