3.

a. VGS = 2.5V and VDS = 2.5V , therefore saturation.

ID =k

′

2

W

L(VGS − VT )2(1 + λVDS)

=115x10−6

2(2.5− 0.43)2(1 + 0.06 · 2.5)

= 283.3µA

VGS = −0.5V and VDS = −1.25V , therefore saturation (again).

ID =k

′

2

W

L(VGS − VT )2(1 + λVDS)

=30x10−6

2(0.5− 0.6)2(1 + 0.1 · 1.25)

= 0.17µA

b. VGS = 3.3V and VDS = 2.2V , therefore linear/triode.

ID = k′W

L

((VGS − VT )VDS −

V 2DS

2

)= 115x10−6

((3.3− 0.43)2.2− 2.22

2

)= 447.8µA

VGS = −2.5V and VDS = −1.8V , therefore linear/triode (again).

ID = k′W

L

((VGS − VT )VDS −

V 2DS

2

)= 30x10−6

((2.5− 0.4)1.8− 1.82

2

)= 64.8µA

c. VGS = 0.6V and VDS = 0.1V , therefore linear/triode.

ID = k′W

L

((VGS − VT )VDS −

V 2DS

2

)= 115x10−6

((0.6− 0.63)0.1− 0.12

2

)= 1.38µA

VGS = −2.5V and VDS = −0.7V , therefore linear/triode (again).

ID = k′W

L

((VGS − VT )VDS −

V 2DS

2

)= 30x10−6

((2.5− 0.4)0.7− 0.72

2

)= 36.75µA

1

6. For a short channel device,

ID = k′W

L

[(VGS − VT )Vmin −

V 2min

2

](1 + λVDS)

Vmin = min [(VGS − VT ), VDS , VDSAT ]

To begin with, the operating regions need to be determined.For any of these data to be in saturation, VT should be:

VGS − VT < VDSAT

2− VT < 0.6

⇓VT > 1.4V

This is quite a high value in our process, thus we can assume that all data are taken in velocity saturation.We will check this assumption later.In velocity saturation:

ID = k′W

L

[(VGS − VT )VDSAT −

V 2DSAT

2

](1 + λVDS)

Using rows 1 and 2 from the table we get:

1812 = k′W

L

[(2.5− VT0) 0.6− 0.62

2

](1 + λ1.8)

1297 = k′W

L

[(2− VT0) 0.6− 0.62

2

](1 + λ1.8)

⇓VT0 = 0.44V

Note that VT0 < 1.4V and therefore 1, 2 and 3 are in velocity saturation. Now, using rows 2 and 3:

1297

1361=

1 + λ1.8

1 + λ2.5⇓

λ = 0.08V −1

Now using rows 2 and 4:VT = 0.587V

and using rows 2 and 5:VT = 0.691V

Both of these values are less than 1.4V, so all the data in our table were taken in velocity saturation.

VT = TT0 + γ

(√|Vsg|+ |2φp| −

√2|φf |

)The two values of VT we found previously, and knowing that VT0 = 0.44V , we can nd:

|2φf | = 0.6V

γ = 0.3V12

Now we can substitute in data and nd:W

L= 15

2

9.

a. Device is always in saturation.

−VXR

=k

′

p

2

W

L(VX − |Vtp|)2

b. This is an approximate graph.

Figure 1: Load lines for M1.

c.

1V

20kΩ=

30x10−6

2

W

L× (1.5− 0.4)

2

50µA = 1.5x10−6W

L× 1.21

2.755 =W

L⇓

W ≈ 0.69µm

d. Voltage at Node X would go up since the current drive of PMOS is lower.

Figure 2: Load lines with M1 velocity saturated.

3

13.

Vin = 0.2⇒ IDS = 3x10−8A (1)

or

Vin = 0.2⇒ IDS = 5x10−9A (2)

∆t = C∆V

I

∆t(1) = 1pF × 1

3x10−8

= 33.3µs

∆t(2) = 1pF × 1

5x10−9

= 200µs

17.

Cox = 6fF

µm2

LD = 0.5µm

WD = 1µm

Cg is dened by the following relationships:

cut− off CoxWL+ 2CoW

linear CoxWL+ 2CoW

saturation2

3CoxWL+ 2CoW

Diusion capacitance, Cd, is given by:

Cd = CjLDWD + Cjsw(2LD +WD)

Cj =Cjo

(1 + VDS

φ )mj

Cjsw =Cjswo

(1 + VDS

φ )mjsw

a. Vin = 2.5V , Vout = 2.5V . Velocity saturation.

Cg = 1.62fF

Q = 4.05fC = 4.05x10−15C

Cd = 0.827fF

Vin = 2.5V , Vout = 0.5V . Linear region.

Cg = 2.12fF

Q = 5.3fC = 5.3x10−15C

Cd = 1.263fF

Vin = 2.5V , Vout = 0V . Linear region.

Cg = 2.12fF

Q = 5.3fC = 5.3x10−15C

Cd = 1.56fF

4

b. Vin = 0⇒ Cutt off

Cg = GoxWL

= 2.12pF

Q = 0

Cd are the same as they were in part a.

22.

a.

s =0.25

0.1= 2.5

Speed scales inversely to tp which scales as 1s2 therefore speed scale with s2 so:

f = 625MHz

Power scales with αs⇒ P = 25W .

b. In full scaling, speed scales with:f = 250MHz

Power scales as 1s2 and thus:

P = 1.6W

c. We want to keep power constant:

s

u3= 1⇒ u = s

13 = (2.5)

13 = 1.36

Voltage becomes V = 1.842V , speed scales as s2

u = 4.6 so:

f = 460MHz

5