μ

n

= 300 sq. cm / v.sec

t

ox

= 200 x

10

m

Ɛ o . Ɛ r = 3.9 x 8.85 x 10 -14 F/cm

W = 10 μ m, L = 1 μ m

V GS = 1V, V T = 0.5V

λ ~ 0 (device is long channel, so V A and r o are high)

C 0v ~ 0

V DD = 1.8

R L = 10K









Let us calculate:





I D = Bias current or DC current = 1/2 μ n C ox .

(

V

GS

-

V

T

)

2

(1 +

λ

V

DS

) =

1/2 μ n C ox .

(

V

GS

-

V

T

)

2





Since lambda value is 0.









Calculating: C

ox

=

Ɛ o .

Ɛ r

/ t

ox





=>

t

ox

=

200 x

10

m = 200 x 10

cm





Hence,

C

ox

=

Ɛ o .

Ɛ r

/ t

ox

=

3.9 x 8.85 x

10

/

200 x 10





=>

C

ox

= 0.172 x

10

F/cm









Now,

μ

n

= 300 cm

/ V.sec





=>

I

D

= 1/2 . 300 . 0.17 x

10

. (10/1) . (1 - 0.52)

= 63.75

μ

A

= 63.75 x 10

A









Now, let us calculate g m = 2I D / (V GS - V T ) = 2 x 63.75

μ / (1/2) = 255

μs









Our transistor is working in the saturation region. To show that, V

DS

> V

GS

- V

T









To find

V

DS

:

V

DS

=

V

DD

- (voltage drop over this transistor which is R

L

x I

D

)





Since, V

DD

= 1.8V





V

DS

= 1.2 v which is higher than

V

GS

- V

T 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 = g m (r o || R L ) = g m . R L = 255

μ x 10k ~ 2.5 appx.





C

GS = 2/3 WL C ox + WC ov (Here, WC OV is zero)





C

GS = 10 x

10

F = 10fF









At last, we will calculate f T





f

T

=

g

m

/ 2

π

C

GS = 4GHz for this transistor.





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