4.4 (a)


14
3.9o 
cm2 3.9 8.854x10 F / cm
K  nC  n
 n
 500

9
Tox
Tox
 V  sec  50x10 m100cm / m
'
n
ox
"
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
A 50nm

Kn'  34.5

4.8
a 0  0.8V  I D = 0

2
V 20nm
 86.3
A
V
2
| Kn'  34.5
A 50nm
 173
2
V 10nm
A
V
2
| Kn'  34.5
A 50nm
V
2
5nm
(b) VGS - VTN = 0.2V , VDS = 0.25V  Saturation region
200 A  5m 
2
K ' W 
V 
I D  n VGS  VTN  DS VDS  
1
0.8
 40.0 A





2
2 L 
2 
 2 V 0.5m 
(c) VGS - VTN = 1.2V , VDS = 0.25V  triode region
 A  5m 
W 
V 
0.25 
I D  Kn' VGS  VTN  DS VDS  200 2 
2  0.8 
0.25 538 A
L 
2 
2 
 V 0.5m 
(d ) VGS - VTN = 2.2V , VDS = 0.25V  triode region
 A  5m 
W 
V 
0.25 
I D  Kn' VGS  VTN  DS VDS  200 2 
3  0.8 
0.25 1.04 mA
L 
2 
2 
 V 0.5m 
W  A  5m 
mA
e Kn  Kn' L  200 V 2 0.5m  2.00 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
+0.2 V
10
1
D
G
+5
+
S
S
10
1
V
G
+5
W
=
L
I
+
V
-
-0 .2 V
V GS
DS
+
GS
(a)
I
B
V
DS
B
D
+
(b)
(a)
VGS  VG  VS  5V VDS  VD  VS  0.2V  Triode region operation
 A 10 
W 
V 
0.2 
I = I D  Kn' VGS  VTN  DS VDS  100 2  5  0.70 
0.2  840 A
L 
2 
2 
 V  1 
(b)

 345
A
V2
VGS  VG  VS  5  0.2 = 5.2V VDS  VD  VS  0  0.2 0.2V

 A 10 
0.2 
I = I S  100 2  5.2  0.70 
0.2  880 A
2 
 V  1 
4.15
1
1
a  Ron 

 23.0 
' W
 6  100 
K n VGS  VTN  100 x10 
5  0.65
L
 1 
VGS  4.35V
or VGS  0.44V
b  Ron
1
 35.7 
 6  100 
100 x10 
3.3  0.50
 1 
 2.8V
or VGS  0.28V

VGS
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 
375 A  5m 
A  5m 
mA
2 2
' W
 375 2 
 3.75 2
2 1 V  1.88 mA | K n  K n
2 
2 V 0.5um 
L
V 0.5um 
V
d  VGS  VTN
ID 

= 2 -1 = 1V and VDS  3.3 | VDS  VGS  VTN  so the saturation region is correct
= 3 -1 = 2V and VDS  3.3 | VDS  VGS  VTN  so the saturation region is correct
375 A  5m 
2 2
3 1 V  7.50 mA
2 
2 V 0.5m 
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
4.30
VDS > VGS - VTN so the transistor is saturated.
Kn
500 A
2
2
4 1 1 0.025  2.48 mA
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
4.33
(a) Since VDS = VGS and VTN > 0 for both transistors, both devices are saturated.
K n' W
K n' W
2
2
I

V

V
and
I

 GS1 TN 
VGS 2  VTN  .
Therefore
D1
D2
2 L
2 L
From the circuit, however, ID2 must equal ID1 since IG = 0 for the MOSFET:
K n' W
K n' W
2
2
I  ID1  ID 2 or
VGS1  VTN  
VGS2  VTN 
2 L
2 L

which requires VGS1 = VGS2. Using KVL:
(a) ID 



VDD  VDS1  VDS2  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 L
2 V 1
(b) The current simply scales by a factor of two (see last equation above), and ID = 18.1
mA.
4.40 See figures in previous problem but use W/L = 20/1.
25x106 20  2
I

 1  250 A but this would
(a) If the transistor were saturated, then D
2  1 
require 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 V DS
= VGS and since the transistor is a depletion-mode device, it is still operating in the triode
region.
20 
V 
VDS  10  10 5 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
ID  2000x106 1
 99.5 A
0.04858  99.5 A Checking :

2 
10 5

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 .
4.49
(a) VGS  VTP  1.1 0.75  0.35V | VDS  0.2V  Triode region
0.20.2  40.0 A
40A 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
40A 20 

ID 
1.3

0.75

 
  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 40A 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
10A 10 
0.20.2  32.8 A
ID 
1.3

0.995








V 2  1 
2 
4.85
100k
12V  3.75V | Assume saturation
a VGG 
100k  220k
100x106 5 
2
3.75  VGS  24 x10 3 ID  VGS  24 x10 3 
 VGS 1
 2
1 
2
6VGS
11VGS  2.25  0  VGS  1.599V and ID  89.7A
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
100x106 10 
2
3.75  VGS  24 x10 ID  VGS  24 x10 
 VGS 1
 2
 1 
3
3
12VGS2  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 
4.100

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
4.103

a
The transistor is saturated by connection.
VGS  12 10 5 ID and
ID 
100x106 10  A 
2
  2 VGS  0.75V 
 1 V 
2
2
50VGS
 74VGS  16.13  0  VGS  1.214V,  0.266V  VGS  1.214 V since V GS must
100x106 10  A 
2
  2 1.214  0.75V 
 1 V 
2
12 1.21
ID  104A | Checking : ID 
 108 A | Q - Point : 108 A,1.21 V
10 5
exceed the threshold voltage.
| ID 
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.
+1 2 V
10
ID 
W =
L
1
10 M 
33 0 k 
IDS
I
+
G
VDS
+
V GS
-
-
100 A 10 
K n' W
2
2
V  0.75
VGS  VTN   
2   GS
 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 V GS  1.008V, 0.486V
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) Checking: ID =(12-1.01)V/330k = 33.3 A
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 
V 1 
2

Collecting terms : 16.5x10 4 ID  40  ID  242 A
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  242A
40k
4.123
(a) The transistor is saturated by connection. For this circuit,
VGS  VDD  ID R  15  75000ID
4 x105 1
2
 15  75000ID  0.75  153 A
2 1
VGS  15  75000ID  3.525V
ID 
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.
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VDS  15
V 
 4 x105 15  0.75  DS VDS  VDS  0.347 V and ID  195 A.
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75000
2 
15  0.347
Checking : ID 
V  195A
Q - point : 195 A,-0.347 V 
785k
Checking the region of operation: VDS  0.347V  VGS VTP  15  0.75  14.25V
ID 
Triode region is correct
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