V. Chapter 3. Exercise Solutions. EX3.1. ( ). 1 ,. 3 ,. 4.5. 4.5. 3 1 2. TN. GS. DS. DS
.... and Design, 4th edition. Chapter 3. By D. A. Neamen. Exercise Solutions ...
Microelectronics: Circuit Analysis and Design, 4th edition Chapter 3 By D. A. Neamen Exercise Solutions ______________________________________________________________________________________
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: 2 I D = ( 0.2 ) ⎡ 2 ( 3 − 1)(1) − (1) ⎤ ⇒ I D = 0.6 mA ⎣ ⎦ ______________________________________________________________________________________ EX3.2
0.5 = K p (3 − 1.2) ⇒ K p = 0.154 mA/V 2
2
(a) i D = K p (υ SG + VTP ) = 0.154(2 − 1.2 ) = 0.0986 mA 2
2
[
]
[
]
2 (b) i D = K p 2(υ SG + VTP )υ SD − υ SD = 0.154 2(5 − 1.2 )(2) − (2 ) = 1.72 mA 2
______________________________________________________________________________________ EX3.3 ⎛ R2 VGS = ⎜⎜ ⎝ R1 + R2
⎞ ⎛ 245 ⎞ ⎟⎟ ⋅ V DD = ⎜ ⎟(2.2) = 0.8983 V ⎝ 245 + 355 ⎠ ⎠
I D = (25)(0.8983 − 0.35) = 7.52 μ A V DS = 2.2 − (0.00752)(100 ) = 1.45 V ______________________________________________________________________________________ 2
EX3.4 I DQ = 0.5 mA, V SDQ = 2.0 V
3.3 − 2.0 = 2.6 k Ω 0.5 2 I D = K p VSGQ + VTP RD =
( 0.5 = 0.2(V
) − 0.6 )
2
SGQ
⇒ VSGQ = 2.18 V
VG = 3.3 − 2.18 = 1.12 V ⎛ R2 ⎞ 1 ⎟⎟ ⋅ V DD = (R1 R2 )(3.3) VG = ⎜⎜ R1 ⎝ R1 + R2 ⎠ 1 1.12 = (300)(3.3) ⇒ R1 = 884 k Ω R1
Microelectronics: Circuit Analysis and Design, 4th edition Chapter 3 By D. A. Neamen Exercise Solutions ______________________________________________________________________________________ Then 884 R2 = 300 ⇒ R2 = 454 k Ω 884 + R2 ______________________________________________________________________________________
EX3.5
⎛ 30 ⎞ (a) VGSQ = ⎜ ⎟(5) = 1.667 V ⎝ 30 + 60 ⎠ 2 I DQ = (0.5)(1.667 − 0.6) = 0.5689 mA V DSQ = 5 − (0.5689 )(4 ) = 2.724 V
(b) K n (+ 5% ) = 0.525 mA/V 2 , VTN (− 5% ) = 0.57 V
I D = 0.525(1.667 − 0.57) = 0.6314 mA V DS = 5 − (0.6314 )(4 ) = 2.474 V 2
K n (− 5% ) = 0.475 mA/V 2 , VTN (+ 5% ) = 0.63 V
I D = 0.475(1.667 − 0.63) = 0.5105 mA V DS = 5 − (0.5105)(4) = 2.958 V Then 0.5105 ≤ I D ≤ 0.6314 mA 2.474 ≤ V DS ≤ 2.958 V ______________________________________________________________________________________ 2
EX3.6
⎛ 345 ⎞ (a) VG = ⎜ ⎟(4.4) − 2.2 = 0.330 V ⎝ 345 + 255 ⎠ 2.2 = I DQ R S + V SGQ + VG
(
2.2 = K p RS VSGQ + VTP
(
)
2
+ VSGQ + 0.330
)
2 1.87 = (0.035)(6 ) VSGQ − 0.6VSGQ + 0.09 + VSGQ
0.21V
2 SGQ
+ 0.874VSGQ − 1.8511 = 0 ⇒ VSGQ = 1.545 V
We find 2 I DQ = 35(1.545 − 0.3) = 54.22 μ A
V SDQ = 4.4 − (0.05422 )(6 + 42 ) = 1.797 V
(b) For VTP = −0.315 V We have 2 1.87 = (0.035)(6) VSGQ − 0.63VSGQ + 0.099225 + VSGQ
(
)
2 0.21VSGQ + 0.8677VSGQ − 1.849 = 0 ⇒ VSGQ = 1.550 V
Then 2 I DQ = 35(1.550 − 0.315) = 53.36 μ A For VTP = −0.285 V We have 2 1.87 = (0.035)(6 ) VSGQ − 0.57VSGQ + 0.081225 + VSGQ
(
)
2 0.21VSGQ + 0.8803VSGQ − 1.8529 = 0 ⇒ VSGQ = 1.5395 V
Microelectronics: Circuit Analysis and Design, 4th edition Chapter 3 By D. A. Neamen Exercise Solutions ______________________________________________________________________________________ Then 2 I DQ = 35(1.5395 − 0.285) = 55.08 μ A Therefore 53.36 ≤ I DQ ≤ 55.08 μ A ______________________________________________________________________________________ EX3.7 4.4 = VSD (sat ) + I D (6 + 42 )
4.4 = VSG + VTP + (0.035)(48)(VSG + VTP )
2
(
2 4.4 = VSG − 0.3 + 1.68 VSG − 0.6VSG + 0.09
)
1.68V − 0.008VSG − 4.5488 = 0 ⇒ VSG = 1.648 V We find 2 I D = 35(1.648 − 0.3) = 63.59 μ A 2 SG
VSD = 4.4 − (0.0636)(48) = 1.348 V Note: VSD (sat ) = 1.648 − 0.3 = 1.348 V ______________________________________________________________________________________
EX3.8 2 (a) I DQ = 60 = 30 VSGQ − 0.4 ⇒ VSGQ = 1.814 V
(
I DQ =
)
3 − 1.814 = 0.060 ⇒ RS = 19.77 k Ω RS
V D = 1.814 − 2.5 = −0.686 V − 0.686 − (− 3) RD = = 38.57 k Ω 0.060 (b) VTP (+ 5% ) = 0.42 V, K p (− 5% ) = 28.5 μ A/V 2
(
)
2 3 = I D RS + VSG = (0.0285)(19.77) VSG − 0.84VSG + 0.1764 + VSG which yields V SG = 1.849 V
I D = (28.5)(1.849 − 0.42) = 58.2 μ A V SD = 6 − (0.0582)(19.77 + 38.57 ) = 2.605 V 2
VTP (− 5%) = 0.38 V, K p (+ 5% ) = 31.5 μ A/V
(
2
)
2 3 = (0.0315)(19.77 ) VSG − 0.76VSG + 0.1444 + VSG which yields VSG = 1.780 V
I D = 31.5(1.780 − 0.38) = 61.72 μ A VSD = 6 − (0.06172)(19.77 + 38.57 ) = 2.399 V Then 58.2 ≤ I D ≤ 61.72 μ A 2.399 ≤ V SD ≤ 2.605 V ______________________________________________________________________________________ 2
Microelectronics: Circuit Analysis and Design, 4th edition Chapter 3 By D. A. Neamen Exercise Solutions ______________________________________________________________________________________ EX3.9 (a)
VI = 4 V, Driver in Non ⋅ Sat. 2 K nD ⎡⎣ 2 (VI − VTND ) VO − VO2 ⎤⎦ = K nL [VDD − VO − VTNL ] 5 ⎡⎣ 2 ( 4 − 1) VD − VD2 ⎤⎦ = ( 5 − VD − 1) = ( 4 − VO ) = 16 − 8VO + VO2 2
2
6VD2 − 38VO + 16 = 0
VD =
38 ± 1444 − 384 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 ______________________________________________________________________________________ EX3.10 (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 2 ( 5 − 1)( 0.25 ) − ( 0.25 ) ⎤ = 4 ⇒ nD = 2.06 ⎦ 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
______________________________________________________________________________________ EX3.11 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 ______________________________________________________________________________________
EX3.12 Transistor in nonsaturation (a) V DS = 0.2 V, VGS = 5 V
[
2 VO = VDD − I D RD = VDD − K n RD 2(VGS − VTN )VDS − VDS
[
]
]
0.2 = 5 − K n (0.5) 2(5 − 1)(0.2) − (0.2) ⇒ K n = 6.154 mA/V 2 2
Microelectronics: Circuit Analysis and Design, 4th edition Chapter 3 By D. A. Neamen Exercise Solutions ______________________________________________________________________________________
[
]
(b) I D = (6.154) 2(5 − 1)(0.2) − (0.2) = 9.60 mA P = I DV DS = (9.60 )(0.2 ) = 1.92 mW ______________________________________________________________________________________ EX3.13
a.
2
V1 = 5 V, V2 = 0, M 2 cutoff ⇒ I D 2 = 0
I D = K n ⎡⎣ 2 (VI − VTN ) VO − VO2 ⎤⎦ =
5 − VO RD
( 0.05 )( 30 ) ⎡⎣ 2 ( 5 − 1)V0 − V02 ⎤⎦ = 5 − V0 1.5V02 − 13V0 + 5 = 0 V0 =
(13) − 4 (1.5)( 5) ⇒ V0 = 0.40 V 2 (1.5 )
13 ±
2
5 − 0.40 ⇒ I R = I D1 = 0.153 mA 30 V1 = V2 = 5 V
I R = I D1 = b.
5 − VO = 2 K n ⎡⎣ 2 (VI − VTN ) VO − VO2 ⎤⎦ RD
{
}
5 − V0 = 2 ( 0.05 )( 30 ) ⎣⎡ 2 ( 5 − 1)V0 − V02 ⎦⎤ 3V02 − 25V0 + 5 = 0 V0 =
25 ±
( 25) − 4 ( 3)( 5 ) ⇒ V0 = 0.205 V 2 ( 3) 2
5 − 0.205 ⇒ I R = 0.160 mA 30 = I D 2 = 0.080 mA
IR = I D1
______________________________________________________________________________________ EX3.14 VGS 3 =
I REF1 120 + VTN = + 0.4 = 1.814 V K n3 60
VGS 2 = VGS 3 = 1.814 V
I Q1 = K n 2 (VGS 2 − VTN ) = 30(1.814 − 0.4) = 60 μ A 2
VGS1 =
I Q1 K n1
+ VTN =
2
60 + 0.4 = 1.495 V 50
______________________________________________________________________________________
Microelectronics: Circuit Analysis and Design, 4th edition Chapter 3 By D. A. Neamen Exercise Solutions ______________________________________________________________________________________ EX3.15 2 ⎛ 0.04 ⎞ 0.1 = ⎜ ⎟ (15 )(VSGC − 0.6 ) ⎝ 2 ⎠ VSGC = 1.177 V = VSGB 2 ⎛ 0.04 ⎞ ⎛ W ⎞ 0.2 = ⎜ ⎟ ⎜ ⎟ (1.177 − 0.6 ) ⎝ 2 ⎠ ⎝ L ⎠B ⎛W ⎞ ⎜ ⎟ = 30 ⎝ L ⎠B 2 ⎛ 0.04 ⎞ 0.2 = ⎜ ⎟ ( 25 )(VSGA − 0.6 ) ⎝ 2 ⎠ VSGA = 1.23 V ______________________________________________________________________________________
EX3.16 I REF = K n 3 (VGS 3 − VTN ) = K n 4 (VGS 4 − VTN ) VGS 3 = 2 V ⇒ VGS 4 = 3 V 2
( 2 − 1) (a)
2
=
2
Kn4 K 1 2 ( 3 − 1) ⇒ n 4 = K n3 K n3 4
I Q = K n 2 (VGS 2 − VTN ) But VGS 2 = VGS 3 = 2 V
2
0.1 = K n 2 ( 2 − 1) ⇒ K n 2 = 0.1 mA / V 2 2
(b)
0.2 = K n 3 ( 2 − 1) ⇒ K n 3 = 0.2 mA / V 2 2
0.2 = K n 4 ( 3 − 1) ⇒ K n 4 = 0.05 mA / V 2 (c) ______________________________________________________________________________________ 2
EX3.17 VS 2 = 5 − 5 = 0
RS 2 =
I D 2 = K n 2 (VGS 2 − VTN 2 )
5 = 16.7 K 0.3
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 = 25.8 K 0.1 VS 1 = VG 2 − VDSQ1 = 2.425 − 5 = −2.575 V RD1 =
RS 1 =
−2.575 − ( −5 ) 0.1
⇒ RS 1 = 24.3 K
I D1 = K n1 (VGS 1 − VTN 1 )
2
0.1 = 0.5 (VGS 1 − 1.2 ) ⇒ VGS 1 = 1.647 V VG1 = VGS 1 + VS 1 = 1.647 + ( −2.575 ) ⇒ VG1 = −0.928 V 2
⎛ R2 ⎞ 1 VG1 = ⎜ ⎟ (10 ) − 5 = ⋅ RTN ⋅ (10 ) − 5 + R R R ⎝ 1 2 ⎠ 1
Microelectronics: Circuit Analysis and Design, 4th edition Chapter 3 By D. A. Neamen Exercise Solutions ______________________________________________________________________________________ 1 −0.928 = ( 200 )(10 ) − 5 ⇒ R1 = 491 K R1 491 R2 = 200 ⇒ R2 = 337 K 491 + R2 ______________________________________________________________________________________ EX3.18 VS1 = I D RS − 5 = (0.25)(16) − 5 = −1 V I DQ = K n (VGS 1 − 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) ⇒ R3 = 50.7 K ⎟ (5) ⇒ 0.507 = 500 ⎝ R1 + R2 + R3 ⎠ VS 2 = VS 1 + VDS 1 = −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) + + R R R ⎝ 500 ⎠ 2 3 ⎠ ⎝ 1 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 = 4 K 0.25 ______________________________________________________________________________________ RD =
EX3.19 VDS ( sat ) = VGS − VP = −1.2 − ( −4.5 ) ⇒ VDS ( sat ) = 3.3 V ⎛ ( −1.2 ) ⎞ ⎛ V ⎞ I D = I DSS ⎜ 1 − GS ⎟ = 12 ⎜ 1 − ⇒ I D = 6.45 mA ⎜ ( −4.5 ) ⎟⎟ VP ⎠ ⎝ ⎝ ⎠ ______________________________________________________________________________________ 2
2
Microelectronics: Circuit Analysis and Design, 4th edition Chapter 3 By D. A. Neamen Exercise Solutions ______________________________________________________________________________________ EX3.20 I D = 2.5 mA ⎛ V ⎞ I D = I DSS ⎜ 1 − GS ⎟ VP ⎠ ⎝
2
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 = 1.35 kΩ 2.5
RD =
( 20 )
2
R1 + R2
= 2 ⇒ R1 + R2 = 200 kΩ
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
______________________________________________________________________________________ EX3.21
VS = −VGS . I D =
0 − VS VGS = RS RS
⎛ V ⎞ I D = I DSS ⎜1 − GS ⎟ VP ⎠ ⎝
2
⎛ V VGS V2 ⎞ ⎛ V ⎞ = 6 ⎜1 − GS ⎟ = 6 ⎜1 − GS + GS ⎟ 16 ⎠ 1 4 ⎠ 2 ⎝ ⎝ 2 0.375VGS − 4VGS + 6 = 0 2
VGS =
4 ± 16 − 4 ( 0.375 )( 6 ) 2 ( 0.375 )
VGS = 8.86 or VGS = 1.806 V 14243 impossible
ID =
VGS = 1.806 mA RS
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 ) ______________________________________________________________________________________
Microelectronics: Circuit Analysis and Design, 4th edition Chapter 3 By D. A. Neamen Exercise Solutions ______________________________________________________________________________________ EX3.22 RR Rib = R1 R 2 = 1 2 = 100 k Ω R1 + R 2 I DQ = 5 mA, V S = − I DQ R S = −(5)(1.2 ) = −6 V V SDQ = 12 V, V D = V S − V SDQ
= −6 − 12 = −18 V
RD =
−18 − ( −20 ) 5
⇒ RD = 0.4 kΩ 2
I DQ VGS
⎛ V ⎞ ⎛ V ⎞ = I DSS ⎜ 1 − GS ⎟ ⇒ 5 = 8 ⎜ 1 − GS ⎟ 4 ⎠ VP ⎠ ⎝ ⎝ = 0.838 V
2
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
______________________________________________________________________________________
Test Your Understanding Solutions 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
(ii)
VDS = 1 ⇒ Nonsaturation
(iii) VDS = 5 ⇒ Saturation VTN = −1.2 V , VGS = 2 V V DS ( sat ) = VGS − VTN = 2 − ( −1.2 ) = 3.2 V (b) (i) VDS = 0.4 ⇒ Nonsaturation (iii) VDS = 5 ⇒ Saturation ______________________________________________________________________________________
Microelectronics: Circuit Analysis and Design, 4th edition Chapter 3 By D. A. Neamen Exercise Solutions ______________________________________________________________________________________ TYU3.2 Wμ n ∈ox 20 × 10 −4 (500 )(3.9 ) 8.85 × 10 −14 (a) K n = = ⇒ 1.08 mA/V 2 2 Lt ox 2 0.8 × 10 − 4 200 × 10 −8 (b) 2 i D = (1.08) 2(2 − 1.2)(0.4) − (0.4) = 0.518 mA (i)
(
)
(
(
)(
)
[
(ii)
)
i D = (1.08)(2 − 1.2) = 0.691 mA
]
2
(iii)
i D = (1.08)(2 − 1.2) = 0.691 mA
(i)
i D = (1.08) 2(2 + 1.2)(0.4) − (0.4) = 2.59 mA
(ii)
iD
2
[ ] = (1.08)[2(2 + 1.2)(1) − (1) ] = 5.83 mA 2
2
i D = (1.08)(2 + 1.2) = 11.1 mA (iii) ______________________________________________________________________________________ 2
TYU3.3
(a)
VSD (sat) = VSG + VTP = 2 − 1.2 = 0.8 V
(b)
(i) Non Sat (ii) Sat (iii) Sat VSD (sat) = 2 + 1.2 = 3.2 V
(i) Non Sat (ii) Non Sat (iii) Sat ______________________________________________________________________________________ TYU3.4
(a) K p = (b) (i) (ii)
Wμ p ∈ox 2 Lt ox
=
(10 × 10 )(300)(3.9)(8.85 × 10 ) ⇒ 0.324 mA/V 2(0.8 × 10 )(200 × 10 ) −4
−14
−4
[
]
i D = (0.324) 2(2 − 1.2)(0.4) − (0.4) = 0.156 mA 2
i D = (0.324)(2 − 1.2) = 0.207 mA 2
(iii)
i D = (0.324)(2 − 1.2) = 0.207 mA
(i)
i D = (0.324) 2(2 + 1.2)(0.4) − (0.4) = 0.778 mA
(ii)
iD
(iii)
2
−8
2
[ ] = (0.324)[2(2 + 1.2)(1) − (1) ] = 1.75 mA 2
2
i D = (0.324)(2 + 1.2) = 3.32 mA 2
______________________________________________________________________________________ TYU3.5 (a), (i) (ii)
i D = (10)(0.5 − 0.25) = 0.625 μ A 2
(b) (i)
i D = K n (υ GS − VTN ) (1 + λυ DS ) 2
i D = (10)(0.5 − 0.25) [1 + (0.03)(0.5)] = 0.6344 μ A 2
(ii)
i D = (10)(0.5 − 0.25) [1 + (0.03)(1.2)] = 0.6475 μ A 2
Microelectronics: Circuit Analysis and Design, 4th edition Chapter 3 By D. A. Neamen Exercise Solutions ______________________________________________________________________________________ (c) For (a), ro = ∞ 1 1 For (b), ro = = = 53.3 M Ω λI DQ (0.03)(0.625) ______________________________________________________________________________________ TYU3.6 (a) VTN = VTNO = 0.4 V
[ 2φ + υ − 2φ ] = 0.4 + 0.15[ 2(0.35 ) + 0.5 − 2(0.35) ] ⇒ V = 0.4 + 0.15[ 2(0.35 ) + 1.5 − 2(0.35) ] ⇒ V
(b) VTN = VTNO + γ
f
SB
f
TN
(c)
VTN
TN
= 0.439 V = 0.497 V
______________________________________________________________________________________ TYU3.7 V DS = 2.2 − (0.07 )R D = 1.2 ⇒ RD = 14.3 k Ω
VGS =
I DQ Kn
70 + 0.25 = 1.778 V 30
+ VTN =
⎛ R2 ⎞ ⎛ R ⎞ ⎟⎟ ⋅ V DD = ⎜ 2 ⎟(2.2) VGS = 1.778 = ⎜⎜ R + R ⎝ 500 ⎠ 2 ⎠ ⎝ 1 We find R2 = 404 k Ω , R1 = 96 k Ω ______________________________________________________________________________________
TYU3.8
3.3 − 1.6 = 0.17 mA 10 W ⎛ 0.1 ⎞⎛ W ⎞ 2 0.17 = ⎜ = 2.36 ⎟⎜ ⎟(1.6 − 0.4) ⇒ L ⎝ 2 ⎠⎝ L ⎠ ______________________________________________________________________________________ ID =
TYU3.9 (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
______________________________________________________________________________________
Microelectronics: Circuit Analysis and Design, 4th edition Chapter 3 By D. A. Neamen Exercise Solutions ______________________________________________________________________________________ TYU3.10 (a) V DS > [0 − (− 1.2)] ⇒ Saturation 3.3 − 1.8 = 0.1875 mA 8 W ⎛ 0.08 ⎞⎛ W ⎞ 2 0.1875 = ⎜ = 3.26 ⎟⎜ ⎟[0 − (− 1.2 )] ⇒ L L 2 ⎝ ⎠⎝ ⎠ ID =
(b) V DS < [0 − (− 1.2)] ⇒ Nonsaturation 3.3 − 0.8 ID = = 0.3125 mA 8 W ⎛ 0.08 ⎞⎛ W ⎞ 2 0.3125 = ⎜ = 6.10 ⎟⎜ ⎟ 2(0 − (− 1.2))(0.8) − (0.8) ⇒ 2 L L ⎝ ⎠⎝ ⎠ ______________________________________________________________________________________
[
]
TYU3.11 (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 ) VDSL ( sat ) = VGSL − VTNL = −VTNL ⇒ VDSL = VDD − VOt = 2 V 2
2
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
______________________________________________________________________________________ TYU3.12 Transistor biased in nonsaturation 2 2 I D = K n 2(VGS − VTN )VDS − VDS = (4) 2(10 − 0.7 )(0.2) − (0.2) = 14.72 mA 10 − 0.2 RD = = 0.666 k Ω 14.72 ______________________________________________________________________________________
[
]
[
]
TYU3.13 (a) Transistor biased in the nonsaturation region 5 − 1.5 − VDS ID = = 12 R 2 ⎤⎦ 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
5 − 1.5 − 0.374 ⇒ R = 261 Ω 12 Then ______________________________________________________________________________________ R=
Microelectronics: Circuit Analysis and Design, 4th edition Chapter 3 By D. A. Neamen Exercise Solutions ______________________________________________________________________________________ TYU3.14 5 − VO ID = = K n ⎡⎣ 2 (V2 − VTN ) VO − VO2 ⎤⎦ RD 5 − ( 0.10 ) 25
a.
2 = K n ⎡ 2 ( 5 − 1)( 0.10 ) − ( 0.10 ) ⎤ ⇒ K n = 0.248 mA / V 2 ⎣ ⎦
5 − V0 = 2 ( 0.248 ) ⎡⎣ 2 ( 5 − 1) V0 − V02 ⎤⎦ 25 5 − V0 = 12.4 ⎡⎣8V0 − V02 ⎤⎦ 12.4V02 − 100.2V0 + 5 = 0 100.2 ±
(100.2 ) − 4 (12.4 )( 5 ) ⇒ V0 = 0.0502 V 2 (12.4 ) 2
V0 = b. ______________________________________________________________________________________
TYU3.15 VSGC =
I REF 2 − VTP = K pC
40 + 0.3 ⇒ VSGC = VSGB = 1.30 V 40
I Q 2 = K pB (VSGB + VTP ) = 60(1.30 − 0.3) = 60 μ A 2
V SGA =
I Q2 K pA
2
60 + 0.3 = 1.19 V 75
− VTP =
______________________________________________________________________________________ TYU3.16 2 I Q = K n1 (VGS1 − VTN )
120 = K n1 (1.5 − 0.7 ) ⇒ K n1 = 187.5 μ A/V 2 VGS 2 = VGS 3 = 2 V 2
120 = K n 2 (2 − 0.7 ) ⇒ K n 2 = 71.0 μ A/V 2 2
I REF = K n3 (VGS 3 − VTN )
2
80 = K n3 (2 − 0.7 ) ⇒ K n3 = 47.3 μ A/V 2 VGS 4 = 5 − 2 = 3 V 2
80 = K n 4 (3 − 0.7 ) ⇒ K n 4 = 15.12 μ A/V 2 ______________________________________________________________________________________ 2
TYU3.17 2 2 I DQ = K (VGS − VTN ) ⇒ 5 = 50 (VGS − 0.15 ) ⇒ VGS = 0.466 V 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
______________________________________________________________________________________
Microelectronics: Circuit Analysis and Design, 4th edition Chapter 3 By D. A. Neamen Exercise Solutions ______________________________________________________________________________________ TYU3.18 2 ⎤⎦ I D = K ⎡⎣ 2 (VGS − VTN ) VDS − VDS 2 = 100 ⎡ 2 ( 0.7 − 0.2 )( 0.1) − ( 0.1) ⎤ ⎣ ⎦ ID = 9 μA
2.5 − 0.1 ⇒ RD = 267 kΩ 0.009 ______________________________________________________________________________________ RD =
