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V
Tn
V
TO
:=
V
Tn
1.00 volt
=
k
n
100
µ
A
⋅
volt
2
−
⋅
:=
WLratio
6
1
:=
K
n
k
n
WLratio
⋅
:=
K
n
600
µ
A volt
2
−
⋅
=
The voltage on the gate is set by the voltage divider of R
1
and R
2
:
V
G
V
DD
R
1
R
1
R
2
+
⋅
:=
V
G
3.125 volt
=
Since this is greater than V
Tn
, the MOSFET is ON.
Start by first guessing the MOSFET to be SATURATED, since this is easier to analyze.
V
GS
1.0 volt
⋅
:=
(Start with a guess for the root finder.)
V
GS
root V
G
V
GS
−
1
2
K
n
⋅
R
3
⋅
V
GS
V
Tn
−
()
2
⋅
−
V
GS
,
⎡
⎣
⎤
⎦
:=
V
GS
1.478 volt
=
I
D
1
2
K
n
⋅
V
GS
V
Tn
−
2
⋅
:=
I
D
68.6
µ
A
=
V
BS
I
D
−
R
3
⋅
:=
V
BS
1.647
−
volt
=
V
DS
V
DD
I
D
R
3
R
4
+
⋅
−
:=
V
DS
7.53 volt
=
V
DSsat
V
GS
V
Tn
−
:=
V
DSsat
0.478 volt
=
Since V
DS
> V
DSsat
, the guess for saturation was correct.
Also, since
γ
= 0, no body bias corrections to the threshold voltage are needed, even though this
circuit does produce a body bias of
V
SB
= 2.151 Volts.
EE331 Devices and Circuits 1
Prof. R. B. Darling
Homework # 8 Solutions
First, define some useful units and constants:
µ
m1
0
6
−
m
⋅
:=
eV
1.602 10
19
−
⋅
joule
⋅
:=
ms
10
3
−
sec
⋅
:=
µ
s1
0
6
−
sec
⋅
:=
q
1.602 10
19
−
⋅
coul
⋅
:=
k
B
8.62 10
5
−
⋅
eV
⋅
K
1
−
⋅
:=
nm
10
9
−
m
⋅
:=
ε
rSiO2
3.9
:=
ε
o
8.854 10
14
−
⋅
farad
⋅
cm
1
−
⋅
:=
n
i
10
10
cm
3
−
⋅
:=
V
T
0.025 volt
⋅
:=
ε
rSi
11.7
:=
Ω
ohm
:=
k
Ω
10
3
ohm
⋅
:=
M
Ω
10
6
ohm
⋅
:=
Problem 1:
Problem 4.93 (a) of Jaeger and Blalock.
V
DD
10.0 volt
⋅
:=
R
1
100 k
Ω
⋅
:=
R
2
220 k
Ω
⋅
:=
R
3
24 k
Ω
⋅
:=
R
4
12 k
Ω
⋅
:=
V
TO
1.0 volt
⋅
:=
EE331 HW#8 Solutions
Page 1
R. B. Darling
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View Full DocumentV
Tp
0.75
−
volt
⋅
:=
k
n
100
µ
A
⋅
volt
2
−
⋅
:=
k
p
40
µ
A
⋅
volt
2
−
⋅
:=
Equate the current through both transistors (which are both SATURATED) to give:
1
2
k
n
W
n
L
n
⋅
V
o
V
Tn
−
()
2
⋅
1
2
k
p
⋅
W
p
L
p
⋅
V
DD
V
o
−
V
Tp
+
2
⋅
:=
Simplifying gives:
V
DD
V
o
−
V
Tp
+
V
o
V
Tn
−
k
n
k
p
:=
since the W/L ratios are equal and cancel out.
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 Winter '08
 Taicheng

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