Equations:
k T
- - - - - - - - - - - L N
V b i
=
2
s i
- - - - - - - - - - - - - - - - - - - - - - - - - -
x d m a x
=
q N D
G 0
=
q N D
⎛
⎛
2
⎜
⎜
- - -
I D
=
G 0
V D S
–
3
⎝
⎝
⎛
⎜
V b i V G S
–
+
V D S
⎝
2
q N D a
- - - - - - - - - - - - - - - - - - - - - - - - - -
V D s a t
=
–
V b i V G S
2
s i
0
⎛
2
⎜
- - - - - - - - - - - - - - - - -- - - - - - - - -
g m
=
G 0
1
–
⎝
q N D a
Example:
Given: ND=1E16_1/cm^3, W=6_ , a=1_ , L=2_ , n= 1248_cm^2/ (V s), VGS= -4_V, VDS=4_V,
T=26.85 C.
Solution: Vbi=0.3493_V, xdmax=1.0479_ , G0=5.9986E–4_S, ID=0.2268_mA, VDsat=3.2537_V,
Vt= -7.2537_V, gm=0.1462_mA/V.
Stress Analysis (14)
Variable
1
5-54 Equation Reference
N D
⎛
⎞
- - - - - - - - -
⎝
⎠
q
n i
0
V b i V G S
–
+
V DS
a W
⎛
⎞
- - - - - - - - - - - -
n
⎝
⎠
L
⎞
2
s i
0
- - - - - - - - - - - - - - - - - - - - - - - - - -
2
⎠
q N D a
3
3
- - -
- - -
⎞
2
2
–
V b i V G S
–
⎠
V t
=
Vbi
–
⎞
s i
0
V b i V G S
–
2
⎠
Description
Elongation
Normal strain
Shear strain
Angle of twist
Normal stress
Maximum principal normal stress
2
q ND a
------------------------- -
–
2
si
0