Kenneth R. Laker, University of Pennsylvania, updated 5Feb15
ESE 570 MOS TRANSISTOR THEORY – Part 1
2Kenneth R. Laker, University of Pennsylvania, updated 5Feb15
2
GATE
Mass Action Law
Two-Terminal MOS Structure
VB
n+ n+
Si – Oxide interface
3Kenneth R. Laker, University of Pennsylvania, updated 5Feb15
Acc
epto
rs
Don
ors
MetalloidsChemical Periodic Table
American Chemical Society (ACS)
4Kenneth R. Laker, University of Pennsylvania, updated 5Feb15
Gate
Oxi
deVG
= 0
Si surface
qΦM
Ideal Equilibrium MOS Capacitor Energy Bands
NOTE: 1. qΦ and E are in units of energy = electron-volts (eV); where 1 eV = 1.6 x 10 -19 J. 2. 1 eV corresponds to energy acquired by a free electron that is accelerated by an electric potential of one volt. 3. Φ and V corresponds to potential difference in volts.
Work Functions qΦM, qΦSi = energy required to move an electron from EF to Evacuum for metal gate, Si respectively.
EC, EFm
p doped Si
met
al
EFp
q/Si=q3Si(,E C−EV -
E i=EC−EV
2
5Kenneth R. Laker, University of Pennsylvania, updated 5Feb15
MOS Capacitor with External BiasThree Regions of Operation:1. Accumulation Region – VG < 02. Depletion Region – VG > 0, small3. Inversion Region – VG ≥ VT, large
6Kenneth R. Laker, University of Pennsylvania, updated 5Feb15
VG < 0
Surface potential: q/S=q/,0-)0
Band bending due to VG < 0
Si surface
Accumulation
qΦFp
q/Fp=E Fp−E i−bulk)0
q/, x -=E Fp−E i , x -
qΦ(x)qΦS
Band bending: .E i ,x -=E i ,x -−E i−bulk*0
x
Energy Bands - Accumulation Region
EFm
EFp
0
qV G=E Fp−E Fm
7Kenneth R. Laker, University of Pennsylvania, updated 5Feb15
MOS Capacitor - Depletion Region
tox
mobile holes- - - - -
8Kenneth R. Laker, University of Pennsylvania, updated 5Feb15
DepletionVG > 0 (small)
xd
Surface Potential:
Si surface
Band bending due to VG > 0
qΦFp
qΦS
qΦ(x)
q/Fp=E Fp−E i−bulk)0q/, x -=E Fp−E i , x -
q/S=q/,0-*0
Band bending: .E i ,x -=E i ,x -−E i−bulk)0
x
Energy Bands - Depletion Region
EFm
EFp
0
qV G=E Fp−E Fm
9Kenneth R. Laker, University of Pennsylvania, updated 5Feb15
Bulk or Fermi potential
MOS Capacitor - Depletion Region
2Fp=2F=kTq
lnni
N A)0
26 mV at room T
22S
2Fp
Mobile hole charge density (per unit area) in thin layer parallel to Si-Oxide interfaceDepletion region potential needed to displace dQ by distance x into bulk (Poisson Eq.)
tox
∣2S−2F∣∣2S−2F∣
2Fp−2S2S
2Fp
- - - -
NOTE
surface potential (Fermi potential at surface)
Q=−q N A xd=−+2 q N A1Si∣2Fp−2S∣xd=+ 21Si∣2Fp−2S∣q N A
10Kenneth R. Laker, University of Pennsylvania, updated 5Feb15
Si surface
/S=−/Fp @ V G=V T0
InversionVG ≥ VT0 > 0
xdm
qΦS
qΦFp
q/Fp=E Fp−E i−bulk)0q/, x -=E Fp−E i , x -
Surface Potential: q/S=q/,0-=−q/Fp*0Band bending: .E i ,x -=E i ,x -−E i−bulk)0
x
Energy Bands - Inversion Region
0
EFm
EFpqV G=E Fp−E Fm
11Kenneth R. Laker, University of Pennsylvania, updated 5Feb15
MOS Capacitor - Inversion Region
tox
∣−22F∣
∣2S−2F∣2S=−2F
2S=−2F,2S=−2F-
2Fp=2F=kTq
lnni
N A
2Fn=2F=kTq
lnN D
ni
xdm=xd l2S=−2F=+ 21Si∣−22F∣
qN A
- - - -- - - - -
VG = VT for φS = - φFp
VG ≥ VT (threshold voltage)
2S=−2Fp
xdm=+ 21Si∣22Fp∣q N A
.=−+q N A1Si∣22Fp∣
12Kenneth R. Laker, University of Pennsylvania, updated 5Feb15
MOS Capacitor - Inversion Region
INVERSION CONDITION – Key Equations2S=−2F
2F=2Fp=kTq
lnni
N A
n-Sub2F=2Fn=kTq
lnN D
ni
QB0=−+qN A1Si∣22F∣ c/cm2p-Sub
Where 1Si≈31ox F/cm
when n = NA Depletion region charge density
V
V
VG ≥ VT (threshold voltage)
VG = VFB for φS = φF → flat band (FB) condition, i.e. no band bending.
13Kenneth R. Laker, University of Pennsylvania, updated 5Feb15
nMOS layout
14Kenneth R. Laker, University of Pennsylvania, updated 5Feb15
G
15Kenneth R. Laker, University of Pennsylvania, updated 5Feb15
Two-Terminal MOS Structure -> nMOS Transistor
depletion region
-
- --
- - - - -
- - -
-
VGV
S
VD
16Kenneth R. Laker, University of Pennsylvania, updated 5Feb15
nMOS Transistor = MOS Capacitor + source/drain
where
VSB
= 0
xdm=+ 21Si∣22Fp−V SB∣q N A
where
NOTE: In CadenceSPICE = Spectre
NOTE: Since NA >> n
i : φ
Fp < 0
-
- - -- - - -
- -
-
1Si≈31ox
VG
VD
VS
m
17Kenneth R. Laker, University of Pennsylvania, updated 5Feb15
+
)
l
for p-sub 2F=2Fp
[VT0
-> VT0 in SPICE]VT0n,p
work function between gate and channel
with VSB
= 0.V
FBV
FB
-+- for nMOS and pMOS
V FB=/GC−Qox
C ox≈/GC
V T0=/GC−Qox
C ox−22F−
QB0
Cox
VFB = flat band voltage
QB0=−+q N A1Si∣22F∣
18Kenneth R. Laker, University of Pennsylvania, updated 5Feb15
Adjusting VT0
Using and an Added Channel Implant
V T0=V FB−22F−QB0
C ox
Intrinsic VT0
- no channel implant adjustment
.V T0=±q N I
Cox
Adjusted V'T0
– due to channel implant adjustment with carrier concentration N
I
V T0' =V T0(.V T0=V FB−22F−
QB0
C ox±
q N I
Cox
for p-type implant(q N I
C ox
−q N I
C ox
for n-type implant
NOTE: When channel implant adjustment NI is done as a step in the CMOS process,
the SPICE parameter VT0 refers to the adjusted threshold voltage V'T0
.
19Kenneth R. Laker, University of Pennsylvania, updated 5Feb15
−QB−QB0
Cox=+2q N A1Si
C ox,+∣22F−V SB∣−+∣22F∣-
V T=V FB−22F−QB
Cox(
QB0
Cox−
QB0
Cox
V T=V FB−22F−QB0
Cox−
QB−QB0
Cox
VT0
V T0=V FB−22F−QB0
C ox
for VSB
= 0
units = V1/2
QB=−+q N A1Si∣22F−V SB∣ QB0=−+q N A1Si∣22F∣
V T=V T0(0,+∣22F−V SB∣−+∣22F∣-
20Kenneth R. Laker, University of Pennsylvania, updated 5Feb15
VSB
is ≥ 0 in nMOS, ≤ 0 in pMOS
∣0∣● VT0 is positive in nMOS (V
T0n) , negative in pMOS (V
T0p)QOX is negative for nMOS and pMOS
21Kenneth R. Laker, University of Pennsylvania, updated 5Feb15
|VSB
|
22Kenneth R. Laker, University of Pennsylvania, updated 5Feb15
for
1
1
φF
2Fp=k Tq
lnni
N A=0.26V ln ,1.45 x 1010
1016 -=−0.35V
V T0n=V FB−22Fp−QB0
Cox
1 A=10−10 m
23Kenneth R. Laker, University of Pennsylvania, updated 5Feb15
1 V T0n=V FB−22Fp−QB0
Cox
2Fp=−0.35 V
QB0=−+2q N A1Si∣22Fp∣
F = C/V
.=−+2 ,1.6 x 10−19 C -,1016 cm−3-,1.06 x 10−12 F cm−1-∣−0.70 V∣
V T0n=−1.04V −2 ,−0.35V -−,−0.72 V -=0.38V
24Kenneth R. Laker, University of Pennsylvania, updated 5Feb15
Example 12
+C 2 cm−4 V−1
C cm−2V −1 =V 1/2
Units Calc.
V1/2
2Fp=−0.35V
V Tn=V T0n(0,+∣22Fp−V SB∣−+∣22F∣-
bulk potential
.=5.824 x 10−8 C /,V 1/2 cm2-6.8 x 10−8 C /,V cm2-
=0.85V 1/2
25Kenneth R. Laker, University of Pennsylvania, updated 5Feb15
2
1
V Tn=V T0n(0,+∣22Fp−V SB∣−+∣22F∣-
V Tn=0.38V (0,+∣0.70V −V SB∣−+∣0.70V∣-
V T0n=0.38 V0=0.85V 1/2
2Fp=−0.35V
where