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1/4

= . =

/

= ./ = = /

WL

BL

CC

Tuesday, November 20, 12

7/30/2019 bono2sol

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ANSWER:

iave = CV

t= 0.1pF

40mV

1ps= 4mA

CBLvBL = CCvC vC =100f F

20f F40mV = 0.2V

At point 1, vC = 0, vDS = 2.5V and vGS Vt = 5V 1V = 4V > vDS, so

i1 = (0.15mA)(W/L)

2(4)2.5 2.52

= 2.0625mA

WL

At point 2, vC = 0.2V = vS, vDS = 2.5V .04V 0.2V = 2.26V, and vGS Vt = 5V 0.2V 1V = 3.8V, so

i2 = (0.15mA)(W/L) 2(3.8)2.26 2.262

= 1.81mA

W

L

and

iave = 4mA =2.0625mA + 1.81mA

2

W

L

W

L= 2.1

Tuesday, November 20, 12

7/30/2019 bono2sol

3/4

/ / / /

= = ./ = = . = . (/) = (/) = /

VDD

M1 M3

M2 M4

M5M6

QQ

S R

Tuesday, November 20, 12

7/30/2019 bono2sol

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ANSWER: Consider the node Q. For point 1, when vQ = VDD, M4 is not conducting currentbecause vSD,4 = 0, and the current discharging the capacitor is

i1 = 0.15mA

W

L

(1.8V 0.5V)2 = (253.5A)

W

L

At point 2, vQ = VDD/2, and both M4 and M6 have current. Both transistors are in trioderegion and

iM4 =

1

2

0.3mA

4 (6/1)

2(1.8V

0.5V)0.9V

(0.9V)2

= 344.25A

iM6 = 0.15mA

W

L

2(1.8V 0.5V)0.9V (0.9V)2

= (229.5A)

W

L

Thus,

i2 = iM6 iM4 = (229.5A)

W

L

344.25A

and

iave =1

2(i1 + i2) = (241.5A)

W

L

172.125A = 10pF

0.9V

1ns= 9mA

W

L= 9000A + 172.125A241.5A 38

Tuesday, November 20, 12