1. a) Since there is no gate current for MOS transistor,
I
G
=
0, therefore:
V
G
=
1
.
2
·
100
k
50
k
+
100
k
=
0
.
8
V
Assuming the transistor is in saturation mode, the drainsource current is given by (having in mind
λ
=
0
,
V
BS
=
0
)
:
I
DS
=
k
n
2
(
V
GS

V
t
)
2
(
1
+
λ
V
DS
) =
k
n
2
(
V
GS

V
t
0
)
2
(1)
At the same time, the drainsource current is related to the voltage drop across
R
S
as follows:
I
DS
=
V
S
R
S
(2)
From (1) and (2), we can solve for
V
S
:
k
n
2
(
V
G

V
S

V
t
0
)
2
=
V
S
R
S
⇒
4
m
2
(
0
.
8

V
S

0
.
4
)
2
=
V
S
1
k
⇒
V
2
S

1
.
3
V
S
+
0
.
16
=
0
Solving this quadratic equation, we keep the solution
V
S
=
V
S
1
=
0
.
14
V
and discard the other one,
V
S
2
=
1
.
16
V
, for which
V
S
<
V
G
.
⇒
I
DS
=
V
S
R
S
=
0
.
14
1
k
=
0
.
14
mA
=
I
S
=
I
D
V
D
=
V
CC

I
D
R
D
=
1
.
8

0
.
14
m
×
5
k
=
1
.
1
V
Check for the saturation conditions:
V
GS
=
0
.
66
V
>
V
t
=
0
.
4
V
V
OV
=
V
GS

V
t
=
0
.
8

0
.
14

0
.
4
=
0
.
26
V
<
V
DS
=
1
.
1

0
.
14
=
0
.
96
V
Therefore, the transistor is indeed in the saturation regime of operation.
b) Edge of triode
⇒
V
DS
=
V
OV
. Changing
V
CC
will only affect the drain voltage:
V
DS
= (
V
CC

I
D
R
D
)

V
S
=
V
CC

0
.
14
m
×
5
k

0
.
14
=
V
CC

0
.
84
=
V
OV
=
0
.
26
⇒
V
CC
=
0
.
26
+
0
.
84
=
1
.
1
V
c) Edge of triode
⇒
V
DS
=
V
OV
. Changing
R
D
will only affect the drain voltage:
V
DS
= (
V
CC

I
D
R
D
)

V
S
=
1
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 Spring '08
 Staff
 Gate, Transistor, Vgs

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