4
4.1 CHAPTER (a) VG > VTN corresponds to the inversion region (b) VG << VTN corresponds to the accumulation region (c) VG < VTN corresponds to the depletion region 4.2 (a)
" ox -14 3.9o 3.9 8.854x10 F / cm F nF C = = = = 6.91x10-8 2 = 69.1 2 -9 Tox Tox 50x10 m( 100cm / m) cm cm
ox
(
)
(b), (c) & (d): Scaling the result from part (a) yields
" Cox = 69.1
nF 50nm nF nF 50nm nF nF 50nm nF " " = 173 2 | Cox = 69.1 2 = 346 2 | Cox = 69.1 2 = 691 2 2 cm 20nm cm cm 10nm cm cm 20nm cm 11.7 8.854x10-14 F / cm 2( ) 8.854x10-14 F / cm 11.7 1.602x10
-19
4.3 Cd =
s
2s (0.75V ) qN B
=
(
(
(10
15
/ cm
3
)
)(0.75V )
)
= 10.5x10-9 F / cm 2
4.4 (a)
' n " ox
-14 3.9o cm 2 3.9 8.854x10 F / cm K = nC = n = n = 500 Tox Tox V - sec 50x10-9 m( 100cm / m)
ox
(
)
F A A = 34.5 x 10-6 2 = 34.5 2 V - sec V V (b) & (c) Scaling the result from part (a) yields
' Kn = 34.5x10-6
' Kn = 34.5
A 50nm
V 20nm
2
= 86.3
A
V
2
' | Kn = 34.5
A 50nm
V 10nm
2
= 173
A
V
2
' | Kn = 34.5
A 50nm
V
2
5nm
= 345
A
V2
4.5
(a) Q (b) Q
"
= C ( GS - VTN )= V
" ox
ox
Tox
(V (V
GS
- VTN ) = - VTN )=
3.9 8.854x10-14 F / cm
-9
25x10 m( 100cm / m) 3.9 8.854x10-14 F / cm
-9
(
)(1V )= 1.38x10
-7
C cm 2 C cm 2
"
= C ( GS - VTN )= V
" ox
ox
Tox
GS
10x10 m( 100cm / m)
(
)(2V )= 6.91x10
-7
75
4.6
V cm2 6 cm a ) vn = -n E = - 500 2000 = -1.00x10 ( cm s V - s V cm2 cm (a) vn = -nE = -400 V - s 4000 cm = -1.60x106 s
4.7 The carrier velocity must increase as the carriers travel down the channel to compensate for the decrease in carrier density 4.8 (a) 0 < 0.8V I D = 0
(b) VGS - VTN = 0.2V , VDS = 0.25V Saturation region 200 A 5m 2 K' W V 1- I D = n VGS - VTN - DS VDS = ( 0.8) = 40.0 A 2 2 L 2 2 V 0.5m (c) VGS - VTN = 1.2V , VDS = 0.25V triode region 0.25 A 5m V ' W I D = Kn VGS - VTN - DS VDS = 200 2 2 - 0.8 - (0.25) = 538 A 2 2 L V 0.5m (d ) VGS - VTN = 2.2V , VDS = 0.25V triode region 0.25 A 5m V ' W I D = Kn VGS - VTN - DS VDS = 200 2 3 - 0.8 - (0.25)= 1.04 mA 2 2 L V 0.5m mA W A 5m (e) Kn = Kn' L = 200 V 2 0.5m = 2.00 V 2
4.9
mA W A 60m = 200 2 = 4.00 2 L V V 3m mA mA A 3m A 10m (b) Kn = 200 V 2 0.15m = 4.00 V 2 | (c) Kn = 200 V 2 0.25m = 8.00 V 2
(a) K
n
' = Kn
4.10 (a) 0 < 1V cutoff region, I D = 0
(b) 1V = 1V cutoff region, I
D
=0
(c) VGS - VTN = 1V , VDS = 0.1V triode region 0.1 A 10m V ' W I D = Kn VGS - VTN - DS VDS = 250 2 2 -1- (0.1) = +231 A 2 2 L V 1m
76
(d) VGS - VTN = 2V , VDS = 0.1V triode region A 10m V 0.1 ' W I D = Kn VGS - VTN - DS VDS = 250 2 3 -1- (0.1)= +488 A 2 L 2 V 1m W A 10m mA (e) Kn = Kn' L = 250 V 2 1m = 2.50 V 2
4.11 Identify the source, drain, gate and bulk terminals and find the current I in the transistors in Fig. P-4.3.
W = L 10 1 D G +5 + VGS (a) S (b) +0.2 V I B V + V DS
GS
-0.2 V S I V B D +
DS
W = L
10 1
+ +5 G
(a) VGS = VG - VS = 5V VDS = VD - VS = 0.2V Triode region operation A 10 V 0.2 ' W I = I D = Kn VGS - VTN - DS VDS = 100 2 5 - 0.70 - 0.2 = +840 A 2 L 2 V 1 (b) VGS = VG - VS = 5 - (-0.2) = 5.2V VDS = VD - VS = 0 - (-0.2)= 0.2V
0.2 A 10 I = -I S = -100 2 5.2 - 0.70 - 0.2 = -880 A 2 V 1
4.12
0.5 A 10 (a) I = 100 2 5 - 0.75 - 0.5 = 2.00 mA 2 V 1 0.2 A 10 (b) I = 100 2 3 - 0.75 - 0.2 = 430 A 2 V 1
VDS = VD - VS = 0 - (-0.5)= 0.5 V
4.13 (a) VGS = VG - VS = 5 - (-0.5) = 5.5 V
(b) V
0.5 A 10 I = -I S = -100 2 5.5 - 0.75 - 0.5 = -2.25 mA 2 V 1
GS
= VG - VS = 3 - (-1) = 4 V
VDS = VD - VS = 0 - (-1) = 1 V
1 A 10 I = -I S = -100 2 4 - 0.75 - 1 = -2.75 mA 2 V 1
77
4.14
' Kn = Kn
W L
(a) W =
1
Kn 4mA /V 2 L= (0.5m)= 20 m ' Kn 100A /V 2
(b) W =
800A /V 2 (0.5m)= 4 m 100A /V 2
4.15
(a) R (b) R
4.16
on
=
' Kn
W (VGS - VTN ) L
=
1 = 23.0 -6 100 100x10 (5 - 0.65) 1
on
=
1 = 35.7 100 -6 100x10 (3.3 - 0.50) 1 1 W 1 = ' L Kn ( GS - VTN )Ron V
W (VGS - VTN ) L W 1 4.71 W 1 7.84 (a) = = | (b) = = -6 -6 L 100x10 (5 - 0.75)(500) 1 L 100x10 (3.3 - 0.75)(500) 1
' Kn
Ron =
or
4.17
Ron
0.1V ID 5 A = 0.020 = 20m | Kn = = = 17.0 2 5A V (VGS - VTN - 0.5VDS )VDS 5 - 2 - 0.5(0.1) (0.1)
(
)
Note that this will require
W Kn 17.0 170 = ' = = L Kn 0.1 1
4.18 Picking two values in saturation : K K 2 2 395A = n (4 - VTN ) and 140A = n (3 - VTN ) 2 2 Taking the ratio of these two equations :
A 125 2 W V = 1.25 = L 100 A 1 2 V From the graph, VTN is somewhat less than 2 V. VTN > 0 enhancement - mode transistor
395 (4 - VTN ) A = 2 VTN = 1.5 V K n = 125 140 (3 - VTN ) V2
2
78
4.19 Using the parameter values from problem 4.22:
800uA
VGS = 5 V
600uA
VGS = 4.5 V
Drain Current (A)
400uA
VGS = 4 V
VGS = 3.5 V
200uA
VGS = 3 V VGS = 2.5 V VGS = 2 V
0 A 0V
1.0V
2.0V
3.0V
4.0V
5.0V
6.0V
Drain-Source Voltage (V)
4.20 (a) For VGS = 0, VGS VTN and ID = 0 (b) For VGS = 1 V, VGS = VTN and ID = 0
(c ) VGS - VTN
ID =
= 2 -1 =1V and VDS = 3.3 | VDS > (VGS - VTN ) so the saturation region is correct
mA 375 A 5m A 5m 2 2 ' W = 375 2 = 3.75 2 (2 -1) V = 1.88 mA | K n = K n 2 V 2 V 0.5um L V 0.5um = 3 -1 = 2V and VDS = 3.3 | VDS > (VGS - VTN ) so the saturation region is correct 375 A 5m 2 2 (3 -1) V = 7.50 mA 2 2 V 0.5m
(d) VGS - VTN
ID =
4.21 (a) For VGS = 0, VGS < VTN and ID = 0 (b) For VGS = 1 V, VGS < VTN and ID = 0
(c ) VGS - VTN
ID =
= 2 -1.5 = 0.5V and VDS = 4 | VDS > (VGS - VTN ) so the saturation region is correct
200 A 10m A 10m mA 2 2 ' W = 200 2 (2 -1.5) V = 250 A | K n = K n = 2.00 2 2 2 V 1um L V 1um V = 3 -1.5 =1.5V and VDS = 4 | VDS > (VGS - VTN ) so the saturation region is correct 200 A 10m 2 2 (3 -1.5) V = 2.25 mA 2 2 V 1m
(d) VGS - VTN
ID =
79
4.22 (a)
VGS - VTN = 2 - 0.75 = 1.25 V and VDS = 0.2 V. VDS < VGS - VTN so the transistor is operating in the triode region.
' ID = K n
V 0.2 W A 10 VGS - VTN - DS VDS = 200 2 2 - 0.75 - 0.2 = 460 A 2 2 L V 1 (b) VGS - VTN = 2 - 0.75 = 1.25 V and VDS = 2.5 V. VDS > VGS - VTN so the transistor is operating in the saturation region. 200 A 10 K' W 2 2 ID = n (2 - 0.75) = 1.56 mA (VGS - VTN ) = 2 2 V 1 2 L (c) VGS < VTN so the transistor is cutoff with ID = 0. 300 ' ID K n so (a) ID = (d) 460A = 690A (b) ID = 2.34 mA (c) ID = 0 200
4.23 (a) VGS - VTN = 4 V, VDS = 6 V. VDS > VGS - VTN --> Saturation region
(b) VGS < VTN --> Cutoff region (c) VGS - VTN = 1 V, VDS = 2 V. VDS > VGS - VTN --> Saturation region (d) VGS - VTN = 0.5 V, VDS = 0.5 V. VDS = VGS - VTN --> Boundary between triode and saturation regions (e) The source and drain of the transistor are now reversed because of the sign change in VDS. Assuming the voltages are defined relative to the original S and D terminals as in Fig. P4.11(b), VGS = 2 - (-0.5) = 2.5 V, VGS - VTN = 2.5 - 1 = 1.5 V, and VDS = 0.5 V --> triode region (f) The source and drain of the transistor are again reversed because of the sign change in VDS. Assuming the voltages are defined relative to the original S and D terminals as in Fig. P4.11(b), VGS = 3 - (-6) = 9 V, VGS - VTN = 9 - 1 = 8.0 V, and VDS = 6 V --> triode region
D --> 'S' G + 2V (e) + S --> 'D' VG'S' = +2.5 V VD'S' = +0.5 V G 0.5 V + 3V (f) + S --> 'D' V G'S' = +9.0 V V D'S' = +6.0 V D --> 'S' -
6.0 V
80
4.24 (a) VGS - VTN = 2.6 V, VDS = 3.3 V. VDS > VGS - VTN --> Saturation region
(b) VGS < VTN --> Cutoff region (c) VGS - VTN = 1.3 V, VDS = 2 V. VDS > VGS - VTN --> Saturation region (d) VGS - VTN = 0.8 V, VDS = 0.5 V. VDS < VGS - VTN --> triode region (e) The source and drain of the transistor are now reversed because of the sign change in VDS. Assuming the voltages are defined relative to the original S and D terminals as in Fig. 4.54(b), VGS = 2 - (-0.5) = 2.5 V, VGS - VTN = 2.5 0.7 = 1.8 V, and VDS = 0.5 V --> triode region (f) The source and drain of the transistor are again reversed because of the sign change in VDS. Assuming the voltages are defined relative to the original S and D terminals as in Fig. 4.54(b), VGS = 3 - (-3) = 6 V, VGS - VTN = 6 0.7 = 5.3 V, and VDS = 3 V --> triode region
4.25
R
2
R
4
D G B S R R
V
DD
+ -
1
3
4.26
D G
+V I B S
DD
+V D I B S
DD
D G S D B G S B
G
D B G S
(a)
4.27
(b)
81
VDS = 3.3V, VGS VTN = 1.3 V; VDS > VGS - VTN so the transistor is saturated.
(a) gm = K n (VGS - VTN ) = 250
A 20m (2 - 0.7) = 6.50 mS V 2 1m A 20m (3.3 - 0.7) = 13.0 mS V 2 1m
(b) gm = K n (VGS - VTN ) = 250
4.28
(a) gm =
iD 760 -140 A = = 310 S | As a check, we can use the results from Problem 4.22. vGS 5-3 V
gm = K n (VGS - VTN ) = 125
(b) gm =
iD vGS
(4 -1.5)V = 313 S V2 A 390 -15 A = = 188 S | Checking : gm = 125 2 (3 -1.5)V = 188 S V 4-2 V
A
4.29 VDS > VGS - VTN so the transistor is saturated.
Kn 250 A 2 2 5 - 0.75) ( 0.025(6)) = 2.60 mA 1+ (VGS - VTN ) (1+ VDS ) = 2 ( 2 2 V K 250 A 2 2 (b) ID = n (VGS - VTN ) = 5 - 0.75) = 2.26 mA 2 ( 2 2 V (a) ID =
4.30 VDS > VGS - VTN so the transistor is saturated.
Kn 500 A 2 2 4 -1) ( 0.02(5)) = 2.48 mA 1+ (VGS - VTN ) (1+ VDS ) = 2 ( 2 2 V K 500 A 2 2 (b) ID = n (VGS - VTN ) = 4 -1) = 2.25 mA 2 ( 2 2 V (a) ID =
4.31 (a) The transistor is saturated by connection.
ID = 12V - VGS 100x10-6 10 A 2 = 2 (VGS - 0.75V ) 5 1 V 10 2
2 12.5VGS -17.8VGS - 4.97 = 0 VGS = 0.266V , 1.214V VGS = 1.214 V since it must exceed 0.75V
ID =
12 -1.214 = 108 A 10 5
Checking :
100x10-6 10 A 2 2 (1.214 - 0.75V ) = 108 A 1 V 2
(b) ID =
12V - VGS 1000x10-6 A 2 = V - 0.75V ) (1+ 0.025VGS ) 5 2 ( GS 10 2 V
82
Starting with the solution from part (a) and solving iteratively yields VGS = 1.20772 V and ID = 108 A, essentially no change. (c) 12V - VGS 100x10-6 25 A 2 ID = = 2 (VGS - 0.75V ) 5 1 V 10 2
2 62.5VGS - 91.75VGS + 11.16 = 0 VGS = 0.446V , 1.046V VGS = 1.046 V since VGS must exceed the threshold voltage.
12 -1.046 ID = = 110 A 10 5
100x10-6 25 A 2 Checking : ID = 2 (1.046 - 0.75V ) = 110 A 1 V 2
4.32 (a) The transistor is saturated by connection. 12V - VGS 100x10-6 10 A 2 ID = = 2 (VGS - 0.75V ) 4 1 V 5x10 2
2 31.25VGS - 45.88VGS + 5.58 = 0 VGS = 0.0588V , 1.401V VGS = 1.401 V since VGS must exceed the threshold voltage.
12 -1.401 100x10-6 10 A 2 = 212 A Checking : ID = 2 (1.401- 0.75V ) = 212 A 4 1 V 5x10 2 -6 12V - VGS 1000x10 A 2 = V - 0.75V ) (1+ 0.02VGS ) (b) ID = 4 2 ( GS 5x10 2 V ID =
Starting with the solution from part (a) and solving iteratively yields VGS = 1.3925 V and IDS = 212 A, essentially no change
4.33 (a) Since VDS = VGS and VTN > 0 for both transistors, both devices are saturated. K' W K' W 2 2 ID1 = n VGS1 - VTN ) and ID 2 = n ( (VGS 2 - VTN ) . Therefore 2 L 2 L From the circuit, however, ID2 must equal ID1 since IG = 0 for the MOSFET: ' ' Kn W Kn W 2 2 I = ID1 = ID 2 or (VGS1 - VTN ) = (VGS 2 - VTN ) 2 L 2 L which requires VGS1 = VGS2. Using KVL:
VDD = VDS1 + VDS 2 = VGS1 + VGS 2 = 2VGS 2 V VGS1 = VGS 2 = DD = 5V 2 ' K W 100 A 10 2 2 I= n (VGS1 - VTN ) = (5 - 0.75) V 2 = 9.03 mA 2 2 V 1 2 L (b) The current simply scales by a factor of two (see last equation above), and ID = 18.1 mA. (c) For this case,
83
' Kn W K' W 2 2 VGS1 - VTN ) (1 + VDS1 ) and ID 2 = n ( (VGS 2 - VTN ) (1 + VDS2 ). 2 L 2 L Since VGS = VDS for both transistors ' ' Kn W Kn W 2 2 ID1 = (VGS1 - VTN ) (1 + VGS1 ) and ID 2 = (VGS 2 - VTN ) (1 + VGS2 ) 2 L 2 L and ID1 = ID2 = I ' K' W Kn W 2 2 (VGS1 - VTN ) (1 + VGS1 ) = n (VGS 2 - VTN ) (1 + VGS2 ) 2 L 2 L which again requires VGS1 = VGS2 = VDD/2 = 5V. K' W 100 A 10 2 2 I= n (VGS1 - VTN ) (1+ VDS ) = (5 - 0.75) V 2 (1+ (.04 )5) = 10.8 mA 2 2 L 2 V 1
ID1 =
4.34 (a) Since VDS = VGS and VTN > 0 for both transistors, both devices are saturated ("by
connection").
' ' Kn W Kn W 2 2 ID1 = (VGS1 - VTN ) and ID 2 = (VGS 2 - VTN ) . Therefore 2 L 1 2 L 2 From the circuit, however, ID2 must equal ID1 since IG = 0 for the MOSFET: ' K n 10 K ' 40 2 2 VGS1 - VTN ) = n (VGS 2 - VTN ) I = ID1 = ID 2 or ( 2 1 2 1 which requires VGS1 = 2VGS2 - VTN. Using KVL:
VDD = VDS1 + VDS 2 = VGS 2 + VGS1 = 3VGS 2 - VTN V + VTN 10 + 0.75 VGS 2 = DD = = 3.583V VGS1 = 6.417 3 3 100 A 10 K' W 2 2 I= n (VGS1 - VTN ) = (6.417 - 0.75) V 2 = 16.1 mA 2 2 V 1 2 L 100 A 40 2 Checking : I = (3.583 - 0.75) V 2 = 16.1 mA which agrees. 2 2 V 1 (b) For this case with VGS = VDS for both transistors and ID1 = ID2, K' W K' W 2 2 ID1 = n (VGS1 - VTN ) (1 + VGS1 ) and ID 2 = n (VGS 2 - VTN ) (1 + VGS2 ) 2 L 1 2 L 2
where VGS2 = VDD VGS1. Therefore, ' K ' 40 K n 10 2 2 1 (VGS1 - VTN ) (1 + VGS1 ) = n (10 - VGS1 - VTN ) ( + (10 - VGS1 )) 2 1 2 1 VGS1 = 6.3163, VGS2 = 3.6837, ID1 = 20.4 mA, Checking: ID2 = 20.4 mA which agrees.
84
4.35 (a) Since VDS = VGS and VTN > 0 for both transistors, both devices are saturated ("by
connection").
' Kn W K' W 2 2 (VGS1 - VTN ) and ID 2 = n (VGS 2 - VTN ) . Therefore 2 L 1 2 L 2 From the circuit, however, ID2 must equal ID1 since IG = 0 for the MOSFET: ' ' K n 25 K n 12.5 2 2 I = ID1 = ID 2 or (VGS1 - VTN ) = (VGS 2 - VTN ) 2 1 2 1 Solving for VGS2 yields: VGS 2 = 2VGS1 - 2 -1 VTN
ID1 =
(
)
Also, VDD = VDS1 + VDS 2 or VGS1 = 10 - VGS 2 VGS1 = I= 10 +
( 2 -1)V
1+ 2
TN
= 4.271V VGS 2 = 5.729V
' 100 A 25 Kn W 2 2 2 (4.271- 0.75) V = 15.5 mA (VGS1 - VTN ) = 2 2 V 1 2 L 100 A 12.5 2 2 Checking : I = (5.729 - 0.75) V = 15.5 mA - agrees. 2 2 V 1 (b) For this case with VGS = VDS for both transistors and ID1 = ID2, K' W K' W 2 2 ID1 = n (VGS1 - VTN ) (1 + VGS1 ) and ID 2 = n (VGS 2 - VTN ) (1 + VGS2 ) 2 L 1 2 L 2
where VGS2 = VDD VGS1. Therefore, ' ' K n 40 K n 10 2 2 1 (VGS1 - VTN ) (1 + VGS1 ) = (10 - VGS1 - VTN ) ( + (10 - VGS1 )) 2 1 2 1 VGS1 = 4.3265 V, VGS2 = 5.6735 V, ID1 = 19.4 mA, Checking: ID2 = 19.4 mA both agree
4.36 VGS - VTN = 5 - (-2) = 7 V > VDS = 6 V so the transistor is operating in the triode region. 6 -6 (a) ID = 250x10 5 - (-2) - 6 = 6.00 mA 2 (b) Our triode region model is independent of , so ID = 6.00 mA. 4.37 Since VDS = VGS, and VTN < 0 for an NMOS depletion mode device, VGS - VTN will be greater than VDS and the transistor will be operating in the triode region.
85
4.38 (a) VDS = 6V | VGS - VTN = 0 - (-3) = 3V so the transistor is saturated
2 Kn 250 A 2 0 - (-3V )] = 1.13 mA (VGS1 - VTN ) = 2 [ 2 V 2 2 250 A 0 - (-3V )] ( 0.025(6)) = 1.29 mA 1+ (b) ID = 2 [ 2 V
ID =
4.39
+10 V 100 k D G S (a) (b) I DS W = 10 L 1 G D S -10 V 100 k I DS W = 10 L 1
(a) If the transistor were saturated, then 100x10-6 10 2 ID = (-2) = 2.00 mA 1 2 but this would require a power supply of greater than 200 V (2 mA x 100 k). Thus the transistor must be operating in the triode region.
10V - VDS V = 10-3 0 - (-2) - DS VDS 10 5 2 10 - VDS = 50VDS (4 - VDS ) and VDS = 0.0504V using the quadratic equation. 0.0504 10V - VDS ID = 10-3 2 - = 99.5 A 0.0504 = 99.5 A Checking : 2 10 5 (b) For R = 50 k and W/L = 20/1, 10V - VDS V = 2x10-3 0 - (-2) - DS VDS 4 5x10 2
10 - VDS = 50VDS (4 - VDS ), the same as part (a). 0.0504 10V - VDS ID = 2x10-3 2 - = 199 A 0.0504 = 199 A Checking : 5x10 4 2 (c) In this circuit, the drain and source terminals of the transistor are reversed because of the power supply voltage, and the current direction is also reversed. However, now VDS = VGS and since the transistor is a depletion-mode device, it is still operating in the triode region.
86
10V - VDS V = 1000x10-6 VDS - (-2) - DS VDS 5 10 2 10 - VDS = 50VDS (4 + VDS ) and VDS = 0.04915V using the quadratic equation. 0.04915 10V - VDS ID = 10-3 0.04915 - (-2) - = 99.5 A 0.04915 = 99.5 A Checking : 10 5 2 (d) In this circuit, the drain and source terminals of the transistor are reversed because of the power supply voltage, and the current direction is also reversed. However, now VDS = VGS and since the transistor is a depletion-mode device, it is still operating in the triode region. 10V - VDS V = 2000x10-6VDS - (-2) - DS VDS 4 5x10 2 10 - VDS = 50VDS (4 + VDS ) Same as part (c). VDS = 0.04915V using the quadratic equation. 0.04915 10V - VDS ID = 10-3 0.04915 - (-2) - = 99.5 A 0.04915 = 99.5 A Checking : 10 5 2
4.40 See figures in previous problem but use W/L = 20/1. 25x10-6 20 2 ID = (-1) = 250 A but this would require (a) If the transistor were saturated, then 2 1 a power supply of greater than 25 V. Thus the transistor must be operating in the triode region.
20 10V - VDS V = 100x10-6 0 - (-1) - DS VDS 5 1 10 2 10 - VDS = 100VDS (2 - VDS ) and VDS = 0.05105V using the quadratic equation. 0.05105 10 - 0.0510 ID = 2.00x10-3 1- V = 99.5 A 0.05105 = 99.5 A Checking : 2 10 5 (b) In this circuit, the drain and source terminals of the transistor are reversed because of the power supply voltage, and the current direction is also reversed. However, now VDS = VGS and since the transistor is a depletion-mode device, it is still operating in the triode region. 20 V VDS = 10 - ( 5 )( 10 100x10-6 ) VDS - (-1) - DS VDS 1 2 V VDS = 10 - 200VDS 1+ DS and VDS = 0.04858V using the quadratic equation. 2 0.04858 10 - 0.04858 V = 99.5 A ID = 2000x10-6 1+ 0.04858 = 99.5 A Checking : 2 10 5
87
4.41 (a) VTN = 0.75 + 0.75 1.5 + 0.6 - 0.6 = 1.26V
(
)
VGS - VTN = 2 -1.26 = 0.74V > VDS = 0.2V Triode region 10 0.2 ID = 200x10-6 2 -1.26 - 0.2 = 256 A (compared to 460 A) 1 2 (b) VGS - VTN = 2 -1.26 = 0.74V < VDS = 2.5V Saturation region 200x10-6 10 2 ID = (2 -1.26) = 548 A (compared to 1.56 mA) 1 2 (c) VGS < VTN so the transistor is cut off, and ID = 0. 300 ' (d) ID K n so (a) ID = 256A = 384 A (b) ID = 822 A (c) ID = 0 200
4.42 (a) VTN = 1.5 + 0.5
( 4 + 0.75 -
0.75 = 2.16V | VGS < VTN Cutoff & ID = 0
)
(b) ID = 0. The result is independent of VDS .
4.43 (a) VTN = 1+ 0.7
( 3 + 0.6 -
0.6 = 1.79V
)
VGS - VTN = 2.5 -1.79 = 0.71V < VDS = 5V Saturation region 100x10-6 8 2 (0.71) = 202 A 1 2 (b) 0.5 < 0.71 Triode region 8 0.5 ID = 100x10-6 0.71- 0.5 = 184 A 1 2 ID =
4.44 0.85 = -1.5 + 1.5 VSB + 0.75 - 0.75
(
Checking : VTN = -1.5 + 1.5
( 5.17 + 0.75 -
)
| Solving for VSB yields VSB = 5.17 V 0.75 = 0.85 V
)
4.45 Using trial and error with a spreadsheet yielded:
VTO = 0.74V = 0.84 V 2 F = 0.87V RMS Error = 51.9 mV
88
4.46
-14 3.9o cm 2 3.9(8.854 x10 F /cm) (a) K = p C = p = p = 200 Tox Tox V - sec 50x10-9 m(100cm /m) F A K 'p = 13.8x10-6 = 13.8 2 V - sec V A 50nm A ' (b) Scaling the result from Part (a) yields: K n = 13.8 2 = 34.5 2 V 20nm V A 50nm A ' = 69.0 2 (c) K n = 13.8 2 V 10nm V A 50nm A ' = 138 2 (d) K n = 13.8 2 V 5nm V ' p " ox
ox
4.47
The pinchoff points and threshold voltage can be estimated directly from the graph: e. g. VGS = -3 V curve gives VTP = 2.5 - 3 = - 0.5 V or from the VGS = -5 V curve gives VTP = 4.5 - 5 = 0.5 V. Alternately, choosing two points in saturation, say ID = 1.25 mA for VGS = -3 V and ID = 4.05 mA for VGS = -5 V:
ID1 (VGS1 - VTP ) = or ID 2 (VGS 2 - VTP ) 1.25 (-3 - VTP ) = 4.05 (-5 - VTP ) 400
Solving for VTP yields : 0.8VTP = -0.4V and VTP = -0.500V . Solving for K p : K p = 2ID = 2(1.25mA)
A
(VGS - VTP )
2
(-3 + 0.5)
2
= 0.400
mA W V 2 = 10 | = ' = A V2 L Kp 1 40 2 V Kp
4.48 Using the values from the previous problem
PMOS Output Characteristics
0.0045 0.004 0.0035 0.003 0.0025 0.002 0.0015 0.001 0.0005 0 -6 -5 -4 -3 VDS -2 -1 0 -2 V -3 V -3.5 V -4 V -4.5 V -5 V
(IDSAT, VDSAT): (0.45 mA,-1.5 V) (1.25 mA,-2.5 V) (1.8 mA,-3 V) (2.45 mA,-3.5 V) (3.7 mA,-4V) (4.05 mA,-4 V)
89
4.49 (a) VGS - VTP = -1.1+ 0.75 = -0.35V | VDS = -0.2V Triode region (-0.2) -0.2 = 40.0 A 40A 20 ( ID = -1.1- (-0.75)- ) 2 V 2 1
(b) VGS - VTP = -1.3 + 0.75 = -0.55V | VDS = -0.2V Triode region (-0.2) -0.2 = 72.0 A 40A 20 -1.3 - (-0.75)- ( ID = ) 2 V 2 1 (c) VTP = - 0.75 + .5 1+ .6 - -6 = -0.995V VGS - VTP = -1.1- (-0.995) = -0.105V | VDS = -0.2V saturation region 1 40A 20 2 ID = 2 (-1.1+ 0.995) = 4.41 A 2 V 1 (d) VGS - VTP = -1.3 + 0.995 = -0.305V | VDS = -0.2V triode region 10A 10 (-0.2) -0.2 = 32.8 A ID = -1.3 - (-0.995) - ) ( 2 2 V 1
4.50
[
(
)]
For PMOS : Ron =
1 K 'p
W VGS - VTP L W 1 5810 W 1 2330 (a) = = | (b) = = -6 -6 L 40x10 -5 + 0.70 (1) 1 L 100x10 (5 - 0.70)(1) 1
4.51
or
W 1 = ' L K p VGS - VTN Ron
For PMOS : Ron =
1 K 'p
W VGS - VTP L W 1 2.91 W 1 1.16 (a) = = | (b) = = -6 -6 L 40x10 -5 + 0.70 (2000) 1 L 100x10 (5 - 0.70)(2000) 1
or
W 1 = ' L K p VGS - VTN Ron
90
4.52
For PMOS : Ron = K 'p
1 W VGS - VTP L
(a) Ron =
1 = 11.8 200 -6 100x10 (5 - 0.75) 1 W K' W 500 Checking : = n = 2.5(200) = ' L p K p L n 1 (b) Ron =
1 = 29.4 -6 200 40x10 -5 - (-0.75) 1 W 1 499 (c) = = -6 L 40x10 -5 - (-0.75)(11.8) 1
4.53
+18 V R
2
R4 R S G D B VDD
+ G
S B D
R
1
R
3
(a)
(b)
4.54 (a) For VIN = 0, the NMOS device is on with VGS = 5 and VSB = 0, and the PMOS transistor is off with VGS = 0, VO = 0, and VSB = 0. 1 Ron = = 235 -6 (100x10 )(10)(5 - 0.75)
(b) For VIN = 5V, the NMOS device is off with VGS = 0, and the PMOS transistor is on with VGS = -5V, VO = 5V, and VSB = 0. 1 Ron = = 235 -6 (40x10 )(25)(-5 + 0.75)
4.55
Ron 0.1V 0.5A ID A = 0.2 K p = = = 0.629 2 0.5A V (VGS - VTP - 0.5VDS )VDS [-10V - (-2V ) - 0.5(-0.1V )](-0.1V )
4.56 VTP = -0.75 - 0.5
( 4 + 0.6 -
0.6 = -1.44V
)
VGS - VTP = -1.5 - (-1.44 ) = -0.065 | VDS = -4V Saturation region ID =
2 40x10-6 A 25 [-1.5 - (-1.44 )] = 1.80 A 2 V 1 2
91
4.57 VGS - VTP = -1.5 - (-0.75) = -0.75V | VDS = -0.5V
VGS - VTN < VDS Triode region | ID = 40x10-6
A 40 (-0.5) -0.5 = 400 A -1.5 - (-0.75) - ) ( 2 V 1 2
4.58 The PMOS transistor could be either an enhancement-mode or a depletion-mode device depending upon the specific values of R1, R2 and R4. Thus an enhancement device with VTP < 0 is correct and the symbol is correct. 4.59 If this PMOS transistor is conducting, then its threshold voltage must be greater than zero and it is a depletion-mode device. The symbol is that of an enhancement-mode device and is incorrect. 4.60
R2 S D R1 R4
+18 V R
G
VDD
+
G S D
R
3
(a)
4.61
R2 1.5 M G S R1 1 M R
S
(b)
RD 75 k D
VDD
+
-3 V 10 V
39 k
92
4.62
R
D
75 k R
EQ
G
D -5 V
VDD
+
10 V
600 k V EQ 4V RS
S
39 k
93
4.63
n+
22
Metal Polysilicon 12
4.64
[(2x20)/(12x22)]= 0.152 or 15.2%
n+
14
Metal
Polysilicon 18
[2x10/(18x14)]= 0.079 or 7.9%
94
4.65
Metal 12
n+ Polysilicon 12
(2x10/122)= 0.139 or 13.9%
4.66
Metal
12
n+
Polysilicon 14
[2x10/(14x12)] = 0.119 or 11.9%
95
4.67
(a)
" Cox =
ox
Tox
=
(3.9)8.854 x10-14
5x10 cm
-6
F cm
= 6.906x10-8
F cm 2
F " CGC = CoxWL = 6.906x10-8 2 (20x10-4 cm)(2x10-4 cm)= 27.6 fF cm F " (b) Cox = 1.73 x 10-7 2 | CGC = 69.1 fF cm F " (c) Cox = 3.45 x 10-7 2 | CGC = 138 fF cm F " (d) Cox = 7.90 x 10-7 2 | CGC = 276 fF cm
4.68
" Cox =
ox
Tox
=
(3.9)8.854 x10-14
1x10 cm
-6
F cm
= 3.46x10-7
F cm 2
F " CGC = CoxWL = 3.46x10-7 2 (5x10-4 cm)(5x10-5 cm)= 8.64 fF cm
4.69
' COL
F pF cm = ox L = (0.5x10-4 cm)= 17.3 cm cm Tox 10x10-9 m10 2 m
(3.9)8.854 x10-14
4.70
-15 F 10m)(1m) 1.4 x10 2 ( m C WL F ' + COLW = + 4 x10-15 (a) CGS = CGD = (10m) = 47 fF 2 2 m 2 " 2 ' (b) CGS = COX WL + COLW = 14 fF + 40 fF = 49 fF 3 3 F ' CGD = COLW = 4 x10-15 (10m) = 40 fF m F ' (c ) CGS = CGD = COLW = 4 x10-15 (10m) = 40 fF m
" OX
96
4.71
F ox F cm " Cox = = = 3.453x10-8 2 2 cm Tox cm 100x10-9 m10 m
(3.9)8.854 x10-14
CGC
4.72
-4 cm 2 -8 F 6 2 = C WL = 3.453x10 (50x10 m ) 10 = 17.3 nF m cm 2
" ox
" L = 2 = 1m | W =10L = 5m | Cox =
ox
Tox
=
3.9(8.854 x10-14 F /cm) 150x10 cm
-7
= 0.23 fF / m 2
Triode region :
" (0.23 fF / m 2 )(5m 2 ) + (0.02 fF / m)(5m) = 0.675 fF CoxWL + CGSOW = CGS = CGD = 2 2 2 " Saturation region : CGS = CoxWL + CGSOW = 0.867 fF | CGS = CGSOW = 0.10 fF 3 Cutoff : CGS = CGD = CGSOW = 0.10 fF
4.73
F cm ox F " = 3.453x10-8 2 (a) Cox = T = cm cm ox 100x10-9 m102 m F " CGC = CoxWL = 3.453x10-8 2 10x10-4 cm 1x10-4 cm = 3.45 fF cm F " (b) CGC = CoxWL = 3.453x10-8 cm2 100x10-4 cm 1x10-4 cm = 34.5 fF
(3.9)8.854x10
-14
(
)(
)
(
)(
)
97
4.74 CSB = C j AS + C jsw PS | CDB = C j AD + C jsw PD | AS = 502 = 12.5m 2 | PS = 30 = 15m
Cj =
s
w do
| w do =
N N 10 201016 2s 1 1 + j | j = VT ln A 2 D = 0.025ln = 0.921V 20 q NA NA 10 ni
2(11.7)(8.854 x10-14 ) 1 1 -5 w do = 20 + 16 0.921 = 3.45x10 cm -19 10 10 1.602x10 Cj = CSB
(11.7)(8.854 x10-14 F /cm)
-5
3.45x10 cm = (3.00x10-8 F /cm 2 )( 12.5x10-8 cm 2 )+ 5x10-4 cm(3.00x10-8 F /cm 2 )( 15x10-4 cm)= 26.3 fF
= 3.00x10-8 F /cm 2
CDB = CSB = 26.3 fF
4.75
KP = K 'n = K n
L A 0.25m = 175 2 = 8.75U | VTO = VTN = 0.7 W V 5m
PHI = 2 F = 0.8V | L = 0.25U | W = 5U | LAMBDA = 0.02
4.76 (a) VTO = 0.7 | PHI = 2 F = 0.6 | GAMMA = 0.75
(b) VTO = 0.74
| PHI = 0.87 | GAMMA = 0.84
4.77 KP = K 'n = 50U VTO = VTN = 1 V L = 0.5U W = 2.5U LAMBDA = 0 4.78 KP = K 'n = 10U VTO = VTN = 1 V L = 0.6U W = 1.5U LAMBDA = 0 4.79 KP = K 'p = 10U VTO = VTP = -1 V L = 0.5U
Using the - 3 - V curve, K P = 2 50A
[-3 - (-1)]
2
= 25
A
V2
W = 1.25U LAMBDA = 0
4.80 KP = K 'n = 25U VTO = VTN = 1 V L = 0.6U W = 0.6U LAMBDA = 0
98
NMOS i-v Characteristics for Load-Line Problems
800
5V
600
Drain Current (uA)
400
4V
200 3V
2V 0
0 1 2 3 4 5 6
Drain Voltage (V)
99
4.81
For VDS = 0, ID =
4V = 0.588mA. For ID = 0,VDS = 4V . 6.8k
Also, VGS = 4V. From the graph, the transistor is operating below pinchoff in the triode region and the Q-point is Q-point: (350 A, 1.7V)
800 5V
600
Drain Current (uA)
Q-point (4.82)
400 4V
200
Q-point (4.81)
3V
2V 0
0 1 2 3 4 5 6
Drain Voltage (V)
4.82
For VDS = 0, ID =
5V = 0.602mA. For ID = 0,VDS = 5V . 8.3k
For VGS = 5V, the Q-point is (450 A, 1.25 V). From the graph in Prob. 4.81, the transistor is operating below pinchoff in the triode region.
4.83
800 5V
600
Drain Current (uA)
Q-point (4.84)
400 4V
200
Q-point (4.83)
3V
2V 0
0 1 2 3 4 5 6
Drain Voltage (V)
100
VDD = 3V | 6 =10 4 ID +V DS | VDS = 0, ID = 0.6mA | ID = 0, VDS = 6V 2 From the graph, Q-pt: (140 A, 4.6V) in the saturation region. VGS =
4.84
VDD = 4V | 8 =10 4 ID +V DS | VDS = 6, ID = 0.2mA | VDS = 0, ID = 0.8mA 2 See graph for Problem 4.83: Q-pt: (390 A, 4.1 V) in saturation region. VGS =
4.85
(a)
100k 12V = 3.75V | Assume saturation 100k + 220k -6 5 2 3 3 100x10 3.75 = VGS + 24 x10 ID = VGS + 24 x10 (VGS -1) 2 1 VGG =
2 6VGS -11VGS + 2.25 = 0 VGS = 1.599V and ID = 89.7A
VDS = 12 - 36x10 3 ID = 8.77V | VDS > VGS - VTN Saturation is correct. Checking : VGG = 24 x10 3 ID + VGS = 3.75V which is correct. Q - point : (89.7 A, 8.77 V )
(b)
Assume saturation
3 3
100x10-6 10 2 3.75 = VGS + 24 x10 ID = VGS + 24 x10 (VGS -1) 2 1
2 12VGS - 23VGS + 8.25 = 0 VGS = 1.439V and ID = 96.4 A
VDS = 12 - 36x10 3 ID = 8.53V | VDS > VGS - VTN Saturation is correct. Checking : VGG = 24 x10 3 ID + VGS = 3.75V which is correct. Q - point : (96.4 A, 8.53 V )
4.86
VGG =
10k 12V = | 3.75V Assume saturation 10k + 22k 100x10-6 20 2 3.75 = VGS + 2.4 x10 3 ID = VGS + 24 x10 3 (VGS -1) 2 1
2 2.4VGS - 3.8VGS + 1.35 = 0 VGS = 1.882V and ID = 778A
VDS = 12 - 3.6x10 3 ID = 9.20V | VDS > VGS - VTN Saturation is correct. Checking : VGG = 2.4 x10 3 ID + VGS = 3.75V which is correct. Q - point : (778 A, 9.20 V )
101
4.87
VGG =
1M 12V = 3.75V | Assume saturation 1M + 2.2M 100x10-6 5 2 3.75 = VGS + 2.4 x10 3 ID = VGS + 2.4 x10 5 (VGS -1) 2 1
2 60VGS -119VGS + 56.25 = 0 VGS = 1.206V and ID = 10.6A
VDS = 12 - 3.6x10 5 ID = 8.18V | VDS > VGS - VTN Saturation is correct. Checking : VGG = 2.4 x10 5 ID + VGS = 3.75V which is correct. Q - point : (10.6 A, 8.18 V )
4.88
(a)
100k 15V = 4.69V | Assume Saturation 100k + 220k 100x10-6 5 2 4.69 = VGS + 24 x10 3 ID = VGS + 24 x10 3 (VGS -1) 2 1 VGG =
2 6VGS -11VGS + 1.31 = 0 VGS = 1.705V and ID = 124 A
VDS = 15 - 36x10 3 ID = 10.5 V | VDS > VGS - VTN Saturation is correct. Checking : VGG = 24 x10 3 ID + VGS = 4.68V which is correct. Q - point : (124 A, 10.5 V ) (b) VGG = 4.69V | Assume Saturation
100x10-6 10 2 4.69 = VGS + 24 x10 3 ID = VGS + 24 x10 3 (VGS -1) 2 1
2 12VGS - 23VGS + 7.31 = 0 VGS = 1.514V and ID = 132 A
VDS = 15 - 36x10 3 ID = 10.3 V | VDS > VGS - VTN Saturation is correct. Checking : VGG = 24 x10 3 ID + VGS = 4.68V which is correct. Q - point : (132 A, 10.3 V )
4.89
(a) VGG =
200k 12V = 3.81V | Assume Saturation 200k + 430k 100x10-6 5 2 3.81 = VGS + 47x10 3 ID = VGS + 47x10 3 (VGS -1) 2 1
2 23.5VGS - 45VGS + 15.88 = 0 VGS = 1.448V and ID = 50.3 A
VDS = 12 - 71x10 3 ID = 8.43 V | VDS > VGS - VTN Saturation is correct. Checking : VGG = 47x10 3 ID + VGS = 3.81V which is correct. Q - point : (50.3 A, 8.43 V )
102
(b) VGG = 3.81V
| Assume Saturation
3 3
100x10-6 15 2 3.81 = VGS + 47x10 ID = VGS + 47x10 (VGS -1) 2 1
2 70.5VGS -139VGS + 62.88 = 0 VGS = 1.269V and ID = 54.3 A
VDS = 12 - 71x10 3 ID = 8.15 V | VDS > VGS - VTN Saturation is correct. Checking : VGG = 47x10 3 ID + VGS = 3.82V which is correct. Q - point : (54.3 A, 8.15 V )
4.90
(a) Setting KP = 500U, VTO = 1, and GAMMA = 0 yields ID = 89.6 A, VGS = 1.60 V and VDS = 8.77 V 4 (a) Setting KP = 1000U, VTO = 1, and GAMMA = 0 yields ID = 96.3 A, VGS = 1.44 V and VDS = 8.53 V 4
4.91
(a) Setting KP = 500U, VTO = 1, and GAMMA = 0 yields ID = 124 A, VGS = 1.71 V and VDS = 10.5 V 4 (a) Setting KP = 1000U, VTO = 1, and GAMMA = 0 yields ID = 132 A, VGS = 1.51 V and VDS = 10.2 V 4
4.92
(a) Setting KP = 500U, VTO = 1, and GAMMA = 0 yields ID = 50.2 A, VGS = 1.45 V and VDS = 8.43 V 4 (a) Setting KP = 1000U, VTO = 1, and GAMMA = 0 yields ID = 54.1 A, VGS = 1.27 V and VDS = 8.16 V 4
4.93 (300 k, 700 k) or (1.2 M, 2.8 M). We normally desire the current in the gate bias network to be much less than ID. We also usually like the parallel combination of R1 and R2 to be as large as possible. 4.94
35x10-6 2 (a) ID = (4 -1-1700ID ) and using the quadratic equation, 2 ID = 134A. VDS =10 -134 x10-6 (1700 + 38300) = 4.64V
(b)
25x10-6 2 (4 - 0.75 -1700ID ) and using the quadratic equation, 2 ID = 116A. VDS =10 -116x10-6 (1700 + 38300) = 5.36V ID =
103
4.95
(a) Example 4.3 Setting KP = 25U and VTO = 1 yields ID = 34.4 A, VGS = 2.66 V and VDS = 6.08 V 4 Results agree with hand calculations (b) Example 4.4 Setting KP = 25U and VTO = 1 yields ID = 99.2 A, VGS = 3.82 V and VDS = 6.03 V 4 Results are almost identical to hand calculations
4.96 ' K n = 100A /V 2 | VTN = 0.75V | Choose VDS = VR D = VR S = 4V and VGS - VTN = 1V
RS =
4 4.1 = 40k 39k and VR S = 3.9V | RD = = 41k 43k 100A 100A 2ID 2I W 2 = 1V and K n = D = 200A /V 2 = | 2 Kn 1V L 1
VGS - VTN =
VG = VS + VGS = 3.9 + 1+ 0.75 = 5.65V 5.65V = 5.65V = 12V R1 R R 12V 12V | 5.65V = 1 2 = 530k 560k | R2 = 250k 5.65V R1 + R2 R1 + R2 R2 R1 12V R1 = 500k 510k R1 + 560k W 2 = L 1
R1 = 510k, R2 = 560k, RS = 39k, RD = 43k,
4.97 ' K n = 100A /V 2 | VTN = 0.75V | Choose VDS = VR D = VR S = 3V and VGS - VTN = 1V
3 3 = 12k | RD = = 12k 0.25mA 0.25mA 2ID 2I W 5 VGS - VTN = = 1V and K n = D = 500A /V 2 = 2 Kn 1V L 1 RS = VG = VS + VGS = 3 + 1+ 0.75 = 4.75V 4.75V = 4.75V = 9V R1 R R 9V 9V | 4.75V = 1 2 | R2 = 250k = 473k 470k 4.75V R1 + R2 R1 + R2 R2 R1 9V R1 = 525k 510k R1 + 470k W 5 = L 1
R1 = 510k, R2 = 470k, RS = 12k, RD = 12k,
104
4.98 ' K n = 100A /V 2 | VTN = 0.75V | Choose VDS = VR D = VR S = 5V and VGS - VTN = 1V
5 5 = 10k | RD = = 10k 0.5mA 0.5mA 2ID 2I W 10 VGS - VTN = = 1V and K n = D = 1mA /V 2 = 2 Kn 1V L 1 RS = VG = VS + VGS = 5 + 1+ 0.75 = 6.75V 6.75V = 6.75V = 15V R1 R R 15V 15V | 6.75V = 1 2 = 1.33M 1.5M | R2 = 600k 6.75V R1 + R2 R1 + R2 R2
R1 15V R1 = 1.23M 1.2M R1 + 1.5M W 10 R1 = 1.2M, R2 = 1.5M, RS = 10k, RD = 10k, = L 1
4.99
10-3 2 Assume Saturation. For IG = 0, VGS = -10 4 ID = -10 4 (VGS + 5) 2
2 5VGS + 51VGS + 125 = 0 VGS = -4.10 V and ID = 410 A
VDS = 15 -15000ID = 8.85 V | VDS > VGS - VTN so saturation is ok. Q - Point : (410A, 8.85V)
4.100
6x10-4 2 Assume Saturation. For IG = 0, VGS = -27x10 3 ID = -27x10 4 (VGS + 4 ) 2
2 8.1VGS + 65.8VGS + 129.6 = 0 VGS = -3.36 V and ID = 124 A
VDS = 12 - 78000ID = 2.36 V | VDS > VGS - VTN so saturation is ok. Q - Point : (124 A, 2.36V)
4.101
Kn = 1 mA/V 2 V TN = -5 V ID RD VDD +
IG = 0 + V RG
GS
-
5V -
+
15 V
R
S
105
1mA/V 2 2 Assume Saturation. IG = 0. 250A = (VGS + 5) 2 0.25mA 4.29V = -4.29V | RS = = 17.2k 18k 0.5mA 0.25mA 15 - 5 - 4.29 V VDS = 15 - ID (RD + RS ) RD = = 22.88k 24k 0.25 mA RG is arbitrary but normally fairly large. Choose RG = 510 k. VGS = -5 +
4.102
+ R 2 R
D
5V + 5V + 15 V VDD
+
R1
R
S
5V -
Assume Saturation. IG = 0. Assume power supply is split in thirds: VDS = VR D = VRS = 5V Note that although, this is a depletion - mode device, I D exceeds Kn 2 VTN and this will require VGS > 0. 2 0.25mA/V 2 5V 2 RS = = 2.5k 2.4k and VRS will be 4.8 V | 2mA = (VGS + 2) 2 2mA 2mA = +2.00V | VG = VS + VGS = 4.8 + 2 = 6.8V 0.125mA
VGS = -2 + 6.8 = 15
R1 R1 = 680 k and R2 = 820 k is one convenient possibility. R1 + R2 Another is R1 = 68 k and R2 = 82 k. Both choices have IR 2 << I D . RD =
(15 - 5 - 4.8)V
2mA
= 2.60k 2.70 k.
106
4.103
(a)
The transistor is saturated by connection. 100x10-6 10 A 2 2 (VGS - 0.75V ) 1 V 2
VGS = 12 -10 5 ID and ID =
2 50VGS - 74VGS + 16.13 = 0 VGS = 1.214V, - 0.266V VGS = 1.214 V since VGS must
exceed the threshold voltage. | ID =
100x10-6 10 A 2 2 (1.214 - 0.75V ) 1 V 2 12 -1.21 ID = 104A | Checking : ID = = 108 A | Q - Point : (108 A,1.21 V) 10 5
7
(b) Using KVL, VDS = 10 IG +VGS. But, since IG = 0, VGS = VDS. Also VTN = 0.75 V > 0, so the transistor is saturated by connection.
+12 V W = L 1 10 330 k IDS I
G
ID =
10 M
' 100 A 10 Kn W 2 2 (VGS - 0.75) (VGS - VTN ) = 2 2 V 1 2 L
VGS = 12 - 330k(ID + IG ) -10M(IG ) but I G = 0 VGS = 12 - 330k(ID ) 1.00x10-3 A 2 VGS = 12 - (3.30x10 5 ) V - 0.75) 2 ( GS 2 V
2 165VGS - 246.5VGS + 80.81 = 0 yields VGS = 1.008V , 0.486V
+ VDS
+ V GS -
-
VGS must be 1.008 V since 0.486 V is below threshold.
100 A 10 2 ID = (1.008 - 0.75) = 33.3 A and VDS = VGS 2 2 V 1
Q-Point: (33.3 A, 1.01 V)
4.104
Checking: ID =(12-1.01)V/330k = 33.3 A. 4
(a)
The transistor is saturated by connection. and ID = 100x10-6 20 A 2 2 (VGS - 0.75V ) 1 V 2 100x10-6 20 A 2 2 (1.08 - 0.75V ) = 109A 1 V 2
VGS = 12 -10 5 ID
2 100VGS -149VGS + 44.25 = 0 VGS = 1.08V, 0.410V VGS = 1.08 V since VGS must
exceed the threshold voltage. ID = Checking : ID =
12 -1.08 = 109 A | Q - Point : (109 A, 1.08 V) 10 5
7
(b) Using KVL, VDS = 10 IG +VGS. But, since IG = 0, VGS = VDS. Also VTN = 0.75 V > 0, so the transistor is saturated by connection.
107
+12 V W = L 1 10 330 k IDS I
G
ID =
10 M
' 100 A 20 Kn W 2 2 (VGS - 0.75) (VGS - VTN ) = 2 2 V 1 2 L
VGS = 12 - 330k(ID + IG ) -10M(IG ) but VGS = 12 - 330k(ID ) A 2 VGS = 12 - (3.30x10 5 ) 10-3 2 (VGS - 0.75) V
IG = 0
+ VDS
+ V GS -
-
2 3.3VGS - 4.94VGS + 1.736 = 0 yields VGS = 0.933V, 0.564V
VGS must be 0.933 V since 0.564 V is below threshold.
100 A 20 12 - 0.933 2 ID = V = 33.5A (0.933 - 0.75) = 33.5 A Checking : 2 2 V 1 330k
and VDS = VGS :
4.105
Q-Point: (33.5 A, 0.933 V)
' 100 A 10 Kn W 2 2 (VGS - 0.75) (VGS - VTN ) = 2 2 V 1 2 L
(a) Assume saturation :
ID =
VGS = 15 - 330k(ID + IG ) -10M(IG ) but I G = 0 VGS = 15 - 330k(ID ) and VGS = VDS so saturation regioin operation is correct 10-3 A 2 VGS = 15 - (3.30x10 5 ) V - 0.75) 2 ( GS 2 V
2 3.30VGS - 4.93VGS + 1.556 = 0 yields VGS = 1.041V, 0.453V 100 A 10 15 -1.041 2 ID = V = 42.3A (1.041- 0.75) = 42.3A | Checking : ID = 2 2 V 1 330k
Q - Point : (42.3 A, 1.04 V)
(b) Assume saturation ID =
' 100 A 25 Kn W 2 2 (VGS - 0.75) (VGS - VTN ) = 2 2 V 1 2 L
VGS = 15 - 330k(ID + IG ) -10M(IG ) but I G = 0 VGS = 15 - 330k(ID ) and VGS = VDS so saturation region operation is correct. 2.50x10-3 A 2 V - 0.75) VGS = 15 - (3.30x10 5 ) 2 ( GS V 2
2 8.25VGS -12.355VGS + 4.341 = 0 yields VGS = 0.9345V, 0.563V 100 A 25 15 - 0.9345 2 V = 42.6A ID = (0.9345 - 0.75) = 42.6A | Checking : ID = 2 2 V 1 330k
Q - Point :
(42.6 A, 0.935 V)
108
4.106
(a)
Asssume saturation
ID =
' 100 A 10 Kn W 2 2 (VGS - 0.75) (VGS - VTN ) = 2 2 V 1 2 L
VGS = 12 - 470k(ID + IG ) -10M(IG ) but IG = 0 VGS = 12 - 470k(ID ) and VGS = VDS so saturation region operation is correct. 10-3 A 2 VGS = 12 - (4.70x10 5 ) V - 0.75) 2 ( GS 2 V
2 4.70VGS - 7.03VGS + 2.404 = 0 yields VGS = 0.9666V, 0.529V 100 A 10 12 - 0.967 2 ID = = 23.5A (0.9666 - 0.75) = 23.5A | Checking : ID = 2 2 V 1 470k
Q - Point : (23.5 A, 0.967 V) (b) Since the current in RG is zero, the drain current is independent of RG.
4.107 (a) Create an M-file: function f=bias(id) vtn=1+0.5*(sqrt(22e3*id)-sqrt(0.6)); f=id-(25e-6/2)*(6-22e3*id-vtn)^2; fzero('bias',1e-4) yields ans = 8.8043e-05
(b) Modify the M-file: function f=bias(id) vtn=1+0.75*(sqrt(22e3*id)-sqrt(0.6)); f=id-(25e-6/2)*(6-22e3*id-vtn)^2; fzero('bias',1e-4) yields ans = 8.3233e-05
4.108 Using a spreadsheet similar to Table 4.2 yields: (a) 88.04 A, (b) 83.23 A. 4.109
100k 12V = 3.75V | Assume saturation 100k + 220k 2ID VTN = 1+ 0.6 24 x10 3 ID + 0.6 - 0.6 | VGS = VTN + 5x10-4 3.75 - VGS ID = | Solving iteratively yields ID = 73.1A with VTN = 1.460V 24k V = 12V - ID (24k + 12k) = 9.37 V Transistor is saturated. Q - Point : (73.1 A, 9.37 V) VGG =
(
)
109
4.110
VGG =
100k 12V = 3.75V | Assume saturation 100k + 220k 0.6
(a) VTN = 1+ 0.75( 24 x10 3 ID + 0.6 -
ID = VDS
)|
VGS = VTN +
2ID 5x10-4
3.75 - VGS | Solving iteratively yields ID = 69.7A with VTN = 1.550V 24k = 12V - ID (24k + 12k) = 9.49 V The transistor is saturated. Q - Point : (69.7 A, 9.49 V) 0.6
(b) VTN = 1+ 0.6( 24 x10 3 ID + 0.6 -
ID =
)|
VGS = VTN +
2ID 5x10-4
3.75 - VGS | Solving iteratively yields ID = 73.1A with VTN = 1.460V 24k V = 12V - ID (24k + 24k) = 8.49 V The transistor is saturated. Q - Point : (73.1 A, 8.49 V)
4.111 (a) = 0.6
VTN = 1.46 V VTN = 1.55 V VTN = 1.46 V
ID = 73.1 A ID = 69.7 A ID = 73.1 A
VDS = 9.37 V VDS = 9.49 V VDS = 8.49 V
(b) = 0.75 (c ) = 0.6
These results all agree with the hand calculations. (They should - they are all solving the same sets of equations.)
4.112 (a) = 0
VTN = 1.00 V
ID = 50.2 A ID = 42.2 A ID = 54.1 A ID = 45.2 A
VDS = 8.43 V VDS = 9.01 V VDS = 8.16 V VDS = 8.79 V
= 0.5 VTN = 1.42 V (b) = 0 VTN = 1.00 V = 0.5 VTN = 1.44 V
The = 0 values agree with the hand calculations in the original problem. Including body effect in the simulations reduces the Q-point current by approximately 15%. Although this may sound large, it is within the error that will be introduced by the use of 5% resistors and typical device tolerances. So, we normally omit in our hand calculations, and then refine the results using SPICE.
4.113 (a) = 0
VTN = 1.00 V
ID = 89.6 A ID = 75.5 A ID = 96.3 A ID = 80.8 A
VDS = 8.77 V VDS = 9.28 V VDS = 8.53 V VDS = 9.09 V
= 0.5 VTN = 1.39 V (b) = 0 VTN = 1.00 V = 0.5 VTN = 1.41 V
The = 0 values agree with the hand calculations in the original problem. Including body effect in the simulations reduces the Q-point current by approximately 15%. Although this may sound
110
large, it is within the error that will be introduced by the use of 5% resistors and typical device tolerances. So, we normally omit in our hand calculations, and then refine the results using SPICE.
4.114 =0
= 0.5
VTN = 1.00 V VTN = 1.35 V
ID = 778 A ID = 661 A
VDS = 9.20 V VDS = 9.62 V
The = 0 values agree with the hand calculations in the original problem. Including body effect in the simulations reduces the Q-point current by approximately 15%. Although this may sound large, it is within the error that will be introduced by the use of 5% resistors and typical device tolerances. So, we normally omit in our hand calculations, and then refine the results using SPICE.
4.115 =0
= 0.5
VTN = 1.00 V VTN = 1.45 V
ID = 10.5 A ID = 9.18 A
VDS = 8.03 V VDS = 8.69 V
The = 0 values agree with the hand calculations in the original problem. Including body effect in the simulations reduces the Q-point current by approximately 15%. Although this may sound large, it is within the error that will be introduced by the use of 5% resistors and typical device tolerances. So, we normally omit in our hand calculations, and then refine the results using SPICE.
4.116
(a) Both transistors are saturated by connection and the two drain currents must be equal. K K 2 2 ID1 = n1 (VGS1 - VTN1 ) and ID 2 = n 2 (VGS 2 - VTN 2 ) 2 2 But since the transistors are identical, ID1 = ID2 requires VGS1 = VGS2 = VDD/2 = 2.5V. 100x10-6 20 2 ID1 = ID 2 = (2.5 -1) = 2.25 mA 1 2 (b) For this case, the same arguments hold, and VGS1 = VGS2 = VDD/2 = 5V. 100x10-6 20 2 ID1 = ID 2 = (5 -1) =16.0 mA 1 2 . (c) For this case, the threshold voltages will be different due to the body-effect in the upper transistor. The drain currents must be the same, but the gate-source voltages will be different: VGS1 = VTN1 + VTN1 =1V 2ID 2ID ; VGS 2 = VTN 2 + ; VGS1 + VGS 2 = 5V . Kn Kn VTN 2 = 1+ 0.5 VGS1 + 0.6 - 0.6
(
)
111
Combining these equations yields 5 - 2VGS1 - 0.5 VGS1 + 0.6 - 0.6 = 0 VGS1 = 2.27V ; VGS 2 = 5 - VGS1 = 2.73V ID 2 = ID1 = 100x10-6 20 2 (2.27 -1) = 1.61 mA. 1 2
(
)
Checking : VTN 2 = 1+ 0.5 2.27 + 0.6 - 0.6 = 1.46V ID 2 = 100x10-6 20 2 (2.73 -1.46) = 1.61 mA. 1 2
(
)
4.117 If we assume saturation, we find ID = 234 A and VDS = 0.65 V, and the transistor is not saturated. Assuming triode region operation,
VGS = 10 - 2x10 4 ID | VDS = 10 - 4 x10 4 ID A 2 10 - 4 x10 4 ID 4 ID = 100 2 10 - 2x10 4 ID -1- 10 ( - 4 x10 ID ) 1 2 V VDS = 10 - 4 x10 4 (2.42x10-4 )= 0.320V | Q - Pt : (242 A, 0.320V ) Checking the operating region : VGS - VTN = 4.16V > VDS and the triode region assumption is correct. Checking : ID = 10 - 0.32 V = 242A 40k Collecting terms : 16.5x10 4 ID = 40 ID = 242 A
4.118 If we assume saturation, we find an inconsistent answer. Assuming triode region operation, VGS = 10 - 2x10 4 ID | VDS = 10 - 3x10 4 ID A 4 10 - 3x10 4 ID 4 ID = 100 2 10 - 2x10 4 ID -1- 10 ( - 3x10 ID ) 2 V 1
VDS = 10 - 3x10 4 (3.22x10-4 )= 0.340V | Q - Pt : (322 A, 3.18 V ) Checking the operating region : VGS - VTN = 2.56V > VDS and the triode region assumption is correct. Checking : ID = 10 - 0.34 V = 322A 30k
2 Collecting terms : 1.5x10 8 ID -1.725x10 5 ID + 40 = 0 ID = 322A
112
4.119 For (a) and (b), = 0. The transistor parameters are identical so 3VGS = 15V or VGS = 5V. 20 1 2 (a) ID = ( 100x10-6 ) (5 - 0.75) = 18.1 mA 1 2 50 1 2 100x10-6 ) (5 - 0.75) = 45.2 mA (b) ID = ( 1 2 (c) Now we have three different threshold voltages and need an iterative solution. Using MATLAB: function f=Prob112(id) gamma=0.5; vgs1=.75+sqrt(2*id/2e-3); vtn2=0.75+gamma*(sqrt(vgs1+0.6)-sqrt(0.6)); vgs2=vtn2+sqrt(2*id/2e-3); vtn3=0.75+gamma*(sqrt(vgs1+vgs2+0.6)-sqrt(0.6)); vgs3=vtn3+sqrt(2*id/2e-3); f=15-vgs1-vgs2-vgs3; fzero('Prob112',1e-4) --> ans = 0.0130 ID = 13.0 mA 4.120
W 20 = VTN = 0.75 V ID = 18.1 mA VDS = 5.00 V 1 L W 50 = VTN = 0.75 V ID = 45.2 mA VDS = 5.00 V (b) = 0 L 1 W 20 = VTN 3 = 1.95 V ID 3 = 13.0 mA VDS 3 = 5.56 V (b) = 0.5 L 1 VTN 2 = 1.48 V ID 2 = 13.0 mA VDS 2 = 5.09 V VTN1 = 0.75 V ID1 = 13.0 mA
(a) = 0
VDS1 = 4.36 V
Results are identical to calculations in Prob. 4.119
4.121 For VGS = 5 V and VDS = 0.5 V, the transistor will be in the triode region. (5 - 0.5)V = 54.88A | 54.88x10-6 = 100x10-6 W 5 - 0.75 - 0.5 0.5 | W = 0.274 = 1 ID = L 2 L 1 3.64 82k 4.122 For VGS = 3.3 V and VDS = 0.25 V, the transistor will be in the triode region. (3.3 - 0.25)V = 16.94 A | 16.94 x10-6 = 100x10-6 W 3.3 - 0.75 - 0.25 0.25 | W = 0.280 = 1 ID = L 2 L 1 3.57 180k
113
4.123 (a) The transistor is saturated by connection. For this circuit, VGS = VDD + ID R = -15 + 75000ID
4 x10-5 1 2 ID = (-15 + 75000ID + 0.75) 153 A 2 1 VGS = -15 + 75000ID = -3.525V VDS = VGS = -3.525V | Q - point : (153 A,-3.53 V )
(b) Here the transistor has VGS = -15 V, a large value, so the transistor is most likely operating in the triode region.
VDS - (-15) V = 4 x10-5 -15 - (-0.75) - DS VDS VDS = -0.347 V and ID = 195 A. 75000 2 15 - 0.347 V = 195A Q - point : (195 A,-0.347 V ) Checking : ID = 785k Checking the region of operation: VDS = -0.347V > VGS - VTP = -15 + 0.75 = -14.25V ID =
Triode region is correct
4.124 Set W=1U L=1U KP=40U VTO=-0.75 GAMMA=0 Results are almost identical to hand calculations for both parts. 4.125 (a) IDP = IDN , and both transistors are saturated by connection. 10 = -VGSP + VGSN 1 100A 20 1 40A 20 2 2 2 (-10 + VGSN + 0.75) = (VGSN - 0.75) 2 2 V 1 2 V 1
(9.25 - VGSN ) =
2.5 (VGSN - 0.75) VGSN = 4.04V | VGSP = -5.96V
IDP = IDN = 10.8 mA | VO = VGSN = 4.04V (b) Everything is the same except the currents scale by 80/20: IDP = IDN = 43.2 mA
4.126 For (a) and (b), = 0. The transistor parameters are identical so -3VGS = 15V or VGS = -5V.
1 40 (40x10-6 ) 1 (-5 + 0.75)2 = 14.4 mA 2 1 75 2 (b) ID = (40x10-6 ) (-5 + 0.75) = 27.1 mA 2 1 (a) ID = (c) Now we have three different threshold voltages and need an iterative solution. Using MATLAB: function f=PMOSStack(id) gamma=0.5;
114
vsg1=.75+sqrt(2*id/1.6e-3); vtp2=-0.75-gamma*(sqrt(vsg1+0.6)-sqrt(0.6)); vsg2=-vtp2+sqrt(2*id/1.6e-3); vtp3=-0.75-gamma*(sqrt(vsg1+vsg2+0.6)-sqrt(0.6)); vsg3=-vtp3+sqrt(2*id/1.6e-3); f=15-vsg1-vsg2-vsg3; fzero('PMOSStack',1e-1) --> ans = 0.0104 ID = 10.4 mA.
4.127 (a) W=40U L=1U KP=40U VTO=-0.75 GAMMA=0 (b) W=75U L=1U KP=40U VTO=-0.75 GAMMA=0 (c) W=75U L=1U KP=40U VTO=-0.75 GAMMA=0.5 Results agree with hand calculations. 4.128
4V = 2mA. For ID = 0,VDS = -4V . (VSD = +4V ) 2k 300k VGS = VEQ = -4V = -3V (VSG = +3V ) 300k + 100k From the graph, the transistor is operating below pinchoff in the linear region. For VDS = 0, ID =
5000 VSG = 5 V
4000
A)
3000
V
Drain Current (
SG
=4V
2000
Q-point
1000
VSG = 3 V V =2V
SG
0
-1000
-1
0
1
2
3
4
5
6
Source-Drain Voltage (V)
Q-point: (1.15 mA, 1.7V)
115
PMOS Transistor Output Characteristics
5000
V GS = -5 V
4000
Drain Current ( A)
3000
VGS = -4 V
2000
V GS = -3 V
1000
V GS = -2 V
0
-1000
-1 0 1 2 3 4 5 6
Source-Drain Voltage (V)
116
4.129
(a) VGG =
4 x10-5 20 15V 2 = 7.5V | 7.5 = 10 5 ID - VGS | 7.5 = 10 5 (VGS + 0.75) - VGS 2 2 1
15 VDS = -( - (100k + 50k)ID ) = -5.47V | Q - point : (59.78 A,-5.47 V ) (b) For saturation, VDS VGS - VTP or VSD VSG + VTP 15 - (100k + R)ID 7.5 -100kID - 0.75 R 130 k
4.130 Setting W=20U, L=1U, LEVEL=1, KP=40U, VTO=-0.75 yields results identical to the previous problem. 4.131 (a) Using MATLAB: function f=bias4(id) gamma=0.5; vbs=1e5*id; vgs=-7.5+vbs; vtp=-0.75-gamma*(sqrt(vbs+0.6)-sqrt(0.6)); f=id-(8e-4/2)*(vgs-vtp)^2; fzero('bias4',4e-5) --> ans = 5.5278e-05 --> ID = 55.3 A VDS = -15 + (100k+R) ID (b) VDS VGS - VTP | -15 + (100k+R)ID -1.972 + 1.600 |
2 4VGS + 5.9VGS -1.5 = 0 VGS = -1.148V and ID = 63.5A
R 164 k
4.132 Setting W=20U, L=1U, LEVEL=1, KP=40U, VTO=-0.75 GAMMA=0.5 yields results identical to the previous problem. 4.133 (a)
V
DD
R V + -
GS
S
VDS
G D
I
SD
+
The arrow identifies the transistor as a PMOS device. Since = 0, we do not need to worry about body effect: VTP = VTO. Since VDS = VGS, and VTP < 0, the transistor is saturated.
117
K 'p W 2 ID = (VGS - VTP ) and - VGS = 12 -105 ID 2 L 4 x10-5 10 2 V -VGS = 12 -10 5 ( GS - (-0.75)) 2 1
2 20VGS + 29VGS - 0.75 = 0 yields VGS = -1.475V,+0.0255V We require VGS < VTP = -0.75 V for the transistor to be conducting so 2 4 x10-5 10 A VGS = -1.475V and ID = 2 (-1.475 - (-0.75)) = 105 A 2 1 V Since VDS = VGS, the Q-point is given by: Q-Point = (105 A, -1.475 V).
(b) Using MATLAB for the second part (Set gamma = 0 for part (a)): function f=bias2(id) gamma=1.0; vgs=-12+1e5*id; vsb=-vgs; vtp=-0.75-gamma*(sqrt(vsb+0.6)-sqrt(0.6)); f=id-5e-5*(vgs-vtp)^2; fzero('bias2',1e-4) --> ans = 9.5996e-05 and Q-Point = (96.0 A, 2.40 V).
4.134
-5 2 15V 40 4 4 4x10 = = 7.5V | 7.5 = 5x10 I D - VGS | 7.5 = 5x10 V ( GS + 0.75) - VGS 2 2 1
VGG
VDS = - 15 - (R + 50k)I D = -5V R = 22.3 k
4.135
2 890VGS + 119VGS - 30 = 0 VGS = -1.166V and I D = 138 A
(
)
I D = 138 A
VGG =
4x10-5 40 2 15V = 7.5V | 7.5 = 15 - 5x10 4 I D + VGS | 7.5 = 5x10 4 V ( GS - VTP ) - VGS 2 2 1
VTP = -0.75 - 0.5 5x10 4 I D + 0.6 - 0.6
(
)
VDS = - 15 - (R + 50k)I D = -5V R = 40.1 k
Solving iteratively yields I D = 111 A with VTP = -1.60V
(
)
118
4.136
510k = 9.81V | 9.81 = 15 -105 I D + VGS 510k + 270k 4x10-5 20 2 5.19 = 105 V ( GS + 0.75) - VGS 2 1 (a) VGG = 15V
2 40VGS + 59VGS + 17.31 = 0 VGS = -1.071 V and I D = 41.2 A
-15 + ( 100k + R)I D -1.071+ 0.75 R 256 k
4.137
(b) For saturation, VDS VGS - VTP
510k = 9.81V | 9.81 = 15 -105 I D + VGS 510k + 270k 4x10-5 20 2 5 5.19 = 105 V ( GS - VTP ) - VGS | VTP = -0.75 - 0.5 10 I D + 0.6 - 0.6 2 1 (a) VGG = 15V
(
)
Solving iteratively yields I D = 35.2 A with VTP = -1.38 V and VGS = -1.67 V -15 + ( 100k + R)I D -1.67 + 1.38 R 318 k
4.138 (a) Assume an equal voltage (5V) split between R D , RS and VDS. We need VDS VGS - VTP
or - 5 VGS - VTP . Choose VGS - VTP = -2V. Kn = VGS = -2 - 0.75 = -2.75V. VEQ = 5 - VGS = 7.75V. 7.75 = 15 7.75 = 15 RS = R1 15 R1 R2 15 = = 100k. R2 = 193.5k 200k. R1 + R2 R2 R1 + R2 R2 220k R1 R1 = 214k 220k. VEQ = 15 = 7.86V 220k + 200k R1 + R2
(b) For saturation, VDS VGS - VTP
(V
2I D
GS
- VTP )
2
=
2mA W 12.5 = . 4 L 1
7.86V - 2.75V 15 - 5 - 5.1 = 5.11k 5.1k | RD = = 4.9k 4.7k 1mA 1mA Note that R1 is connected between the gate and + 15 V, and R 2 is connected between the gate and ground. (b) For the NMOS case, choose W/L = 5/1. The resistors now have the same values
except R2 is now connected between the gate and +15 V, R1 is connected between the gate and ground, and RD = 15 - 6 - 5.1 = 3.9k 1mA
119
4.139 (a) Assume an equal voltage (3V) split between R D , RS and VDS. We need VDS VGS - VTP
or - 3 VGS - VTP . Choose VGS - VTP = -1V. Kn = VGS = -1- 0.75 = -1.75V. VEQ = 3 - VGS = 4.75V. 4.75 = 9 4.75 = 9 RS = R1 9 R1 R2 9 = = 1M. R2 = 1.7 M 1.8 M. R1 + R2 R2 R1 + R2 R2 2 M R1 R1 = 2.01M 2 M | VEQ = 9 = 4.74V 1.8 M + 2 M R1 + R2
(V
2I D
GS
- VTP )
2
=
1mA W 25 = . 1 L 1
4.74V -1.75V 9 -3-3 = 5.97k 6.2k | RD = = 6.0k 6.2k 0.5mA 0.5mA Note that R1 is connected between the gate and + 9 V, and R 2 is connected between the gate and ground. R1 = 2 M, R2 = 1.8 M, RS = RD = 6.2k, W / L = 25/1
(b) For the NMOS case, choose W/L
the gate and ground.
4.140
= 40/1. The resistors now have the same values
except R2 is now connected between the gate and + 9 V, and R1 is connected between
40 A 10 4 VGS = 10 4 I D | Assume saturation : I D = 10 I D - 4 2 2 V 1
(
)
2
VDS = - 15 -10 4 I D = -12.2V | Q - Pt : (281 A,-12.2 V )
2 Collecting terms : 108 I D - 8.5x10 4 I D + 16 = 0 I D = 281 A
(
)
Checking : VGS - VTP = 2 - 4 = -1.19 V | VDS = -12.2 | Saturation is correct.
4.141
VGS = 10 4 I D | VTP = 4 - 0.25 VGS + 0.6 - 0.6
(
)
| ID =
2 4x10-4 (VGS - VTP ) 2
VDS = - 15 -10 4 I D = -12.4V | Q - Pt : (260 A,-12.4 V )
Solving these equations iteratively yields I D = 260 A
(
)
120
4.142 Note: The answers are very sensitive to round-off error and are best solved iteratively using MATLAB, a spreadsheet, HP solver, etc. Hand calculations using the quadratic equation will generally yield poor results.
Saturated by connection with V = -1 TP -5 2 4x10 10 5 6 11 2 ID = 3.3x10 I D -12 - (-1) 121- 7.265x10 I D + 1.089x10 I D = 0 2 1
[
]
I D = 34.6A, 32.1A | VDS = 3.3x105 I D -12 = -0.582V ,-1.407V | Q - point : (32.1 A,-1.41 V )
since the transistor would not be conducting for V = -0.582V. GS
4.143 Note: The answers are very sensitive to round-off error and are best solved iteratively using MATLAB, a spreadsheet, HP solver, etc. Hand calculations using the quadratic equation will generally yield poor results. Saturated by connection with V = -3 TP 4x10-5 30 2 5 6 11 2 ID = 3.3x10 I D -12 - (-3) 81- 5.941x10 I D + 1.089x10 I D = 0 2 1
[
]
I D = 27.07A | VDS = 3.3x105 I D -12 = -3.067V | Q - point : (27.1 A,-3.07 V )
4.144 Note: The answers are very sensitive to round-off error and are best solved iteratively using MATLAB, a spreadsheet, HP solver, etc. Hand calculations using the quadratic equation will generally yield poor results. (a) Large VGS - Assume triode region.
VDS = 12 - 3.3x105 I D
Q - po int : (36.1 A, 80.6 mV) | VDS < VGS - VTN so triode region is correct. 2 10x10-6 25 5 b) Saturated by connection : I D = 3.3x10 I D -12 + 0.75 ( 2 1
10 V | I D = 40x10-6 12 - 0.75 - DS VDS 2 1
(
)
Q - point : (32.4A,-1.32V)
(c) V
ID =
TP
= -0.75 - 0.5 3.3x105 I D + 0.6 - 0.6
(
)
( )
Q - point : (28.8 A,-2.49 V)
40x10-6 25 5 3.3x10 I D -12 - VTP 2 1
(
)
2
| VDS = - 12 - 3.3x105 I D
121
4.145
(a) Kn = n
ID =
ox W
Tox
-14 cm2 3.9 8.854x10 F / cm = 500 V -s L 40x10-7 cm
(
) 20m = 432 A 2m V2
2 432A (4 -1) = 1.94 mA 2
(b) K
4.146
' n
= 2Kn
V | V = 2
'
2 864A 4 1 | I = - = 0.972 mA 2 2 2 ' D
(a) C
GC
3.9 8.854x10-14 F / cm = C WL = 20x10-7 cm
" OX ' | CGC =
(
) 20x10 (
-4
cm 10-4 cm = 34.5 fF
)(
)
(b) = 2
4.147
CGC
= 17.3 fF
fT = fT =
1 gm 2 CGC
| gm =
iD " W " = KP ( GS - VTP ) = pCOX V | CGC = COX WL vGS L
1 p (VGS - VTP ), but (VGS - VTP )< 0 for PMOS transistor. 2 L2 1 p Since f T should be positive, f T = (VGS - VTP ) 2 L2
4.148
1 V ( GS - VTN ) 2 L2 1 400cm2 /V - s fTN = (1V )= 6.37 GHz | fTP = 0.4 fTN = 2.55 GHz 2 10-4 cm 2 1 400cm 2 /V - s (b) fTN = 2 -5 2 (1V )= 637 GHz | fTP = 0.4 fTN = 255 GHz 10 cm
(a) f
T
=
(
)
(
)
122
4.149
(a) Kn = n
ID =
ox W
Tox
-14 cm2 3.9 8.854x10 F / cm = 400 V - s L 80x10-7 cm
(
) 2m = 345 A 0.1m V2
345A 2 (2) = 690A 2
" ox
3.9 8.854x10-14 F / cm W (b) I D = C 2 (VGS - VTN )vsat = 80x10-7 cm
4.150
(
) 2x10
cm 7 (2V ) 10 cm / s = 86.3A 2
-4
(
)
For VGS = 0, ID = 10-22 A. For VTN = 0.5 V and VGS = 0, ID = 10-15 A.
123