1. Chapter 3
Problem Solutions
3.1
⎛ W ⎞ ⎛ k ′ ⎞ ⎛ 10 ⎞ ⎛ 0.08 ⎞
Kn = ⎜ ⎟ ⎜ n ⎟ = ⎜ ⎟⎜ ⎟ = 0.333 mA/V
2
⎝ L ⎠ ⎝ 2 ⎠ ⎝ 1.2 ⎠ ⎝ 2 ⎠
For VDS = 0.1 V ⇒ Non Sat Bias Region
(a) VGS = 0 ⇒ I D = 0
VGS = 1 V I D = 0.333 ⎡ 2 (1 − 0.8 )( 0.1) − ( 0.1) ⎤ = 0.01 mA
2
(b)
⎣ ⎦
I D = 0.333 ⎡ 2 ( 2 − 0.8 )( 0.1) − ( 0.1) ⎤ = 0.0767 mA
2
(c) VGS = 2 V
⎣ ⎦
I D = 0.333 ⎡ 2 ( 3 − 0.8 )( 0.1) − ( 0.1) ⎤ = 0.143 mA
2
(d) VGS = 3 V
⎣ ⎦
3.2
All in Sat region
⎛ 10 ⎞⎛ 0.08 ⎞
Kn = ⎜ ⎟⎜ ⎟ = 0.333 mA/V
2
⎝ 1.2 ⎠⎝ 2 ⎠
(a) ID = 0
I D = 0.333[1 − 0.8] = 0.0133 mA
2
(b)
I D = 0.333[ 2 − 0.8] = 0.480 mA
2
(c)
I D = 0.333[3 − 0.8] = 1.61 mA
2
(d)
3.3
(a) Enhancement-mode
(b) From Graph VT = 1.5 V
Now
0.03 = K n ( 2 − 1.5 ) = 0.25 K n ⇒ K n = 0.12
2
0.15 = K n ( 3 − 1.5 ) = 2.25 K n
2
K n = 0.0666
0.39 = K n ( 4 − 1.5 ) = 6.25 K n
2
K n = 0.0624
0.77 = K n ( 5 − 1.5 ) = 12.25 K n
2
K n = 0.0629
From last three, K n (Avg) = 0.0640 mA/V 2
(c) iD (sat) = 0.0640(3.5 − 1.5) 2 ⇒ iD (sat) = 0.256 mA for VGS = 3.5 V
iD (sat) = 0.0640(4.5 − 1.5) 2 ⇒ iD (sat) = 0.576 mA for VGS = 4.5 V
3.4
a. VGS = 0
VDS ( sat ) = VGS − VTN = 0 − ( −2.5 ) = 2.5 V
i. VDS = 0.5 V ⇒ Biased in nonsaturation
I D = (1.1) ⎡ 2 ( 0 − (−2.5) )( 0.5 ) − ( 0.5 ) ⎤ ⇒ I D = 2.48 mA
2
⎣ ⎦
ii. VDS = 2.5 V ⇒ Biased in saturation
I D = (1.1) ( 0 − ( −2.5 ) ) ⇒ I D = 6.88 mA
2
iii. VDS = 5 V Same as (ii) ⇒ I D = 6.88 mA
b. VGS = 2 V
VDS ( sat ) = 2 − ( −2.5 ) = 4.5 V
i. VDS = 0.5 V ⇒ Nonsaturation
I D = (1.1) ⎡ 2(2 − (−2.5))(0.5) − (0.5) 2 ⎤ ⇒ I D = 4.68 mA
⎣ ⎦

2. ii. VDS = 2.5 V ⇒ Nonsaturation
I D = (1.1) ⎡ 2(2 − (−2.5))(2.5) − (2.5) 2 ⎤ ⇒ I D = 17.9 mA
⎣ ⎦
iii. VDS = 5 V ⇒ Saturation
I D = (1.1) ( 2 − ( −2.5 ) ) ⇒ I D = 22.3 mA
2
3.5
VDS > VGS − VTN = 0 − ( −2 ) = 2 V
Biased in the saturation region
k′ W
I D = n ⋅ (VGS − VTN )
2
2 L
⎛ 0.080 ⎞ ⎛ W ⎞ W
⎟ ⎜ ⎟ ⎡ 0 − ( −2 ) ⎤ ⇒
2
1.5 = ⎜ ⎣ ⎦ = 9.375
⎝ 2 ⎠⎝ L ⎠ L
3.6
μ n ∈ox ( 600 )( 3.9 ) (8.85 ×10−14 ) 2.071× 10−10
′
kn = μ n Cox = = =
tox tox tox
(a) 500 A ′
kn = 41.4 μ A/V 2
(b) 250 ′
kn = 82.8 μ A/V 2
(c) 100 ′
kn = 207 μ A/V 2
(d) 50 ′
kn = 414 μ A/V 2
(e) 25 ′
kn = 828 μ A/V 2
3.7
a.
∈ox ( 3.9 ) ( 8.85 × 10 )⇒∈
−14
Cox = = ox
= 7.67 ×10−8 F/cm 2
t0 x 450 × 10−8 t0 x
μ n Cox W
Kn = ⋅
2 L
( 650 ) ( 7.67 ×10−8 ) ⎛ ⎞
1 64
= ⎜ ⎟
2 ⎝ 4 ⎠
K n = 0.399 mA / V 2
b. VGS = VDS = 3 V ⇒ Saturation
I D = K n (VGS − VTN ) = ( 0.399 )( 3 − 0.8 ) ⇒ I D = 1.93 mA
2 2
3.8
⎛ ω ⎞⎛ k′ ⎞
I D = ⎜ ⎟ ⎜ n ⎟ (VGS − VTN )
2
⎝ 2 ⎠⎝ 2 ⎠
⎛ ω ⎞ ⎛ 0.08 ⎞
⎟ ( 2.5 − 1.2 ) ⇒ ω = 23.1 μ m
2
1.25 ⎜ ⎟⎜
⎝ 1.25 ⎠ ⎝ 2 ⎠
3.9
∈ ( 3.9 ) (8.85 ×10−14 )
Cox = ox =
t0 x 400 × 10−8
= 8.63 × 10−8 F/cm 2

3. μ n Cox W
Kn = ⋅
2 L
⎛W ⎞
= ( 600 ) ( 8.63 × 10−8 ) ⎜
1
⎟
2 ⎝ 2.5 ⎠
K n = (1.036 × 10−5 ) W
I D = K n (VGS − VTN )
2
1.2 × 10 −3 = (1.036 × 10 −5 ) W ( 5 − 1) ⇒ W = 7.24 μ m
2
3.10
Biased in the saturation region in both cases.
′
kp W
I D = ⋅ (VSG + VTP )
2
2 L
⎛ 0.040 ⎞⎛ W ⎞
⎟⎜ ⎟ ( 3 + VTP )
2
(1) 0.225 = ⎜
⎝ 2 ⎠⎝ L ⎠
⎛ 0.040 ⎞ ⎛ W ⎞
⎟ ⎜ ⎟ ( 4 + VTP )
2
(2) 1.40 = ⎜
⎝ 2 ⎠⎝ L ⎠
Take ratio of (2) to (1):
1.40 (4 + VTP ) 2
= 6.222 =
0.225 (3 + VTP ) 2
4 + VTP
6.222 = 2.49 = ⇒ VTP = −2.33 V
3 + VTP
⎛ 0.040 ⎞ ⎛ W ⎞ W
⎟ ⎜ ⎟ ( 3 − 2.33) ⇒
2
Then 0.225 = ⎜ = 25.1
⎝ 2 ⎠⎝ L ⎠ L
3.11
VS = 5 V, VG = 0 ⇒ VSG = 5 V
VTP = −0.5 V ⇒ VSD ( sat ) = VSG + VTP = 5 − 0.5 = 4.5 V
a. VD = 0 ⇒ VSD = 5 V ⇒ Biased in saturation
I D = 2 ( 5 − 0.5 ) ⇒ I D = 40.5 mA
2
b. VD = 2 V ⇒ VSD = 3 V ⇒ Nonsaturation
I D = 2 ⎡ 2 ( 5 − 0.5 )( 3) − ( 3) ⎤ ⇒ I D = 36 mA
2
⎣ ⎦
c. VD = 4 V ⇒ VSD = 1 V ⇒ Nonsaturation
I D = 2 ⎡ 2 ( 5 − 0.5 )(1) − (1) ⎤ ⇒ I D = 16 mA
2
⎣ ⎦
d. VD = 5 V ⇒ VSD = 0 ⇒ I D = 0
3.12
(a) Enhancement-mode
(b) From Graph VTP = + 0.5 V
0.45 = k p ( 2 − 0.5 ) = 2.25 K p ⇒ K p =
2
0.20
1.25 = k p ( 3 − 0.5 ) = 6.25 K p
2
0.20
2.45 = k p ( 4 − 0.5 ) = 12.25 K p
2
0.20
4.10 = k p ( 5 − 0.5 ) = 20.25 K p
2
0.202
Avg K p = 0.20 mA/V 2
(c) iD (sat) = 0.20 (3.5 − 0.5) 2 = 1.8 mA
iD (sat) = 0.20 (4.5 − 0.5) 2 = 3.2 mA

4. 3.13
VSD ( sat ) = VSG + VTP
(a) VSD ( sat ) = −1 + 2 ⇒ VSD ( sat ) = 1 V
(b) VSD ( sat ) = 0 + 2 ⇒ VSD ( sat ) = 2 V
(c) VSD ( sat ) = 1 + 2 ⇒ VSD ( sat ) = 3 V
′
kp W k′ W
⋅ (VSG + VTP ) = ⋅ ⋅ ⎡VSD ( sat ) ⎤
2 2
ID =
p
2 L 2 L ⎣ ⎦
⎛ 0.040 ⎞
⎟ ( 6 )(1) ⇒ I D = 0.12 mA
2
(a) ID = ⎜
⎝ 2 ⎠
⎛ 0.040 ⎞
⎟ ( 6 )( 2 ) ⇒ I D = 0.48 mA
2
(b) ID = ⎜
⎝ 2 ⎠
⎛ 0.040 ⎞
⎟ ( 6 )( 3) ⇒ I D = 1.08 mA
2
(c) ID = ⎜
⎝ 2 ⎠
3.14
VSD (sat) = VSG + VTP = 3 − 0.8 = 2.2 V
⎛ 15 ⎞⎛ 0.04 ⎞
KP = ⎜ ⎟⎜ ⎟ = 0.25 mA/V
2
⎝ 1.2 ⎠⎝ 2 ⎠
VSD = 0.2 Non Sat I D = 0.25 ⎡ 2 ( 3 − 0.8 )( 0.2 ) − ( 0.2 ) ⎤ = 0.21 mA
2
a)
⎣ ⎦
VSD = 1.2 V Non Sat I D = 0.25 ⎡ 2 ( 3 − 0.8 )(1.2 ) − (1.2 ) ⎤ = 0.96 mA
2
b)
⎣ ⎦
c) VSD = 2.2 V Sat I D = 0.25(3 − 0.8) = 1.21 mA
2
d) VSD = 3.2 V Sat ID = 1.21 mA
e) VSD = 4.2 V Sat ID = 1.21 mA
3.15
μ p ∈ox ( 250 )( 3.9 ) (8.85 ×10−14 ) 8.629 × 10−11
′
k p = μ p Cox = = =
t0 x t0 x t0 x
(a) tox = 500Å ⇒ k ′ = 17.3 μ A/V 2
p
(b) 250Å ⇒ k ′ = 34.5 μ A/V 2
p
(c) 100Å ⇒ k ′ = 86.3 μ A/V 2
p
(d) ′
50Å ⇒ k p = 173 μ A/V 2
(e) 25Å ⇒ k ′ = 345 μ A/V 2
p
3.16
∈ox ( 3.9 ) ( 8.85 × 10 )
−14
Cox = = −8
= 6.90 × 10−8 F/cm 2
t0 x 500 × 10
kn = ( μ n Cox ) = ( 675 ) ( 6.90 × 10−8 ) ⇒ 46.6 μ A/V 2
′
k ′ = ( μ p Cox ) = ( 375 ) ( 6.90 × 10−8 ) ⇒ 25.9 μ A/V 2
p
PMOS:

5. k′ ⎛ W ⎞
⎜ ⎟ (VSG + VTP )
2
ID =
p
2 ⎝ L ⎠p
⎛ 0.0259 ⎞⎛ W ⎞ ⎛W ⎞
⎟⎜ ⎟ ( 5 − 0.6 ) ⇒ ⎜ ⎟ = 3.19
2
0.8 = ⎜
⎝ 2 ⎠⎝ L ⎠ p ⎝ L ⎠p
L = 4 μ m ⇒ W p = 12.8 μ m
⎛ 0.0259 ⎞
Kp = ⎜ ⎟ ( 3.19 ) ⇒ K p = 41.3 μ A/V = K n
2
⎝ 2 ⎠
Want Kn = Kp
′
kn ⎛ W ⎞ k′ ⎛ W ⎞
⎜ ⎟ = ⎜ ⎟ = 41.3
p
2 ⎝ L ⎠N 2 ⎝ L ⎠p
⎛ 46.6 ⎞ ⎛ W ⎞ ⎛W ⎞
⎜ ⎟ ⎜ ⎟ = 41.3 ⇒ ⎜ ⎟ = 1.77
⎝ 2 ⎠ ⎝ L ⎠N ⎝ L ⎠N
L = 4 μ m ⇒ WN = 7.09 μ m
3.17
VGS = 2 V, I D = ( 0.2 )( 2 − 1.2 ) = 0.128 mA
2
1 1
r0 = = ⇒ r0 = 781 kΩ
λ I D ( 0.01)( 0.128 )
VGS = 4 V, I D = ( 0.2 )( 4 − 1.2 ) = 1.57 mA
2
1
r0 = ⇒ r = 63.7 kΩ
( 0.01)(1.57 ) 0
1 1
VA = = ⇒ VA = 100 V
λ ( 0.01)
3.18
⎛ 0.080 ⎞
⎟ ( 4 )( 3 − 0.8 ) = ( 0.16 )( 3 − 0.8 ) ⇒ I D = 0.774 mA
2 2
ID = ⎜
⎝ 2 ⎠
1 1 1
r0 = ⇒λ = = ⇒ λ (max) = 0.00646 V −1
λ ID r0 I D ( 200 )( 0.774 )
1 1
VA ( min ) = = ⇒ VA ( min ) = 155 V
λ ( max ) 0.00646
3.19
VTN = VTNO + γ ⎡ 2φ f + VSB − 2φ f ⎤
⎣ ⎦
ΔVTN = 2 = ( 0.8 ) ⎡ 2φ f + VSB − 2 ( 0.35 ) ⎤
⎣ ⎦
2.5 + 0.837 = 2 ( 0.35 ) + VSB ⇒ VSB = 10.4 V
3.20
VTN = VTNo + r ⎡ 2φ f + VSB − 2φ f ⎤
⎣ ⎦
= 0.75 + 0.6 ⎡ 2 ( 0.37 ) + 3 − 2 ( 0.37 ) ⎤
⎣ ⎦
= 0.75 + 0.6 [1.934 − 0.860]
VTN = 1.39 V
VDS (sat) = 2.5 − 1.39 = 1.11 V

6. ⎛ 0.08 ⎞
Sat Region I D = (15 ) ⎜ ⎟ ( 2.5 − 1.39 )
2
(a)
⎝ 2 ⎠
I D = 0.739 mA
⎛ 0.08 ⎞ ⎡
Non-Sat I D = (15 ) ⎜ ⎟ 2 ( 2.5 − 1.39 )( 0.25 ) − ( 0.25 ) ⎤
2
(b)
⎝ 2 ⎠⎣ ⎦
I D = 0.296 mA
3.21
a.
VG = %ox t0 x = ( 6 × 106 )( 275 × 10−8 )
VG = 16.5 V
16.5
b. VG = ⇒ VG = 5.5 V
3
3.22
Want VG = ( 3)( 24 ) = %ox t0 x = ( 6 × 106 ) t0 x
t0 x = 1.2 ×10−5 cm = 1200 Angstroms
3.23
⎛ R2 ⎞ ⎛ 18 ⎞
VG = ⎜ ⎟ VDD = ⎜ ⎟ (10 ) = 3.6 V
⎝ R1 + R2 ⎠ ⎝ 18 + 32 ⎠
Assume transistor biased in saturation region
V V − VGS
= K n (VGS − VTN )
2
ID = S = G
RS RS
3.6 − VGS = ( 0.5 )( 2 )(VGS − 0.8 )
2
= VGS − 1.6VGS + 0.64
2
VGS − 0.6VGS − 2.96 = 0
2
( 0.6 ) + 4 ( 2.96 )
2
0.6 ±
VGS = ⇒ VGS = 2.046 V
2
VG − VGS 3.6 − 2.046
ID = = ⇒ I D = 0.777 mA
RS 2
VDS = VDD − I D ( RD + RS )
= 10 − ( 0.777 )( 4 + 2 ) ⇒ VDS = 5.34 V
VDS > VDS ( sat )
3.24

7. ID(mA)
4
(a)
Q-pt
Q-pt
1.67
(b)
4 5 V (V)
DS
(a) VGS = 4 V VDS (sat) = 4 − 0.8 = 3.2 V
If Sat I D = 0.25 ( 4 − 0.8 ) = 2.56
2
VDS = 1.44 ×
Non-Sat
4 = I D RD + VDS = K n RD ⎣ 2 (VGS − VT ) VDS − VDS ⎦ + VDS
⎡ 2
⎤
4 = ( 0.25 )(1) ⎡ 2 ( 4 − 0.8 ) VDS − VDS ⎤ + VDS
⎣
2
⎦
4 = 2.6VDS − 0.25VDS
2
0.25VDS − 2.6VDS + 4 = 0
2
2.6 ± 6.76 − 4
VDS = = 1.88 V
2 ( 0.25 )
4 − 1.88
ID = = 2.12 mA
1
(b) Non-Sat region
5 = I D RD + VDS = K n RD ⎡ 2 (VGS − VT )VDS − VDS ⎤ + VDS
⎣
2
⎦
5 = ( 0.25 )( 3) ⎡ 2 ( 5 − 0.8 ) VDS − VDS ⎤ + VDS
⎣
2
⎦
5 = 7.3VDS − 0.75VDS
2
0.75 VDS − 7.3VDS + 5 = 0
2
7.3 ± 53.29 − 15
VDS =
2 ( 0.75 )
VDS = 0.741 V
5 − 0.741
ID = = 1.42 mA
3
3.25
ID(mA)
2.92
(a)
Q-pt
1.25
(b)
3.5 5 V (V)
SD

8. (a) VSG = VDD = 3.5 VSD ( sat ) = 3.5 − 0.8 = 2.7 V
If biased in Sat region, I D = ( 0.2 )( 3.5 − 0.8 )
2
= 1.46 mA
VSD = 3.5 − (1.46 )(1.2 ) = 1.75 V ×
Biased in Non-Sat Region.
3.5 = VSD + I D RD = VSD + K p RD ⎡ 2 (VSG + VTP ) VSD − VSD ⎤
⎣
2
⎦
3.5 = VSD + ( 0.2 )(1.2 ) ⎡ 2 ( 3.5 − 0.8 ) VSD − VSD ⎤
⎣
2
⎦
3.5 = VSD + 1.296 VSD − 0.24 VSD
2
0.24 VSD − 2.296 VSD + 3.5 = 0
2
+2.296 ± 5.272 − 3.36
VSD = use − sign VSD = 1.90 V
2 ( 0.24 )
I D = ( 0.2 ) ⎡ 2 ( 3.5 − 0.8 )(1.9 ) − (1.9 ) ⎤ = 0.2 [10.26 − 3.61]
2
⎣ ⎦
3.5 − 1.90
ID = = 1.33 mA
1.2
I D = 1.33 mA
(b) VSG = VDD = 5 V VSD ( sat ) = 5 − 0.8 = 4.2 V
If Sat Region I D = ( 0.2 )( 5 − 0.8 ) = 3.53 mA, VSD < 0
2
Non-Sat Region.
5 = VSD + K p RD ⎡ 2 (VSG + VTP ) VSD − VSD ⎤
⎣
2
⎦
5 = VSD + ( 0.2 )( 4 ) ⎡ 2 ( 5 − 0.8 ) VSD − VSD ⎤
⎣
2
⎦
5 = VSD + 6.72 VSD − 0.8 VSD
2
0.8 VSD − 7.72 VSD + 5 = 0
2
7.72 ± 59.598 − 16
VSD = use − sign VSD = 0.698 V
2 ( 0.8 )
5 − 0.698
ID = ⇒ I D = 1.08 mA
4
3.26
10 − VS
= K p (VSG + VTP )
2
ID =
RS
Assume transistor biased in saturation region
⎛ R2 ⎞
VG = ⎜ ⎟ ( 20 ) − 10
⎝ R1 + R2 ⎠
⎛ 22 ⎞
=⎜ ⎟ ( 20 ) − 10 ⇒ VG = 4.67 V
⎝ 8 + 22 ⎠
VS = VG + VSG
10 − ( 4.67 + VSG ) = (1)( 0.5 )(VSG − 2 )
2
5.33 − VSG = 0.5 (VSG − 4VSG + 4 )
2
0.5VSG − VSG − 3.33 = 0
2
(1) + 4 ( 0.5 )( 3.33)
2
1±
VSG = ⇒ VSG = 3.77 V
2 ( 0.5 )

9. 10 − ( 4.67 + 3.77 )
ID = ⇒ I D = 3.12 mA
0.5
VSD = 20 − I D ( RS + RD )
= 20 − ( 3.12 )( 0.5 + 2 ) ⇒ VSD = 12.2 V
VSD > VSD ( sat )
3.27
VG = 0, VSG = VS
Assume saturation region
I D = 0.4 = K p (VSG + VTP )
2
0.4 = ( 0.2 )(VS − 0.8 )
2
0.4
VS = + 0.8 ⇒ VS = 2.21 V
0.2
VD = I D RD − 5 = ( 0.4 )( 5 ) − 5 = −3 V
VSD = VS − VD = 2.21 − ( −3) ⇒ VSD = 5.21 V
VSD > VSD ( sat )
3.28
VDD = I DQ RD + VDSQ + I DQ RS
⎛ k ′ ⎞⎛ W ⎞
(1) 10 = I DQ ( 5 ) + 5 + VGS and I DQ = ⎜ n ⎟ ⎜ ⎟ (VGS − VTN )
2
⎝ 2 ⎠⎝ L ⎠
⎛ 0.060 ⎞ ⎛ W ⎞
⎟ ⎜ ⎟ (VGS − 1.2 )
2
or (2) I DQ = ⎜
⎝ 2 ⎠⎝ L ⎠
Let VGS = 2.5 V
Then from (1), 10 = I DQ ( 5 ) + 5 + 2.5 ⇒ I D = 0.5 mA
⎛ 0.060 ⎞⎛ W ⎞ W
⎟⎜ ⎟ ( 2.5 − 1.2 ) ⇒
2
Then from (2), 0.5 = ⎜ = 9.86
⎝ 2 ⎠⎝ L ⎠ L
V 2.5
I DQ RS = VGS ⇒ RS = GS = ⇒ RS = 5 k Ω
I DQ 0.5
10
IR = = ( 0.5 )( 0.05 ) = 0.025 mA
R1 + R2
10
Then R1 + R2 = = 400 k Ω
0.025
⎛ R2 ⎞ ⎛ R2 ⎞
⎜ ⎟ (VDD ) = 2VGS ⇒ ⎜ ⎟ (10 ) = 2 ( 2.5 ) ⇒ R1 = R2 = 200 k Ω
⎝ R1 + R2 ⎠ ⎝ 400 ⎠
3.29
⎛ 75 ⎞
K n = ( 25 ) ⎜ ⎟ ⇒ 0.9375 mA/V 2
⎝ 2⎠
⎛ 6 ⎞
VG = ⎜ ⎟ (10 ) − 5 = −2 V
⎝ 6 + 14 ⎠
(VG − VGS ) − ( −5)
= I D = K n (VGS − VTN )
2
RS
−2 − VGS + 5 = ( 0.9375 )( 0.5 )(VGS − 1)
2
3 − VGS = 0.469 (VGS − 2VGS + 1)
2

10. 0.469 VGS + 0.0625 VGS − 2.53 = 0
2
−0.0625 ± 0.003906 + 4.746
VGS = ⇒ VGS = 2.26 V
2 ( 0.469 )
I D = 0.9375 ( 2.26 − 1) ⇒ I D = 1.49 mA
2
VDS = 10 − (1.49 )(1.7 ) ⇒ VDS = 7.47 V
3.30
20 = I DQ RS + VSDQ + I DQ RD
(1) 20 = VSG + 10 + I DQ RD
⎛ k′ ⎞⎛ W ⎞
I DQ = ⎜ ⎟ ⎜ ⎟ (VSG + VTP )
p 2
⎝ 2 ⎠⎝ L ⎠
⎛ 0.040 ⎞ ⎛ W ⎞
⎟ ⎜ ⎟ (VSG − 2 )
2
(2) I DQ = ⎜
⎝ 2 ⎠⎝ L ⎠
For example, let I DQ = 0.8 mA and VSG = 4 V
⎛ 0.040 ⎞ ⎛ W ⎞ W
⎟ ( 4 − 2) ⇒
2
Then 0.8 = ⎜ ⎟⎜ = 10
⎝ 2 ⎠⎝ L ⎠ L
I DQ RS = VSG ⇒ ( 0.8 ) RS = 4 ⇒ RS = 5 k Ω
From (1) 20 = 4 + 10 + ( 0.8 ) RD ⇒ RD = 7.5 k Ω
20
IR = = ( 0.8 )( 0.1) ⇒ R1 + R2 = 250 k Ω
R1 + R2
⎛ R1 ⎞
⎜ ⎟ ( 20 ) = 2VSG = ( 2 )( 4 )
⎝ R1 + R2 ⎠
R1
( 20 ) = 8 ⇒ R1 = 100 k Ω, R2 = 150 k Ω
250
3.31
I Q = 50 = 500 (VGS − 1.2 ) ⇒ VGS = 1.516 V
2
(a) (i)
VDS = 5 − ( −1.516 ) =⇒ VDS = 6.516 V
I Q = 1 = ( 0.5 )(VGS − 1.2 ) ⇒ VGS = 2.61 V
2
(ii)
VDS = 5 − ( −2.61) ⇒ VDS = 7.61 V
(b) (i) Same as (a) VGS = VDS = 1.516 V
(ii) VGS = VDS = 2.61 V
3.32
I D = K n (VGS − VTN )
2
0.25 = ( 0.2 )(VGS − 0.6 )
2
0.25
VGS = + 0.6 ⇒ VGS = 1.72 V ⇒ VS = −1.72 V
0.2
VD = 9 − ( 0.25 )( 24 ) ⇒ VD = 3 V
3.33
(a)

11. ID(mA)
1.0
0.808
Q-pt
0.5
3.81 10 V (V)
DS
5 −1
RD = ⇒ RD = 8 K
0.5
I DQ = 0.5 = 0.25 (VGS − 1.4 ) ⇒ VGS = 2.81 V
2
−2.81 − ( −5 )
RS = ⇒ RS = 4.38 K
0.5
(b) Let RD = 8.2 K, RS = 4.3 K
−VGS − ( −5 )
= I D = 0.25 (VGS − 1.4 )
2
Now
4.3
5 − VGS = 1.075 (VGS − 2.8 VGS + 1.96 )
2
1.075 VGS − 2.01 VGS − 2.89 = 0
2
2.01 ± 4.04 + 12.427
VGS = ⇒ VGS = 2.82 V
2 (1.075 )
I D = 0.25 ( 2.82 − 1.4 ) ⇒ I D = 0.504 mA
2
VDS = 10 − ( 0.504 )( 8.2 + 4.3) ⇒ VDS = 3.70 V
(c) If RS = 4.3 + 10% = 4.73 K
5 − VGS = 1.18 (VGS − 2.8VGS + 1.96 )
2
1.18 VGS − 2.31 VGS − 2.68 = 0
2
2.31 ± 5.336 + 12.65
VGS = = 2.78 V
2 (1.18 )
I D = ( 0.25 )( 2.78 − 1.4 ) ⇒ I D = 0.476 mA
2
If Rs = 4.3 − 10% = 3.87 K
5 − VGS = ( 0.9675 ) (VGS − 2.8VGS + 1.96 )
2
0.9675VGS − 1.71VGS − 3.10 = 0
2
1.71 ± 2.924 + 12.0
VGS = = 2.88 V
2 ( 0.9675 )
I D = ( 0.25 )( 2.88 − 1.4 ) = 0.548 mA
2
3.34
VDD = VSD + I DQ R
9 = 2.5 + ( 0.1) R ⇒ R = 65 k Ω
⎛ k′ ⎞⎛ W ⎞
I DQ = ⎜ ⎟ ⎜ ⎟ (VSG + VTP )
p 2
⎝ 2 ⎠⎝ L ⎠
⎛ 0.025 ⎞ ⎛ W ⎞ W
( 0.1) = ⎜ ⎟ ⎜ ⎟ ( 2.5 − 1.5 ) ⇒
2
=8
⎝ 2 ⎠⎝ L⎠ L
Then for L = 4 μ m, W = 32 μ m

12. 3.35
5 = I DQ RS + VSDQ = I DQ ( 2 ) + 2.5
I DQ = 1.25 mA
10
IR = = (1.25 )( 0.1) ⇒ R1 + R2 = 80 k Ω
R1 + R2
I DQ = K p (VSG + VTP )
2
1.25
1.25 = 0.5 (VSG + 1.5 ) ⇒
2
− 1.5 = VSG
0.5
VSG = 0.0811 V
VG = VS − VSG = 2.5 − 0.0811 = 2.42 V
⎛ R2 ⎞
VG = ⎜ ⎟ (10 ) − 5
⎝ R1 + R2 ⎠
⎛R ⎞
2.42 = ⎜ 2 ⎟ (10 ) − 5 ⇒ R2 = 59.4 k Ω, R1 = 20.6 k Ω
⎝ 80 ⎠
3.36
(a)
ID(mA)
0.429
Q-pt
5 V (V)
SD
VD − ( −5 ) 5−2
RD = = ⇒ RD = 12 K
I DQ 0.25
⎛W ⎞⎛ k′ ⎞
⎟ ⎜ ⎟ (VSG + VTP )
2
ID = ⎜
p
⎝L ⎠⎝ 2 ⎠
⎛ 0.035 ⎞
0.25 = (15 ) ⎜ ⎟ (VSG − 1.2 ) ⇒ VSG = 2.18 V
2
⎝ 2 ⎠
5 − 2.18
RS = ⇒ RS = 11.3 K
0.25
VSD = 2.18 − ( −2 ) = 4.18 V
(b)
k ′ = 35 + 5% = 36.75 μ A/V 2
p
⎛ 0.03675 ⎞ 5 − VSG
I D = (15 ) ⎜ ⎟ (VSG − 1.2 ) =
2
⎝ 2 ⎠ 11.3
3.11(VSG − 2.4VSG + 1.44 ) = 5 − VSG
2
3.11VSG − 6.46VSG − 0.522 = 0
2

13. 6.46 ± 41.73 + 6.49
VSG = = 2.155 V
2 ( 3.11)
5 − 2.155
ID = = 0.252 mA
11.3
VSD = 10 − ( 0.252 )(12 + 11.3) = 4.13 V
k ′ = 35 − 5% = 33.25 μ A/V 2
p
⎛ 0.03325 ⎞ 5 − VSG
I D = (15 ) ⎜ ⎟ (VSG − 1.2 ) =
2
⎝ 2 ⎠ 11.3
2.82 (VSG − 2.4VSG + 1.44 ) = 5 − VSG
2
2.82VSG − 5.77VSG − 0.939 = 0
2
5.77 ± 33.29 + 10.59
VSG = = 2.198 V
2 ( 2.82 )
5 − 2.198
ID = = 0.248 mA
11.3
VSD = 10 − ( 0.248 )(12 + 11.3) = 4.22 V
3.37
−VSD − ( −10 ) −6 + 10
ID = ⇒5= ⇒ RD = 0.8 kΩ
RD RD
I D = K P (VSG + VTP ) ⇒ 5 = 3 (VSG − 1.75 )
2 2
5
VSG = + 1.75 = 3.04 V ⇒ VG = −3.04
3
⎛ R2 ⎞
VG = ⎜ ⎟ (10 ) − 5 = −3.04
⎝ R1 + R2 ⎠
Rin = R1 || R2 = 80 kΩ
1
⋅ ( 80 )(10 ) = 5 − 3.04 ⇒ R1 = 408 kΩ
R1
408 R2
= 80 ⇒ R2 = 99.5 kΩ
408 + R2
3.38
⎛ 60 ⎞
(a) K n1 = ⎜ ⎟ ( 4 ) = 120 μ A/V 2
⎝ 2 ⎠
⎛ 60 ⎞
K n 2 = ⎜ ⎟ (1) = 30 μ A/V 2
⎝ 2 ⎠
For vI = 1 V , M1 Sat. region, M2 Non-sat region.
I D 2 = I D1
30 ⎡ 2 ( −VTNL )( 5 − vO ) − ( 5 − vO ) ⎤ = 120 (1 − 0.8 )
2 2
⎣ ⎦
We find vO − 6.4vO + 7.16 = 0 ⇒ vO = 4.955 V
2
(b) For vI = 3 V , M1 Non-sat region, M2 Sat. region. I D 2 = I D1
30 ⎡ − ( −1.8 ) ⎤ = 120 ⎡ 2 ( 3 − 0.8 ) vO − vO ⎤
2 2
⎣ ⎦ ⎣ ⎦
We find 4vO − 17.6vO + 3.24 = 0 ⇒ vO = 0.193 V
2
(c) For vI = 5 V , biasing same as (b)
30 ⎡ − ( −1.8 ) ⎤ = 120 ⎡ 2 ( 5 − 0.8 ) vO − vO ⎤
2 2
⎣ ⎦ ⎣ ⎦

14. We find 4vO − 33.6vO + 3.24 = 0 ⇒ vO = 0.0976 V
2
3.39
For vI = 5 V , M1 Non-sat region, M2 Sat. region.
I D1 = I D 2
′
⎛ kn ⎞ ⎛ W ⎞ ′
⎛ kn ⎞ ⎛ W ⎞
⎜ 2 ⎟ ⎜ L ⎟ ⎣ 2 (VGS1 − VTN 1 ) VDS 1 − VDS 1 ⎦ = ⎜ 2 ⎟ ⎜ L ⎟ (VGS 2 − VTN 2 )
⎡ ⎤
2 2
⎝ ⎠ ⎝ ⎠1 ⎝ ⎠ ⎝ ⎠2
⎛ W⎞ ⎡
⎜ ⎟ ⎣ 2 ( 5 − 0.8 )( 0.15 ) − ( 0.15 ) ⎤ = (1) ⎡0 − ( −2 ) ⎤
2 2
⎦ ⎣ ⎦
⎝ L ⎠1
⎛W ⎞
which yields ⎜ ⎟ = 3.23
⎝ L ⎠1
3.40
a. M1 and M2 in saturation
K n1 (VGS 1 − VTN 1 ) = K n 2 (VGS 2 − VTN 2 )
2 2
K n1 = K n 2 , VTN 1 = VTN 2 ⇒ VGS1 = VGS 2 = 2.5 V, V0 = 2.5 V
I D = (15 )( 40 )( 2.5 − 0.8 ) ⇒ I D = 1.73 mA
2
b.
⎛W ⎞ ⎛W ⎞
⎜ ⎟ > ⎜ ⎟ ⇒ VGS1 < VGS 2
⎝ L ⎠1 ⎝ L ⎠ 2
40 (VGS1 − 0.8 ) = (15 )(VGS 2 − 0.8 )
2 2
VGS 2 = 5 − VGS 1
1.633 (VGS 1 − 0.8 ) = ( 5 − VGS1 − 0.8 )
2.633VGS 1 = 5.506 ⇒ VGS 1 = 2.09 V
VGS 2 = 2.91 V, V0 = VGS1 = 2.91 V
I D = (15 )(15 )( 2.91 − 0.8 ) ⇒ I D = 1.0 mA
2
3.41
(a)
V1 = VGS 3 = 2.5 V
⎛ W ⎞ ⎛ 0.06 ⎞
⎟ ( 2.5 − 1.2 )
2
I D = 0.5 = ⎜ ⎟ ⎜
⎝ L ⎠3 ⎝ 2 ⎠
⎛W ⎞
⎜ ⎟ = 9.86
⎝ L ⎠3
V2 = 6 V ⇒ VGS 2 = V2 − V1 = 6 − 2.5 = 3.5 V
⎛ W ⎞ ⎛ 0.06 ⎞ ⎛W ⎞
⎟ ( 3.5 − 1.2 ) ⇒ ⎜ ⎟ = 3.15
2
0.5 = ⎜ ⎟ ⎜
⎝ L ⎠2 ⎝ 2 ⎠ ⎝ L ⎠2
VGS1 = 10 − V2 = 10 − 6 = 4 V
⎛ W ⎞ ⎛ 0.06 ⎞ ⎛W ⎞
⎟ ( 4 − 1.2 ) ⇒ ⎜ ⎟ = 2.13
2
0.5 = ⎜ ⎟ ⎜
⎝ L ⎠1 ⎝ 2 ⎠ ⎝ L ⎠1
(b)
′
kn1 = 0.06 + 5% = 0.063 mA/V 2
′ ′
kn 2 = k n3 = 0.6 − 5% = 0.057 mA/V 2
⎛ 0.057 ⎞
For M3: I D = ( 9.86 ) ⎜ ⎟ (V1 − 1.2 )
2
⎝ 2 ⎠
⎛ 0.057 ⎞
For M2: I D = ( 3.15 ) ⎜ ⎟ (V2 − V1 − 1.2 )
2
⎝ 2 ⎠

15. ⎛ 0.063 ⎞
For M1: I D = ( 2.13) ⎜ ⎟ (10 − V2 − 1.2 )
2
⎝ 2 ⎠
0.281(V1 − 1.2 ) = 0.0898 (V2 − V1 − 1.2 ) = 0.0671( 8.8 − V2 )
2 2 2
Take square root.
0.530 (V1 − 1.2 ) = 0.300 (V2 − V1 − 1.2 ) = 0.259 ( 8.8 − V2 )
(1) 0.830V1 = 0.300V2 + 0.276 (2) 0.559V2 = 0.300V1 + 2.64
From (2) ⇒ V2 = 0.537V1 + 4.72
Substitute into (1)
0.830V1 = 0.300 [ 0.537V1 + 4.72] + 0.276 = 0.161V1 + 1.69
V1 = 2.53 V
Then
V2 = 0.537 ( 2.53) + 4.72
V2 = 6.08 V
3.42
ML in saturation
MD in nonsaturation
⎛W ⎞ ⎛W ⎞
⎜ ⎟ (VGSL − VTNL ) = ⎜ ⎟ ⎣ 2 (VGSD − VTND )VDSD − VDSD ⎦
⎡ ⎤
2 2
⎝ L ⎠L ⎝ L ⎠D
⎛W ⎞
(1)( 5 − 0.1 − 0.8) = ⎜ ⎟ ⎡ 2 ( 5 − 0.8)( 0.1) − ( 0.1) ⎤
2 2
⎝ L ⎠D ⎣ ⎦
⎛W ⎞
16.81 = ⎜ ⎟ [ 0.83]
⎝ L ⎠D
⎛W ⎞
⎜ ⎟ = 20.3
⎝ L ⎠D
3.43
ML in saturation
MD in nonsaturation
⎛W ⎞ ⎛W ⎞
⎜ ⎟ (VGSL − VTNL ) = ⎜ ⎟ ⎣ 2 (VGSD − VTND )VDSD − VDSD ⎦
⎡ ⎤
2 2
⎝ L ⎠L ⎝ L ⎠D
(1)(1.8 ) = ⎛ ⎞ ⎡ 2 ( 5 − 0.8 )( 0.05) − ( 0.05) ⎤
2 W 2
⎜ ⎟ ⎣ ⎦
⎝ L ⎠D
⎛W ⎞
3.24 = ⎜ ⎟ [ 0.4175]
⎝ L ⎠D
⎛W ⎞
⎜ ⎟ = 7.76
⎝ L ⎠D
3.44
VDD − V0 5 − 0.1
ID = = = 0.49 mA
RD 10
Transistor biased in nonsaturation
I D = 0.49
⎛W ⎞
= ( 0.015 ) ⎜ ⎟ ⎡ 2 ( 4.2 − 0.8 )( 0.1) − ( 0.1) ⎤
2
⎝L⎠ ⎣ ⎦
⎛W ⎞ W
0.49 = ⎜ ⎟ 0.01005 ⇒ = 48.8
⎝L⎠ L
3.45

16. 5 = I D RD + Vγ + VDS
5 = (12 ) RD + 1.6 + 0.2 ⇒ RD = 267 Ω
⎛ k′ ⎞⎛ W ⎞
I D = ⎜ n ⎟ ⎜ ⎟ (VGS − VTN )
2
⎝ 2 ⎠⎝ L ⎠
⎛ 0.040 ⎞ ⎛ W ⎞ W
⎟ ⎜ ⎟ ( 5 − 0.8 ) ⇒
2
12 = ⎜ = 34
⎝ 2 ⎠⎝ L ⎠ L
3.46
5 = VSD + I D RD + Vγ
5 = 0.15 + (15 ) RD + 1.6 ⇒ RD = 217 Ω
⎛ k′ ⎞⎛ W ⎞
I D = ⎜ ⎟ ⎜ ⎟ (VSG + VTP )
p 2
⎝ 2 ⎠⎝ L ⎠
⎛ 0.020 ⎞⎛ W ⎞ W
⎟⎜ ⎟ ( 5 − 0.8 ) ⇒
2
15 = ⎜ = 85
⎝ 2 ⎠⎝ L ⎠ L
3.47
(a)
VDD − VO ⎛W ⎞⎛ 0.060 ⎞
= 2⎜ ⎟ ⎡( 2 )(VGS − VTN ) VO − VO ⎤
2
⎟⎜ ⎣ ⎦
RD ⎝L ⎠⎝ 2 ⎠
5 − 0.2 ⎛W ⎞
⎟ ( 0.030 ) ⎡ 2 ( 5 − 0.8 )( 0.2 ) − ( 0.2 ) ⎤
2
= 2⎜
20 ⎝L ⎠ ⎣ ⎦
⎛W ⎞ ⎛W ⎞ ⎛W ⎞
0.24 = 0.0984 ⎜ ⎟ ⇒ ⎜ ⎟ = ⎜ ⎟ = 2.44
⎝ L ⎠ ⎝ L ⎠1 ⎝ L ⎠ 2
(b)
5 − VO ⎛ 0.06 ⎞
= ( 2.44 ) ⎜ ⎟ ⎡ 2 ( 5 − 0.8 ) VO − VO ⎤
2
20 ⎝ 2 ⎠⎣ ⎦
5 − VO = 12.30VO − 1.464VO2
1.464VO2 − 13.30VO + 5 = 0
13.30 ± 176.89 − 29.28
VO =
2 (1.464 )
VO = 0.393 V
3.48
⎛W ′
⎞ ⎛ kn ⎞
⎟ ⎜ ⎟ (VGS 1 − VTN )
2
I Q1 = ⎜
⎝L ⎠1 ⎝ 2⎠
⎛W ′
⎞ ⎛ kn ⎞
⎟ ⎜ ⎟ (VDS 2 ( sat ) )
2
I Q1 = ⎜
⎝L ⎠2 ⎝ 2 ⎠
⎛ W ⎞ ⎛ 0.08 ⎞ ⎛W ⎞ ⎛W ⎞
⎟ ( 0.5 ) ⇒ ⎜ ⎟ = 10 = ⎜ ⎟
2
0.1 = ⎜ ⎟ ⎜
⎝ L ⎠2 ⎝ 2 ⎠ ⎝ L ⎠2 ⎝ L ⎠1
⎛ W ⎞ ⎛ 200 ⎞ ⎛ W ⎞
⎜ ⎟ =⎜ ⎟ ⎜ ⎟ = 20
⎝ L ⎠3 ⎝ 100 ⎠ ⎝ L ⎠ 2
M1 & M2 matched.
⎛ 0.08 ⎞
Then 0.1 = (10 ) ⎜ ⎟ (VGS 1 − 0.25 )
2
⎝ 2 ⎠
VGS1 = 0.75 V
VD1 = −0.75 + 2 = 1.25 V
2.5 − 1.25
RD = ⇒ RD = 12.5 K
0.1

17. 3.49
(a)
⎛ W ⎞ ⎛ k′ ⎞
I Q 2 = ⎜ ⎟ ⎜ ⎟ (VSDB ( sat ) )
p 2
⎝ L ⎠B ⎝ 2 ⎠
⎛ W ⎞ ⎛ 0.04 ⎞ ⎛W ⎞ ⎛W ⎞
⎟ ( 0.8 ) ⇒
2
0.25 = ⎜ ⎟ ⎜ ⎜ ⎟ = 19.5 = ⎜ ⎟
⎝ L ⎠B ⎝ 2 ⎠ ⎝ L ⎠B ⎝ L ⎠A
⎛W ⎞ I KQ 2 ⎛ W ⎞
⎜ ⎟ = ⎜ ⎟
⎝ L ⎠C IQ 2 ⎝ L ⎠B
⎛ 100 ⎞
=⎜ ⎟ (19.5 ) = 7.81
⎝ 250 ⎠
′
⎛W ⎞ ⎛ kp ⎞
I Q 2 = ⎜ ⎟ ⎜ ⎟ (VSGA + VTP )
2
⎝ L ⎠A ⎝ 2 ⎠
⎛ 0.04 ⎞
0.25 = (19.5 ) ⎜ ⎟ (VSGA − 0.5 )
2
⎝ 2 ⎠
VSGA = 1.30 V
(b)
VDA = 1.3 − 4 = −2.7 V
−2.7 − ( −5 )
RD = ⇒ RD = 9.2 K
0.25
3.50
⎛W ′
⎞ ⎛ kn ⎞
⎟ ⎜ ⎟ (VDS 2 ( sat ) )
2
IQ = ⎜
⎝L ⎠2 ⎝ 2⎠
⎛W ⎞ ⎛ 0.06 ⎞ ⎛W ⎞ ⎛W ⎞
⎟ ( 0.5 ) ⇒ ⎜ ⎟ = 53.3 = ⎜ ⎟
2
0.4 = ⎜ ⎟ ⎜
⎝L ⎠2 ⎝ 2 ⎠ ⎝ L ⎠2 ⎝ L ⎠1
⎛ W ⎞ ⎛ k′ ⎞
I Q = ⎜ ⎟ ⎜ n ⎟ (VGS 1 − VTN )
2
⎝ L ⎠1 ⎝ 2 ⎠
⎛ 0.06 ⎞
0.4 = ( 53.3) ⎜ ⎟ (VGS 1 − 0.75 )
2
⎝ 2 ⎠
VGS1 = 1.25 V VD1 = −1.25 + 4 = 2.75 V
5 − 2.75
RD = ⇒ RD = 5.625 K
0.4
3.51
VDS ( sat ) = VGS − VP
So VDS > VDS ( sat ) = −VP , I D = I DSS
3.52
VDS ( sat ) = VGS − VP = VGS + 3 = VDS ( sat )
a. VGS = 0 ⇒ I D = I DSS = 6 mA
2
⎛ V ⎞
2
⎛ −1 ⎞
b. I D = I DSS ⎜ 1 − GS ⎟ = 6 ⎜ 1 − ⎟ ⇒ I D = 2.67 mA
⎝ VP ⎠ ⎝ −3 ⎠
2
⎛ −2 ⎞
c. I D = 6 ⎜ 1 − ⎟ ⇒ I D = 0.667 mA
⎝ −3 ⎠
d. ID = 0

18. 3.53
2
⎛ V ⎞
I D = I DSS ⎜ 1 − GS ⎟
⎝ VP ⎠
2
⎛ 1 ⎞
2.8 = I DSS ⎜1 − ⎟
⎝ VP ⎠
2
⎛ 3 ⎞
0.30 = I DSS ⎜1 − ⎟
⎝ VP ⎠
2
⎛ 1 ⎞
⎜1 − ⎟
2.8 ⎝ VP ⎠ = 9.33
= 2
0.30 ⎛ 3 ⎞
⎜1 − ⎟
⎝ VP ⎠
⎛ 1⎞
⎜1 − ⎟
⎝ VP⎠ = 3.055
⎛ 3⎞
⎜1 − ⎟
⎝ VP ⎠
1 9.165
1− = 3.055 −
VP VP
8.165
= 2.055 ⇒ VP = 3.97 V
VP
2
⎛ 1 ⎞
2.8 = I DSS ⎜1 − ⎟ = I DSS ( 0.560 ) ⇒ I DSS = 5.0 mA
⎝ 3.97 ⎠
3.54
VS = −VGS , VSD = VS − VDD
Want VSD ≥ VSD ( sat ) = VP − VGS
VS − VDD ≥ VP − VGS − VGS − VDD ≥ VP − VGS ⇒ VDD ≤ −VP
So VDD ≤ −2.5 V
2
⎛ V ⎞
I D = 2 = I DSS ⎜1 − GS ⎟
⎝ VP ⎠
2
⎛ V ⎞
2 = 6 ⎜ 1 − GS ⎟ ⇒ VGS = 1.06 V ⇒ VS = −1.06 V
⎝ 2.5 ⎠
3.55
I D = K n (VGS − VTN )
2
18.5 = K n ( 0.35 − VTN )
2
86.2 = K n ( 0.5 − VTN )
2
Then
( 0.35 − VTN )
2
18.5
= 0.2146 = ⇒ VTN = 0.221 V
( 0.50 − VTN )
2
86.2
18.5 = K n ( 0.35 − 0.221) ⇒ K n = 1.11 mA / V 2
2
3.56
I D = K (VGS − VTN )
2
250 = K ( 0.75 − 0.24 ) ⇒ K = 0.961 mA / V 2
2

19. 3.57
2
⎛ V ⎞ V V
I D = I DSS ⎜ 1 − GS ⎟ = S = − GS
⎝ VP ⎠ RS RS
2
⎛ V ⎞ V
10 ⎜ 1 − GS ⎟ = − GS
⎝ −5 ⎠ 0.2
⎛ 2V V ⎞
2
2 ⎜1 + GS + GS ⎟ = −VGS
⎝ 5 25 ⎠
2 2 9
VGS + VGS + 2 = 0
25 5
2VGS + 45VGS + 50 = 0
2
( 45 ) − 4 ( 2 )( 50 )
2
−45 ±
VGS = ⇒ VGS = −1.17 V
2 ( 2)
VGS 1.17
ID = − = ⇒ I D = 5.85 mA
RS 0.2
VD = 20 − ( 5.85 )( 2 ) = 8.3 V
VDS = VD − VS = 8.3 − 1.17 ⇒ VDS = 7.13 V
3.58
VDS = VDD − VS
8 = 10 − VS ⇒ VS = 2 V = I D RS = ( 5 ) RS ⇒ RS = 0.4 kΩ
2
⎛ V ⎞
I D = I DSS ⎜ 1 − GS ⎟
⎝ VP ⎠
2
⎛ −1 ⎞
5 = I DSS ⎜1 − ⎟ Let I DSS = 10 mA
⎝ VP ⎠
2
⎛ −1 ⎞
5 = 10 ⎜ 1 − ⎟ ⇒ VP = −3.41 V
⎝ VP ⎠
VG = VGS + VS = −1 + 2 = 1 V
⎛ R2 ⎞ 1
VG = ⎜ ⎟ VDD = ⋅ Rin ⋅ VDD
⎝ R1 + R2 ⎠ R1
1
1 = ( 500 )(10 ) ⇒ R1 = 5 MΩ
R1
5R2
= 0.5 ⇒ R2 = 0.556 MΩ
5 + R2
3.59
2
⎛ V ⎞
I D = I DSS ⎜ 1 − GS ⎟
⎝ VP ⎠
2
⎛ V ⎞
5 = 8 ⎜ 1 − GS ⎟ ⇒ VGS = 0.838 V
⎝ 4 ⎠
VSD = VDD − I D ( RS + RD )
= 20 − ( 5 )( 0.5 + 2 ) ⇒ VSD = 7.5 V

20. VS = 20 − ( 5 )( 0.5 ) = 17.5 V
VG = VS + VGS = 17.5 + 0.838 = 18.3 V
⎛ R2 ⎞ 1
VG = ⎜ ⎟ VDD = ⋅ Rin ⋅ VDD
⎝ R1 + R2 ⎠ R1
1
18.3 = (100 ) ( 20 ) ⇒ R1 = 109 kΩ
R1
109 R2
= 100 ⇒ R2 = 1.21 MΩ
109 + R2
3.60
2
⎛ V ⎞
I D = I DSS ⎜ 1 − GS ⎟
⎝ VP ⎠
2
⎛ V ⎞
5 = 7 ⎜ 1 − GS ⎟ ⇒ VGS = 0.465 V
⎝ 3 ⎠
VSD = VDD − I D ( RS + RD )
6 = 12 − ( 5 )( 0.3 + RD ) ⇒ RD = 0.9 kΩ
VS = 12 − ( 5 )( 0.3) = 10.5 V
VG = VS + VGS = 10.5 + 0.465 = 10.965 V
⎛ R2 ⎞
VG = ⎜ ⎟ VDD
⎝ R1 + R2 ⎠
⎛ R ⎞
10.965 = ⎜ 2 ⎟ (12 ) ⇒ R2 = 91.4 kΩ ⇒ R1 = 8.6 kΩ
⎝ 100 ⎠
3.61
⎛ R2 ⎞ ⎛ 60 ⎞
VG = ⎜ ⎟ VDD = ⎜ ⎟ ( 20 ) ⇒ VG = 6 V
⎝ R1 + R2 ⎠ ⎝ 140 + 60 ⎠
2
⎛ V ⎞ V V − VGS
I D = I DSS ⎜ 1 − GS ⎟ = S = G
⎝ VP ⎠ RS RS
2
⎛ VGS ⎞
(8 )( 2 ) ⎜1 −
⎜ ⎟ = 6 − VGS
⎝ ( −4 ) ⎟
⎠
⎛ V V2 ⎞
16 ⎜ 1 + GS + GS ⎟ = 6 − VGS
⎝ 2 16 ⎠
VGS + 9VGS + 10 = 0
2
(9) − 4 (10 )
2
−9 ±
VGS = ⇒ VGS = −1.30
2
⎛ ( −1.30 ) ⎞
2
I D = 8 ⎜1 −
⎜ ⎟ ⇒ I D = 3.65 mA
⎝ ( −4 ) ⎟⎠
VDS = VDD − I D ( RS + RD )
= 20 − ( 3.65 )( 2 + 2.7 )
VDS = 2.85 V
VDS > VDS ( sat ) = VGS − VP
= −1.30 − ( −4 )
= 2.7 V (Yes)
3.62

21. VDS = VDD − I D ( RS + RD )
5 = 12 − I D ( 0.5 + 1) ⇒ I D = 4.67 mA
VS = I D RS = ( 4.67 ) ( 0.5 ) ⇒ VS = 2.33 V
⎛ R2 ⎞ ⎛ 20 ⎞
VG = ⎜ ⎟ VDD = ⎜ ⎟ (12 ) ⇒ VG = 0.511 V
⎝ R1 + R2 ⎠ ⎝ 450 + 20 ⎠
VGS = VG − VS = 0.511 − 2.33 ⇒ VGS = −1.82 V
2
⎛ V ⎞
I D = I DSS ⎜ 1 − GS ⎟
⎝ VP ⎠
⎛ ( −1.82 ) ⎞
2
4.67 = 10 ⎜ 1 −
⎜ ⎟ ⇒ VP = −5.75 V
⎝ VP ⎟ ⎠
3.63
2
⎛ V ⎞
I D = I DSS ⎜ 1 − GS ⎟ , VGS = 0
⎝ VP ⎠
I D = I DSS = 4 mA
VDD − VDS 10 − 3
RD = = ⇒ RD = 1.75 kΩ
ID 4
3.64
VSD = VDD − I D RS
10 = 20 − (1) RS ⇒ RS = 10 kΩ
VDD 20
R1 + R2 = = = 200 kΩ
I 0.1
2
⎛ V ⎞
I D = I DSS ⎜ 1 − GS ⎟
⎝ VP ⎠
2
⎛ V ⎞
1 = 2 ⎜1 − GS ⎟ ⇒ VGS = 0.586 V
⎝ 2 ⎠
VG = VS + VGS = 10 + 0.586 = 10.586
⎛ R2 ⎞
VG = ⎜ ⎟ VDD
⎝ R1 + R2 ⎠
⎛ R ⎞
10.586 = ⎜ 2 ⎟ ( 20 ) ⇒ R2 = 106 kΩ
⎝ 200 ⎠
R1 = 94 kΩ
3.65

22. VDS = VDD − I D ( RS + RD )
2 = 3 − ( 0.040 )(10 + RD ) ⇒ RD = 15 kΩ
I D = K (VGS − VTN )
2
40 = 250 (VGS − 0.20 ) ⇒ VGS = 0.60 V
2
VG = VGS + VS = 0.60 + ( 0.040 )(10 ) = 1.0 V
⎛ R2 ⎞
VG = ⎜ ⎟ VDD
⎝ R1 + R2 ⎠
⎛ R ⎞
1 = ⎜ 2 ⎟ ( 3) ⇒ R2 = 50 kΩ
⎝ 150 ⎠
R1 = 100 kΩ
3.66
For VO = 0.70 V ⇒ VDS = 0.70 > VDS ( sat ) = VGS − VTN
0.75 − 0.15 = 0.6
Biased in the saturation region
V − VDS 3 − 0.7
I D = DD = ⇒ I D = 46 μ A
RD 50
I D = K (VGS − VTN ) ⇒ 46 = K ( 0.75 − 0.15 ) ⇒ K = 128 μ A / V 2
2 2