1. Chapter 3
Exercise Solutions
EX3.1
VTN = 1 V , VGS = 3 V , VDS = 4.5 V
VDS = 4.5 > VDS ( sat ) = VGS − VTN = 3 − 1 = 2 V
Transistor biased in the saturation region
I D = K n (VGS − VTN ) ⇒ 0.8 = K n ( 3 − 1) ⇒ K n = 0.2 mA / V 2
2 2
(a) VGS = 2 V, VDS = 4.5 V
Saturation region:
I D = ( 0.2 )( 2 − 1) ⇒ I D = 0.2 mA
2
(b) VGS = 3 V, VDS = 1 V
Nonsaturation region:
I D = ( 0.2 ) ⎡ 2 ( 3 − 1)(1) − (1) ⎤ ⇒ I D = 0.6 mA
2
⎣ ⎦
EX3.2
VTP = −2 V , VSG = 3 V
VSD ( sat ) = VSG + VTP = 3 − 2 = 1 V
(a) VSD = 0.5 V ⇒ Nonsaturation
(b) VSD = 2 V ⇒ Saturation
(c) VSD = 5 V ⇒ Saturation
EX3.3
⎛ R2 ⎞ ⎛ 160 ⎞
VG = ⎜ ⎟ (VDD ) = ⎜ ⎟ (10 ) = 3.636 V = VGS
⎝ R1 + R2 ⎠ ⎝ 160 + 280 ⎠
I D = 0.25 ( 3.636 − 2 ) = 0.669 mA
2
VDS = 10 − ( 0.669 )(10 ) = 3.31 V
P = I DVDS = ( 0.669 )( 3.31) = 2.21 mW
EX3.4
I DQ = K P (VSG + VTP )
2
1.2 = 0.4 (VSG − 1.2 ) ⇒ VSG = 2.932 V
2
⎛ R1 ⎞ 1
VSG = ⎜ ⎟ VDD = ⋅ VTTN − VDD
⎝ R1 + R2 ⎠ R2
Note K = kΩ
1
2.932 = ( 200 )(10 ) ⇒ R2 = 682 K
R2
682 R1
= 200 ⇒ R1 = 283 K
682 + R1
10 − 4
RD = =5K
1.2
EX3.5
⎛ R2 ⎞ ⎛ 40 ⎞
(a) VG = ⎜ ⎟ (10 ) − 5 = ⎜ ⎟ (10 ) − 5 = −1 V
⎝ R1 + R2 ⎠ ⎝ 40 + 60 ⎠
V − ( −5 )
= K n (VGS − VTN )
2
ID = S
RS
VS = VG − VGS

2. ( 5 − 1) − VGS = ( 0.5 )(1) (VGS − 2VGS + 1)
2
0.5VGS − 3.5 = 0 VGS = 7 VGS = 2.646 V
2 2
I D = ( 0.5 )( 2.646 − 1) ⇒ I D = 1.354 mA
2
VDS = 10 − (1.354 )( 3) = 5.937 V
4 − VGS = K n (1)(VGS − VTN )
2
(b)
(1) K n = (1.05 )( 0.5 ) = 0.525
(2) K n = ( 0.95)( 0.5 ) = 0.475
(3) VTN = (1.05 )(1) = 1.05 V
(4) VTN = ( 0.95 )(1) = 0.95 V
(1)-(3) 4 − VGS = 0.525 (VGS − 2.1VGS + 1.1025 )
2
0.525VGS − 0.1025VGS − 3.421 = 0
2
0.1025 ± 0.010506 + 7.1841
VGS = = 2.652 V
2 ( 0.525 )
I D = 0.525 ( 2.652 − 1.05 ) = 1.348 mA
2
VDS = 10 − (1.348 )( 3) = 5.957 V
(2)-(4)
4 − VGS = 0.475 (VGS − 1.9VGS + 0.9025 )
2
0.475VGS + 0.0975VGS − 3.5713 = 0
2
−0.0975 ± 0.00950625 + 6.78547
VGS =
2 ( 0.475 )
VGS = 2.641 V
I D = 0.475 ( 2.641 − 0.95 ) = 1.359 mA
2
VDS = 10 − (1.359 )( 3) = 5.924 V
(1)-(4)
4 −VGS = ( 0.525) (VGS −1.9VGS + 0.9025)
2
0.525 VGS + 0.0025VGS − 3.5262 = 0
2
−0.0025 ± 0.00000625 + 7.40502
VGS =
2 ( 0.525)
= 2.5893 V
I D = ( 0.525)( 2.5893 − 0.95) = 1.411
2
VDS = 10 − I D ( 3) = 5.7678 V
(2)-(3)
4 − VGS = 0.475 (VGS − 2.1VGS +1.1025 )
2
0.475VGS + 0.0025VGS − 3.4763 = 0
2
−0.0025 ± 0.00000625 + 6.60499
VGS =
2(0.475)
VGS = 2.7027
I D = (0.475)(2.7027 − 1.05) 2 = 1.2973 mA
VDS = 10 − I D (3) = 6.108 V
1.297 ≤ I DQ ≤ 1.411 mA
5.768 ≤ VDS ≤ 6.108 V
EX3.6

3. ⎛ R2 ⎞
VG = ⎜ ⎟ (10 ) − 5
⎝ R1 + R2 ⎠
⎛ 200 ⎞
=⎜ ⎟ (10 ) − 5 = 0.714 V
⎝ 350 ⎠
VS = 5 − I D RS = 5 − (1.2 ) I D
So
VSG = VS − VG = 5 − (1.2 ) I D − 0.714
= 4.286 − (1.2 ) I D
4.286 − VSG
ID =
1.2
I D = K p (VSG + VTP )
2
(
4.286 − VSG = (1.2 )( 0.25 ) × VSG − 2VSG ( −1) + ( −1)
2 2
)
4.286 − VSG = ( 0.3) V − 0.6VSG + 0.3
2
SG
0.3VSG + 0.4VSG − 3.986 = 0
2
( 0.4 ) + 4 ( 0.3)( 3.986 )
2
−0.4 ±
VSG =
2 ( 0.3)
Must use + sign ⇒ VSG = 3.04 V
I D = ( 0.25 )( 3.04 − 1) ⇒ I D = 1.04 mA
2
VSD = 10 − I D ( RS + RD ) = 10 − (1.04 )(1.2 + 4 ) ⇒ VSD = 4.59 V
VSD > VSD ( sat ) , Yes
EX3.7
VSD = 10 − I DQ ( RS + RP )
VSD = 10 − K P (VSG + VTP ) ( RS + RP )
2
Set VSD = VSG + VTP
VSG + VTP = 10 − ( 0.25 )(VSG + VTP ) ( 5.2 )
2
1.3 (VSG + VTP ) + (VSG + VTP ) − 10 = 0
2
−1 ± 1 + 4 (1.3)(10 )
(VSG + VTP ) =
2 (1.3)
= 2.415 V
VSG = 3.415 V ( 3.42 V )
VSD = 2.415 V ( 2.42 V )
I D = ( 0.25 )( 2.415 ) = 1.46 mA
2
EX3.8
⎛ R2 ⎞ ⎛ 240 ⎞
VG = ⎜ ⎟ (10 ) − 5 = ⎜ ⎟ (10 ) − 5
⎝ R1 + R2 ⎠ ⎝ 240 + 270 ⎠
VG = −0.294 V
VS − ( −5 ) VG − VGS + 5
= K n (VGS − VTN )
2
ID = =
RS RS
⎛ 0.08 ⎞
4.706 − VGS = ⎜ ⎟ ( 4 )( 3.9 ) (VGS − 2.4VGS + 1.44 )
2
⎝ 2 ⎠
0.624VGS − 0.4976VGS − 3.80744 = 0
2

4. 0.4976 ± 0.2476 + 9.50337
VGS =
2 ( 0.624 )
VGS = 2.90 V
⎛ 0.08 ⎞
⎟ ( 4 )( 2.90 − 1.2 ) ⇒ I D = 0.463 mA
2
ID = ⎜
⎝ 2 ⎠
VDS = 10 − I D ( 3.9 + 10 ) ⇒ VDS = 3.57 V
EX3.9
10 − VSG
and I D = K p (VSG + VTP )
2
ID =
RS
0.12 = ( 0.050 )(VSG − 0.8 )
2
VSG = 2.35 V
10 − 2.35
RS = ⇒ RS = 63.75 kΩ
0.12
VSD = 8 = 20 − I D ( RS + RD )
8 = 20 − ( 0.12 )( 63.75 ) − ( 0.12 ) RD
20 − ( 0.12 )( 63.75 ) − 8
RD = ⇒ RD = 36.25 kΩ
0.12
(1) KP = ( 0.05 )(1.05 ) = 0.0525
(2) KP = ( 0.05 )( 0.95 ) = 0.0475
(3) VTP = −0.8 (1.05 ) = −0.84 V
(4) VTP = −0.8 ( 0.95 ) = −0.76 V
10 − VSG
= K P (VSG + VTP )
2
ID =
RS
(1)-(3)
10 − VSG = ( 63.75 )( 0.0525 ) ⎡VSG − 1.68VSG + 0.7056 ⎤
⎣
2
⎦
3.347VSG − 4.623VSG − 7.6384 = 0
2
4.623 ± 21.372 + 102.263
VSG =
2 ( 3.347 )
VSG = 2.352 V ⇒ I D ≅ 0.120 mA
VSD ≈ 8.0 V
(2)-(4)
10 − VSG = ( 63.75 )( 0.0475 ) ⎡VSG − 1.52VSG + 0.5776 ⎤
⎣
2
⎦
3.028VSG − 3.603VSG − 8.251 = 0
2
3.603 ± 12.9816 + 99.936
VSG =
2 ( 3.028 )
VSG = 2.35 V
I D ≈ 0.120
VSD ≈ 8.0
(1)-(4)
10 − VSG = ( 63.75 )( 0.0525 ) ⎡VSG − 1.52VSG + 0.5776 ⎤
⎣
2
⎦
3.347VSG − 4.087VSG − 8.06685 = 0
2

5. 4.087 ± 16.7036 + 107.999
VSG =
2 ( 3.347 )
VSG = 2.279 V
I D = 0.121 mA
VSD = 7.89 V
(2)-(3)
10 − VSG = ( 63.75 )( 0.0475 ) ⎡VSG − 1.68VSG + 0.7056 ⎤
⎣
2
⎦
3.028VSG − 4.0873VSG − 7.8634 = 0
2
4.0873 ± 16.706 + 95.242
VSG =
2 ( 3.028 )
VSG = 2.422 V
I D = 0.119 mA
VSD = 8.11 V
Summary 0.119 ≤ I D ≤ 0.121 mA
7.89 ≤ VSD ≤ 8.11 V
EX3.10
V − VGS
I D = K n (VGS − VTN )
2
I D = DD ,
RS
10 − VGS = (10 )( 0.2 ) (VGS − 2VGSVTN + VTN )
2 2
10 − VGS = 2VGS − 8VGS + 8
2
2VGS − 7VGS − 2 = 0
2
(7) + 4 ( 2) 2
2
7±
VGS =
2 ( 2)
Use + sign: VGS = VDS = 3.77 V
10 − 3.77
ID = ⇒ I D = 0.623 mA
10
Power = I DVDS = ( 0.623)( 3.77 ) ⇒ Power = 2.35 mW
EX3.11
(a) VI = 4 V, Driver in Non ⋅ Sat.
K nD ⎣ 2 (VI − VTND ) VO − VO2 ⎦ = K nL [VDD − VO − VTNL ]
⎡ ⎤
2
5 ⎡ 2 ( 4 − 1) VD − VD ⎤ = ( 5 − VD − 1) = ( 4 − VO ) = 16 − 8VO + VO2
2 2 2
⎣ ⎦
6VD − 38VO + 16 = 0
2
38 ± 1444 − 384
VD =
2 ( 6)
VD = 0.454 V
(b) VI = 2 V Driver: Sat
K nD [VI − VTND ] = K nL [VDD − VO − VTNL ]
2 2
5 [ 2 − 1] = [5 − VO − 1]
2 2
5 = 4 − VO ⇒ VO = 1.76 V
EX3.12
If the transistor is biased in the saturation region

6. I D = K n (VGS − VTN ) = K n ( −VTN )
2 2
I D = ( 0.25 )( 2.5 ) ⇒ I D = 1.56 mA
2
VDS = VDD − I D RS = 10 − (1.56 )( 4 ) ⇒ VDS = 3.75
VDS > VGS − VTN = −VTN
3.75 > − ( −2.5 )
Yes — biased in the saturation region
Power = I DVDS = (1.56 )( 3.75 ) ⇒ Power = 5.85 mW
EX3.13
(a) For VI = 5 V, Load in saturation and driver in nonsaturation.
I DD = I DL
K nD ⎡ 2 (VI − VTND ) VO − VO2 ⎤ = K nL ( −VTNL )
2
⎣ ⎦
K nD ⎡ K
2 5 − 1)( 0.25 ) − ( 0.25 ) ⎤ = 4 ⇒ nD = 2.06
⎣ (
2
K nL ⎦ K nL
(b)
I DL = K nL ( −VTNL ) ⇒ 0.2 = K nL ⎡ − ( −2 ) ⎤
2 2
⎣ ⎦
K nL = 50 μ A / V 2 and K nD = 103 μ A / V 2
EX3.14
For M N
I DN = I DP
K n (VGSN − VTN ) = K p (Vscop + VTP )
2 2
VGSN = 1 + ( 5 − 3.25 − 1) = 1.75 V = VI
Vo = VDSN ( sat ) = 1.75 − 1 ⇒ Vo = 0.75 V
For M P : VI = 1.75 V
VDD − VO = VSD ( sat ) = VSGP + VTP = ( 5 − 3.25 ) − 1 = 0.75 V
So Vot = 5 − 0.75 ⇒ Vot = 4.25 V
EX3.15
For RD = 10 k Ω, VDD = 5 V, and Vo = 1 V
5 −1
ID = = 0.4 mA
10
I D = K n ⎡ 2 (VGS − VTN ) VDS − VDS ⎤
⎣
2
⎦
I D = 0.4 = K n ⎡ 2 ( 5 − 1)(1) − (1) ⎤ ⇒ K n = 0.057 mA / V 2
2
⎣ ⎦
P = I D ⋅ VDS = ( 0.4 )(1) ⇒ P = 0.4 mW
EX3.16
a. V1 = 5 V, V2 = 0, M 2 cutoff ⇒ I D 2 = 0
5 − VO
I D = K n ⎣ 2 (VI − VTN ) VO − VO2 ⎦ =
⎡ ⎤
RD
( 0.05 )( 30 ) ⎡ 2 ( 5 − 1)V0 − V02 ⎤ = 5 − V0
⎣ ⎦

7. 1.5V02 − 13V0 + 5 = 0
(13) − 4 (1.5 )( 5)
2
13 ±
V0 = ⇒ V0 = 0.40 V
2 (1.5 )
5 − 0.40
I R = I D1 = ⇒ I R = I D1 = 0.153 mA
30
b. V1 = V2 = 5 V
5 − VO
RD
{
= 2 K n ⎡ 2 (VI − VTN ) VO − VO2 ⎤
⎣ ⎦ }
5 − V0 = 2 ( 0.05 )( 30 ) ⎡ 2 ( 5 − 1)V0 − V02 ⎤
⎣ ⎦
3V02 − 25V0 + 5 = 0
( 25) − 4 ( 3)( 5 )
2
25 ±
V0 = ⇒ V0 = 0.205 V
2 ( 3)
5 − 0.205
IR = ⇒ I R = 0.160 mA
30
I D1 = I D 2 = 0.080 mA
EX3.17
M 2 & M 3 watched ⇒ I Q1 = I REF 1 = 0.4 mA
0.4 = 0.3 (VGS 3 − 1) ⇒ VGS 3 = VGS 2 = 2.15 V
2
0.4 = 0.6 (VGS 1 − 1) ⇒ VGS1 = 1.82 V
2
EX3.18
⎛ 0.04 ⎞
⎟ (15 )(VSGC − 0.6 )
2
0.1 = ⎜
⎝ 2 ⎠
VSGC = 1.177 V = VSGB
⎛ 0.04 ⎞ ⎛ W ⎞
⎟ ⎜ ⎟ (1.177 − 0.6 )
2
0.2 = ⎜
⎝ 2 ⎠ ⎝ L ⎠B
⎛W ⎞
⎜ ⎟ = 30
⎝ L ⎠B
⎛ 0.04 ⎞
⎟ ( 25 )(VSGA − 0.6 )
2
0.2 = ⎜
⎝ 2 ⎠
VSGA = 1.23 V
EX3.19
I REF = K n 3 (VGS 3 − VTN ) = K n 4 (VGS 4 − VTN )
2 2
(a)
VGS 3 = 2 V ⇒ VGS 4 = 3 V
K K 1
( 2 − 1) = n 4 ( 3 − 1) ⇒ n 4 =
2 2
K n3 K n3 4
I Q = K n 2 (VGS 2 − VTN )
2
(b)
But VGS 2 = VGS 3 = 2 V
0.1 = K n 2 ( 2 − 1) ⇒
2
K n 2 = 0.1 mA / V 2
0.2 = K n 3 ( 2 − 1) ⇒ K n 3 = 0.2 mA / V 2
2
(c)
0.2 = K n 4 ( 3 − 1) ⇒ K n 4 = 0.05 mA / V 2
2
EX3.20

8. 5
VS 2 = 5 − 5 = 0 RS 2 = = 16.7 K
0.3
I D 2 = K n 2 (VGS 2 − VTN 2 )
2
0.3 = 0.2 (VGS 2 − 1.2 ) ⇒ VGS 2 = 2.425 V ⇒ VG 2 = VGS 2 + VS = 2.425 V
2
5 − 2.425
RD1 = = 25.8 K
0.1
VS 1 = VG 2 − VDSQ1 = 2.425 − 5 = −2.575 V
−2.575 − ( −5 )
RS 1 = ⇒ RS 1 = 24.3 K
0.1
I D1 = K n1 (VGS 1 − VTN 1 )
2
0.1 = 0.5 (VGS1 − 1.2 ) ⇒ VGS 1 = 1.647 V
2
VG1 = VGS 1 + VS 1 = 1.647 + ( −2.575 ) ⇒ VG1 = −0.928 V
⎛ R2 ⎞ 1
VG1 = ⎜ ⎟ (10 ) − 5 = ⋅ RTN ⋅ (10 ) − 5
⎝ R1 + R2 ⎠ R1
1
−0.928 = ( 200 )(10 ) − 5 ⇒ R1 = 491 K
R1
491 R2
= 200 ⇒ R2 = 337 K
491 + R2
EX3.21
VS1 = I D RS − 5 = (0.25)(16) − 5 = −1 V
I DQ = K n (VGS1 − VTN ) 2 ⇒ 0.25 = 0.5(VGS 1 − 0.8) 2 ⇒ VGS 1 = 1.507 V
VG1 = VGS 1 + VS 1 = 1.507 − 1 = 0.507 V
⎛ R3 ⎞ R3
VG1 = ⎜ ⎟ (5) ⇒ 0.507 = (5) ⇒ R3 = 50.7 K
⎝ R1 + R2 + R3 ⎠ 500
VS 2 = VS 1 + VDS1 = −1 + 2.5 = 1.5 V
VG 2 = VS 2 + VGS = 1.5 + 1.507 = 3.007 V
⎛ R2 + R3 ⎞ ⎛ R2 + R3 ⎞
VG 2 = ⎜ ⎟ (5) ⇒ 3.007 = ⎜ ⎟ (5)
⎝ R1 + R2 + R3 ⎠ ⎝ 500 ⎠
R2 + R3 = 300.7
R2 = 300.7 − 50.7 ⇒ R2 = 250 K
R1 = 500 − 250 − 50.7 ⇒ R1 = 199.3 K
VD 2 = VS 2 + VDS 2 = 1.5 + 2.5 = 4 V
5−4
RD = ⇒ RD = 4 K
0.25
EX3.22
VDS ( sat ) = VGS − VP = −1.2 − ( −4.5 ) ⇒ VDS ( sat ) = 3.3 V
⎛ ( −1.2 ) ⎞
2 2
⎛ V ⎞
I D = I DSS ⎜ 1 − GS ⎟ = 12 ⎜ 1 − ⇒ I D = 6.45 mA
⎜ ( −4.5 ) ⎟
⎟
⎝ VP ⎠ ⎝ ⎠
EX3.23
Assume the transistor is biased in the saturation region.

9. 2
⎛ V ⎞
I D = I DSS ⎜ 1 − GS ⎟
⎝ VP ⎠
2
⎛ V ⎞
8 = 18 ⎜ 1 − GS ⎟ ⇒ VGS = −1.17 V ⇒ VS = −VGS = 1.17
⎜ ( −3.5 ) ⎟
⎝ ⎠
VD = 15 − ( 8 )( 0.8 ) = 8.6
VDS = 8.6 − (1.17 ) = 7.43 V
VDS = 7.43 > VGS − VP = −1.17 − ( −3.5 ) = 2.33
Yes, the transistor is biased in the saturation region.
EX3.24
I D = 2.5 mA
2
⎛ V ⎞
I D = I DSS ⎜ 1 − GS ⎟
⎝ VP ⎠
2
⎛ V ⎞
2.5 = 6 ⎜1 − GS ⎟ ⇒ VGS = −1.42 V
⎜ ( −4 ) ⎟
⎝ ⎠
VS = I D RS − 5 = ( 2.5 )( 0.25 ) − 5
VS = −4.375
VDS = 6 ⇒ VD = 6 − 4.375 = 1.625
5 − 1625
RD = ⇒ RD = 1.35 kΩ
2.5
( 20 )
2
= 2 ⇒ R1 + R2 = 200 kΩ
R1 + R2
VG = VGS + VS = −1.42 − 4.375 = −5.795
⎛ R2 ⎞
VG = ⎜ ⎟ ( 20 ) − 10
⎝ R1 + R2 ⎠
⎛ R ⎞
−5.795 = ⎜ 2 ⎟ ( 20 ) − 10 ⇒ R2 = 42.05 kΩ → 42 kΩ
⎝ 200 ⎠
R1 = 157.95 kΩ → 158 kΩ
EX3.25
0 − VS VGS
VS = −VGS . I D = =
RS RS
2
⎛ V ⎞
I D = I DSS ⎜ 1 − GS ⎟
⎝ VP ⎠
⎛ V V2 ⎞
2
VGS ⎛ V ⎞
= 6 ⎜ 1 − GS ⎟ = 6 ⎜ 1 − GS + GS ⎟
1 ⎝ 4 ⎠ ⎝ 2 16 ⎠
0.375VGS − 4VGS + 6 = 0
2
4 ± 16 − 4 ( 0.375 )( 6 )
VGS =
2 ( 0.375 )
VGS = 8.86 or VGS = 1.806 V
impossible
VGS
ID = = 1.806 mA
RS

10. VD = I D RD − 5 = (1.81)( 0.4 ) − 5 = −4.278
VSD = VS − V0 = −1.81 − ( −4.276 ) ⇒ VSD = 2.47 V
VSD ( sat ) = VP − VGS = 4 − 1.81 = 2.19
So VSD > VSD ( sat )
EX3.26
R1 R2
Rin = R1 R2 = = 100 kΩ
R1 + R2
I DQ = 5 mA, VS = − I DQ RS = − ( 5 )(1.2 ) = −6 V
VSDQ = 12 V ⇒ VD = VS − VSDQ
= −6 − 12 = −18 V
−18 − ( −20 )
RD = ⇒ RD = 0.4 kΩ
5
2
⎛ V ⎞
2
⎛ V ⎞
I DQ = I DSS ⎜ 1 − GS ⎟ ⇒ 5 = 8 ⎜ 1 − GS ⎟
⎝ VP ⎠ ⎝ 4 ⎠
VGS = 0.838 V
VG = VGS + VS = 0.838 − 6 = −5.162
⎛ R2 ⎞
VG = ⎜ ⎟ ( −20 )
⎝ R1 + R2 ⎠
1
−5.162 = (100 )( −20 ) ⇒ R1 = 387 kΩ
R1
R1 R2
= 100 ⇒ ( 387 ) R2 = 100 ( 387 ) + 100 R2
R1 + R2
( 387 − 100 ) R2 = (100 )( 387 ) ⇒ R2 = 135 kΩ
TYU3.1
(a) VTN = 1.2 V , VGS = 2 V
V DS ( sat ) = VGS − VTN = 2 − 1.2 = 0.8 V
(i) VDS = 0.4 ⇒ Nonsaturation
(ii) VDS = 1 ⇒ Saturation
(iii) VDS = 5 ⇒ Saturation
(b) VTN = −1.2 V , VGS = 2 V
V DS ( sat ) = VGS − VTN = 2 − ( −1.2 ) = 3.2 V
(i) VDS = 0.4 ⇒ Nonsaturation
(ii) VDS = 1 ⇒ Nonsaturation
(iii) VDS = 5 ⇒ Saturation
TYU3.2
W μ n Cox
(a) Kn =
2L
∈ox ( 3.9 ) ( 8.85 × 10 )
−14
Cox = = −8
= 7.67 × 10−8 F / cm
tox 450 × 10
(100 )( 500 ) ( 7.67 ×10−8 )
Kn = ⇒ K n = 0.274 mA / V 2
2 (7)
(b) VTN = 1.2 V, VGS = 2 V

11. (i) VDS = 0.4 V ⇒ Nonsaturation
I D = ( 0.274 ) ⎡ 2 ( 2 − 1.2 )( 0.4 ) − ( 0.4 ) ⎤ ⇒ I D = 0.132 mA
2
⎣ ⎦
(ii) VDS = 1 V ⇒ Saturation
I D = ( 0.274 )( 2 − 1.2 ) ⇒ I D = 0.175 mA
2
(iii) VDS = 5 V ⇒ Saturation
I D = ( 0.274 )( 2 − 1.2 ) ⇒ I D = 0.175 mA
2
VTN = −1.2 V , VGS = 2 V
(i) VDS = 0.4 V ⇒ Nonsaturation
I D = ( 0.274 ) ⎡ 2 ( 2 + 1.2 )( 0.4 ) − ( 0.4 ) ⎤ ⇒ I D = 0.658 mA
2
⎣ ⎦
(ii) VDS = 1 V ⇒ Nonsaturation
I D = ( 0.274 ) ⎡ 2 ( 2 + 1.2 )(1) − (1) ⎤ ⇒ I D = 1.48 mA
2
⎣ ⎦
(iii) VDS = 5 V ⇒ Saturation
I D = ( 0.274 )( 2 + 1.2 ) ⇒ I D = 2.81 mA
2
TYU3.3
(a) VSD (sat) = VSG + VTP = 2 − 1.2 = 0.8 V
(i) Non Sat (ii) Sat (iii) Sat
(b) VSD (sat) = 2 + 1.2 = 3.2 V
(i) Non Sat (ii) Non Sat (iii) Sat
TYU3.4
(a)
⎛ W ⎞ ⎛ μ p Cox ⎞ (3.9)(8.85 × 10−14 )
KP = ⎜ ⎟ ⎜ ⎟ Cox =
⎝ L ⎠⎝ Z ⎠ 350 × 10−8
= 9.861× 10−8
(40) ⎛ ( 300 ) ( 9.861× 10 )⎞
−8
KP = ⎜ ⎟
(2) ⎜ 2 ⎟
⎝ ⎠
K P = 0.296 mA / V
2
(b)
(i) I D = (0.296) ⎡ 2(2 − 1.2)(0.4) − (0.4) 2 ⎤
⎣ ⎦
= 0.142 mA
I D = (0.296) [ 2 − 1.2] ⇒ I D = 0.189 mA
2
(ii)
(iii) ID = 0.189 mA
(i) I D = (0.296) ⎡ 2 ( 2 + 1.2 )( 0.4 ) − ( 0.4 ) ⎤
2
⎣ ⎦
= 0.710 mA
I D = (0.296) ⎡ 2 ( 2 + 1.2 )(1) − (1) ⎤
2
(ii)
⎣ ⎦
=1.60 mA
(iii) I D = ( 0.296 )( 2 + 1.2 )
2
= 3.03 mA
TYU3.5

12. (a) λ = 0, VDS ( sat ) = 2.5 − 0.8 = 1.7 V
For VDS = 2 V , VDS = 10 V ⇒ Saturation Region
I D = ( 0.1)( 2.5 − 0.8 ) ⇒ I D = 0.289 mA
2
(b) λ = 0.02 V −1
I D = K n (VGS − VTN ) (1 + λVDS )
2
For VDS = 2 V
I D = ( 0.1)( 2.5 − 0.8 ) ⎡1 + ( 0.02 )( 2 ) ⎤ ⇒ I D = 0.300 mA
2
⎣ ⎦
VDS = 10 V
I D = ( 0.1) ⎡( 2.5 − 0.8 ) (1 + ( 0.02 )(10 ) ) ⎤ ⇒ I D = 0.347 mA
2
⎣ ⎦
(c) For part (a), λ = 0 ⇒ ro = ∞
For part (b), λ = 0.02 V −1 ,
−1 −1
ro = ⎡λ K n (VGS − VTN ) ⎤ = ⎡( 0.02 )( 0.1)( 2.5 − 0.8 ) ⎤
2 2
or ro = 173 k Ω
⎣ ⎦ ⎣ ⎦
TYU3.6
VTN = VTNO + γ ⎡ 2φ f + VSB − 2φ f ⎤
⎣ ⎦
2φ f = 0.70 V , VTNO = 1 V
(a) VSB = 0 ⇒, VTN = 1 V
(b) VSB = 1 V , VTN = 1 + ( 0.35 ) ⎣ 0.7 + 1 − 0.7 ⎦ ⇒ VTN = 1.16 V
⎡ ⎤
(c) VSB = 4 V , VTN = 1 + ( 0.35 ) ⎡ 0.7 + 4 − 0.7 ⎤ ⇒ VTN = 1.47 V
⎣ ⎦
TYU3.7
I D = K n (VGS − VTN )
2
0.4 = 0.25 (VGS − 0.8 ) ⇒ VGS = 2.06 V
2
⎛ R2 ⎞
VGS = ⎜ ⎟ VDD
⎝ R1 + R2 ⎠
⎛ R ⎞
2.06 = ⎜ 2 ⎟ ( 7.5 ) ⇒ R2 = 68.8 kΩ
⎝ 250 ⎠
R1 = 181.2 kΩ
VDS = 4 = VDD − I D RD
7.5 − 4
RD = ⇒ RD = 8.75 kΩ
0.4
VDS > VDS ( sat ) , Yes
TYU3.8

13. VS − ( −5 )
ID = and VS = −VGS
RS
5 − VGS
So RS =
0.1
I D = K n (VGS − VTN )
2
0.1 = ( 0.080 )(VGS − 1.2 ) ⇒ VGS = 2.32 V
2
5 − 2.32
So RS = ⇒ RS = 26.8 kΩ
0.1
VDS = VD − VS ⇒ VD = VDS + VS = 4.5 − 2.32
VD = 2.18
5 − VD 5 − 2.18
RD = = ⇒ RD = 28.2 kΩ
ID 0.1
VDS > VDS ( sat ) , Yes
TYU3.9
For VDS = 2.2 V
5 − 2.2
ID = ⇒ I D = 0.56 mA
5
I D = K n (VGS − VTN )
2
0.56 = K n ( 2.2 − 1)
2
W μ n Cox
K n = 0.389 mA / V = ⋅
L 2
W ( 389 )( 2 ) W
= ⇒ = 19.4
L ( 40 ) L
TYU3.10
(a) The transition point is
VIt =
(
VDD − VTNL + VTND 1 + K nD / K nL )
1 + K nD /K nL
=
(
5 − 1 + 1 1 + 0.05/ 0.01 )
1 + 0.05/ 0.01
7.236
= ⇒ VIt = 2.236 V
3.236
VOt = VIt − VTND = 2.24 − 1 ⇒ VOt = 1.24 V
(b) We may write
I D = K n D (VGSD − VTND ) = ( 0.05 )( 2.236 − 1) ⇒ I D = 76.4 μ A
2 2
TYU3.11
VIt =
(
VDD − VTNL + VTND 1 + K nD /K nL )
1 + K nD /K nL
2.5 =
(
5 − 1 + 1 1 + K nD /K nL )
1 + K nD /K nL
5 − 2.5
2.5 + 2.5 K nD /K nL = 5 + K nD /K nL ⇒ K nD /K nL = = 1.67 ⇒ K nD /K nL = 2.78
1.5
b. For VI = 5, driver in nonsaturated region.

14. I DD = I DL
K nD ⎡ 2 (VI − VTND ) VO − VO2 ⎤ = K nL (VGSL − VTNL )
2
⎣ ⎦
K nD
⎡ 2 (VI − VTND ) VO − VO2 ⎤ = [VDD − VO − VTNL ]
2
K nL ⎣ ⎦
2.78 ⎡ 2 ( 5 − 1) V0 − V02 ⎤ = [5 − V0 − 1]
2
⎣ ⎦
22.24V0 − 2.78V02 = ( 4 − V0 )
2
= 16 − 8V0 + V02
3.78V02 − 30.24V0 + 16 = 0
( 30.24 ) − 4 ( 3.78 )(16 )
2
30.24 ±
V0 = ⇒ V0 = 0.57 V
2 ( 3.78 )
TYU3.12
We have VDS = 1.2 V < VGS − VTN = −VTN = 1.8 V
Transistor is biased in the nonsaturation region.
V − VDS 5 − 1.2
I D = K n ⎣ 2 (VGS − VTN ) VDS − VDS ⎦ and I D = DD
⎡ 2
⎤ = ⇒ I D = 0.475 mA
RS 8
0.475 = K n ⎡ 2 ( 0 − ( −1.8 ) ) (1.2 ) − (1.2 ) ⎤
2
⎣ ⎦
0.475 = K n ( 2.88 ) ⇒ K n = 0.165 mA/V 2
W μ n Cox
Kn = ⋅
L 2
W ( 165 )( 2 ) W
= ⇒ = 9.43
L 35 L
TYU3.13
(a) Transition point for the load transistor – Driver is in the saturation region.
I DD = I DL
K nD (VGSD − VTND ) = K nL (VGSL − VTNL )
2 2
VDSL ( sat ) = VGSL − VTNL = −VTNL ⇒ VDSL = VDD − VOt = 2 V
Then VOt = 5 − 2 = 3 V , VOt = 3 V
K nD
(VIt − 1) = ( −VTNL )
K nL
0.08
(VIt − 1) = 2 ⇒ VIt = 1.89 V
0.01
(b) For the driver:
VOt = VIt − VTND
VIt = 1.89 V , VOt = 0.89 V
TYU3.14
I D = K n ⎡ 2 (VGS − VTN ) VDS − VDS ⎤
⎣
2
⎦
= ( 0.050 ) ⎡ 2 (10 − 0.7 )( 0.35 ) − ( 0.35 ) ⎤
2
⎣ ⎦
I D = 0.319 mA
VDD − Vo 10 − 0.35
RD = = ⇒ RD = 30.3 kΩ
ID 0.319
TYU3.15
(a) Transistor biased in the nonsaturation region

15. 5 − 1.5 − VDS
ID = = 12
R
I D = K n ⎡ 2 (VGS − VTN ) VDS − VDS ⎤
⎣
2
⎦
12 = 4 ⎡ 2 ( 5 − 0.8 ) VDS − VDS ⎤
⎣
2
⎦
4VDS − 33.6VDS + 12 = 0 ⇒ VDS = 0.374 V
2
5 − 1.5 − 0.374
Then R = ⇒ R = 261 Ω
12
TYU3.16
5 − VO
a. ID = = K n ⎡ 2 (V2 − VTN ) VO − VO2 ⎤
⎣ ⎦
RD
5 − ( 0.10 )
= K n ⎡ 2 ( 5 − 1)( 0.10 ) − ( 0.10 ) ⎤ ⇒ K n = 0.248 mA / V 2
2
25 ⎣ ⎦
5 − V0
b. = 2 ( 0.248 ) ⎡ 2 ( 5 − 1) V0 − V02 ⎤
⎣ ⎦
25
5 − V0 = 12.4 ⎡8V0 − V0 ⎤
⎣
2
⎦
12.4V02 − 100.2V0 + 5 = 0
(100.2 ) − 4 (12.4 )( 5 )
2
100.2 ±
V0 = ⇒ V0 = 0.0502 V
2 (12.4 )
TYU3.17
I DQ = K (VGS − VTN ) ⇒ 5 = 50 (VGS − 0.15 ) ⇒ VGS = 0.466 V
2 2
VS = ( 0.005 )(10 ) = 0.050 V ⇒ VGG = VGS + VS = 0.466 + 0.050 ⇒ VGG = 0.516 V
VD = 5 − ( 0.005 )(100 ) ⇒ VD = 4.5 V
VDS = VD − VS = 4.5 − 0.050 ⇒ VDS = 4.45 V
TYU3.18
I D = K ⎡ 2 (VGS − VTN ) VDS − VDS ⎤
⎣
2
⎦
= 100 ⎡ 2 ( 0.7 − 0.2 )( 0.1) − ( 0.1) ⎤
2
⎣ ⎦
ID = 9 μA
2.5 − 0.1
RD = ⇒ RD = 267 kΩ
0.009