![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
⎣ ⎦](https://image.slidesharecdn.com/ch03s-120608121824-phpapp02/85/Ch03s-1-320.jpg)
![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](https://image.slidesharecdn.com/ch03s-120608121824-phpapp02/85/Ch03s-5-320.jpg)
![(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 )](https://image.slidesharecdn.com/ch03s-120608121824-phpapp02/85/Ch03s-8-320.jpg)
![⎛ 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](https://image.slidesharecdn.com/ch03s-120608121824-phpapp02/85/Ch03s-15-320.jpg)
Ch03s
- 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