Example:
Given: ND=1E22_cm^–3, NA=1E15_1/cm^3, T=26.85_ C, Js=1E–6_ A/cm^2, Va=-20_V,
E1=3.3E5_V/cm, W=10_ , ∆W=1_ , L=10_ , ∆L=1_ , xj=2_ .
Solution: Vbi=.9962_V, xd=5.2551_ , Cj=2005.0141_pF/cm^2, Emax=79908.5240_V/cm,
BV=358.0825_V, J= -1.0E–12_A/cm^2, Aj=3.1993E–6_cm^2, I= -3.1993E–15_mA.
NMOS Transistors (13, 2)
These equations for a silicon NMOS transistor use a two-port network model. They include linear and nonlinear
regions in the device characteristics and are based on a gradual-channel approximation (the electric fields in the
direction of current flow are small compared to those perpendicular to the flow). The drain current and
transconductance calculations differ depending on whether the transistor is in the linear, saturated, or cutoff
region. The equations assume the physical geometry of the device is a rectangle, second-order length-parameter
effects are negligible, shot-channel, hot-carrier, and velocity-saturation effects are negligible, and subthreshold
currents are negligible. ( See "SIDENS" in Chapter 3.)
Equations:
W e
=
W 2
–
W
⎛
W e
⎞
⎛
- - - - - - - -
I D S
=
C o x
n
⎝
⎠
⎝
L e
2
- - - - - - - - - - - - - - - - - - - - - - - -- - - - - - - - - - - - - - - - - - - - - - - - - - -
=
V t
=
V t 0
+
2 A B S
k – T
⎛
p
=
- - - - - - - - - - - - - - L N
⎝
q
W e
⎛
g m
=
C o x
m
- - - - - - - -
⎝
L e
V D s a t
Example:
Given: tox=700_Å, NA=1E15_1/cm^3, n=600_cm^2/ (V s), T=26.85_ C, Vt0=0.75_V, VGS=5_V,
VBS=0_V, VDS=5_V, W=25_ , ∆W=1_ , L=4_m, ∆L=0.75_ , =0.05_1/V.
Solution: We=23_ , Le=2.5_ , Cox=49330.4750_pF/cm^2, =0.3725_V^.5, p= -.2898_V, Vt=0.75_V,
VDsat=4.25_V, IDS=3.0741_mA, gds=1.5370E–4_S, gm=1.4466_mA/V.
5-52 Equation Reference
L e
=
L 2
–
L
Cox
2
V DS
⎞
- - - - - --------- -
V G S V t
–
V D S
–
⎠
2
s i
0 q N A
C o x
p
–
A B S V B S
–
2 ABS
N A
⎞
- - - - - - - - -
g d s
=
IDS
⎠
n i
⎞
1 +
V D S
2 IDS
⎠
=
V G S V t
–
ox
0
------------------- -
=
tox
1 +
VDS
p