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EE_331F_2007_-_HW8S
Course: EE 331, Spring 2008School: Washington
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331 Devices EE and Circuits I Problem Set #8 Solution Autumn 2007 6.71 I DD = 0.20mW = 60.1A VTNL = -1+ 0.5 3.3 + 0.6 - 0.6 = -0.400V VH = 3.3V 3.3V W W 1.16 0.20 | For VO = VL = 0.2V , 60.1A = 100A 3.3 - 0.6 - 0.20 = 2 1 L S L S W 1.36 2 100A W VTNL = -1+ 0.5 0.20 + 0.6 - 0.6 = -0.940V | 60.1A = (-0.940) | = 2 L L 1 L L Using Kn' = 60 A; (W/L)s = 2.43/1, and (W/L)L = 2.86/1 ( ) (...
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331
Devices EE and Circuits I Problem Set #8 Solution
Autumn 2007
6.71 I DD =
0.20mW = 60.1A VTNL = -1+ 0.5 3.3 + 0.6 - 0.6 = -0.400V VH = 3.3V 3.3V W W 1.16 0.20 | For VO = VL = 0.2V , 60.1A = 100A 3.3 - 0.6 - 0.20 = 2 1 L S L S W 1.36 2 100A W VTNL = -1+ 0.5 0.20 + 0.6 - 0.6 = -0.940V | 60.1A = (-0.940) | = 2 L L 1 L L Using Kn' = 60 A; (W/L)s = 2.43/1, and (W/L)L = 2.86/1
(
)
(
)
6.82
W W 0.1 4.32 80A = 100A 2.5 - 0.6 - 0.1 = 2 1 L A L A
VTNB = 0.6 + 0.5 0.1+ 0.6 - 0.6 = 0.631 W W 0.1 4.65 80A = 100A 2.5 - 0.1- 0.631- 0.1 = 2 1 L B L A Using Kn' = 60 A; (W/L)A = 7.20/1, and (W/L)B = 7.75/1
6.90
W 1.81 Y = (A + B)(C + D)(E + F ) | = 1 L L W 2.22 6.66 | =3 = 1 L A-F 1
(
)
7.1
' n
-14 3.9o cm2 (3.9) 8.854x10 F / cm K = nC = n = n = 500 Tox Tox V - sec 10x10-9 m( 100cm / m) " ox
ox
(
)
F A A = 173 x 10-6 2 = 173 2 V - sec V V p ' 200 A A " K 'p = pCox = Kn = 173 2 = 69.1 2 n V V 500
' Kn = 173x10-6
7.8
(a) V
H
= 2.5V | VL = 0V | For MN , VGS = 0, so M N is cut off. , VGS = 2.5, VDS = 0V and VTN = 0.60V. For VDS < VGS - VTN , MN is in the triode region. , VGS = 1.25, VDS = 1.25 V VTN and = 0.60V. For VDS > VGS - VTN , MN is saturated.
For MP , VGS = -2.5, VDS = 0V and VTP = -0.60V. For VDS < VGS - VTP , MP is in the triode region.
(b) For M (c) For M
N
For MP , VGS = 0, so MP is cut off.
N
For MP , VGS = -1.25, VDS = -1.25V and VTP = -0.75V. For VDS > VGS - VTP , M P is saturated.
7.11 For vI = vO, both transistors will be saturated since vGS = vDS for each device. Equating the drain currents with Kn = Kp yields: (a) Both transistors are saturated with VDS = VGS
K 2 2 Kn (vI - VTN ) = 2p (vI - VDD - VTP ) so vI - VTN = VDD - vI + VTP 2 V + VTN + VTP 2.5 + 0.6 - .6 vO = vI = DD = = 1.25V 2 2 2 2 K 100A 2 (b) I DN = 2n (vI - VTN ) = 2 1 (1.25 - 0.6) = 42.3A K 2 2 40A 5 1.25 - 2.5 + 0.6) = 42.3A Checking I DP = p (vI - VDD - VTp ) = ( 2 1 2 Using Kn' = 60 A; Kp' should be 24 A; Current = 25.3 A
(c) For K
n
= 2.5K p ,
2.5K p K 2 2 (vI - VTN ) = 2p (vI - VDD - VTP ) or 1.58(vI - VTN )= VDD - vI + VTP 2 V + 1.58VTN + VTP 2.5 + 1.58(0.6)+ (-0.6) vO = vI = DD = = 1.104V 2.58 2.58 2 100A 2 (d ) I DN = 2 1 (1.104 - 0.6) = 25.4A | Check by finding IDP : 2 40A 2 I DP = 1.104 - 2.5 + 0.6) = 25.3A ( 2 1 Using Kn = 60 A; Current = 15.24 A
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