DP
=
10
m
A
2
2
1
1.448

3.3
+
0.75
(
29
2
=
12.1
m
A
(b)
For v
I
= v
O
, both transistors will be saturated since v
GS
= v
DS
for each device.
Equating the drain
currents with K
n
= K
p
yields:
i
( 29
K
n
2
v
I

V
TN
(
29
2
=
K
p
2
v
I

V
DD

V
TP
(
29
2
and
v
I

V
TN
=
V
DD

v
I
+
V
TP
v
O
=
v
I
=
V
DD
+
V
TN
+
V
TP
2
=
2.5
+
0.6

0.6
2
=
1.25
V
ii
( 29
I
DN
=
K
n
2
v
I

V
TN
(
29
2
=
25
m
A
2
2
1
1.25

0.6
(
29
2
=
10.6
m
A
For K
n
= 2.5 K
p
,
iii
( 29
2.5
K
p
2
v
I

V
TN
(
29
2
=
K
p
2
v
I

V
DD

V
TP
(
29
2
and
1.58
v
I

V
TN
(
29
=
V
DD

v
I
+
V
TP
v
O
=
v
I
=
V
DD
+
1.58
V
TN
+
V
TP
2.58
=
2.5
+
1.58 0.60
(
29
+ 
0.60
(
29
2.58
=
1.104
V
iv
( 29
I
DN
=
25
m
A
2
2
1
1.104

0.60
(
29
2
=
6.35
m
A

Check by finding I
DP
:
I
DP
=
10
m
A
2
2
1
1.104

2.5
+
0.60
(
29
2
=
6.34
m
A
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