EE-612:

Lecture 1
1D MOS Electrostatics
Mark Lundstrom
Electrical and Computer Engineering
Purdue University
West Lafayette, IN USA
Fall 2006

NCN
www.nanohub.org
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 outline

1) Review of some fundamentals

2) Identify next steps

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 energy bands

E n0 = N C e(EF − EC ) kB T

EC

EG n0 p0 = ni2
EF
EV

p0 = NV e(EV − EF ) kB T
x
Assumptions:
i) equilibrium, ii) Boltzmann carrier statistics, iii) uniform electrostatic potential
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 electrostatic potential

ψ (x ) EC ( x ) = constant − qψ ( x )

the energy bands will bend

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 band bending

E n0 ( x ) = N C e(EF − EC ( x )) kB T
slope = electric field
EC ( x ) = C − qψ ( x )

EF
EV (x) = EC (x) − EG

p0 (x) = NV e(EV ( x )− EF ) kB T
x

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1D MOS electrostatics (L >> Tox)

VG
V=0 V=0

V=0

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1D MOS electrostatics (L >> Tox)

V=0 V=0

VG

x
V=0
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 ‘flat band’ conditions

S iO 2 ψ (x) = constant = 0
EFM = EFM
0
− qVG

VG′ = 0 EC

EFM EF
EV

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 electrostatic potential

qψ (x )
qψ S
EC
ψ (x ) = 0
EF
qVG′ > 0 EV
EFM
x
EC ( x ) = constant − qψ ( x )

EC (∞) − EC (x)
ψ (x) =
q
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 accumulation
• bands bend up

• surface potential < 0

ψS < 0
• hole density increases
EC exponentially
qVG′ < 0
EF p0 (x) = NV e(EV ( x )− EF ) kB T
EV
∞

p-type Si ( )
QS = + ∫ q p0 (x) − N A− dx C/cm 2
0

QS ~ e− qψ S kB T
(not quite, but close)

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 depletion

• bands bend down

ψS > 0 • surface potential > 0

EC • For x < W:
EF
p0 (x) = NV e(EV ( x )− EF ) kB T ≈ 0
qVG′ > 0 EV
n0 (x) = N C e(EF − EC ( x )) kB T ≈ 0
x
W ρ(x) ≈ −qN A (x < W ) C cm 3

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 depletion (ii)
ρ(x) = −qN A

d (ε Si E )
ψS > 0 = ρ(x) = −qN A
dx
EC
dE −qN A
=
EF dx ε Si
qVG′ > 0 EV
qN A
E (x ) = (W − x )
x εS i

W qN AW −QS
ES = =
εS
i
εSi

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 depletion (iii)
E S = qN A W ε S i
E
ES dE
= − qN A ε S i
dx
ψS > 0
EC
x
W
EF
qVG′ > 0 EV ψ (x ) = − ∫ E(x)dx
1
x ψ S = E SW
2
W
W = 2ε Siψ S qN A
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 depletion (iv)

W = 2ε Siψ S qN A
ψS > 0
QS = −qN AW = − 2qN Aε Siψ S
EC

EF QS ~ ψ S (very close)
qVG′ > 0 EV

x
W

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 inversion

k BT ⎛ N A ⎞
ψ S > 2ψ B ψB = ln ⎜
q ⎝ ni ⎟⎠
ψ S >> 0 (surface potential to make n(x = 0) = NA)

EC p0 (x < W ) ≈ 0
EF (EF − EC ( x )) kB T
n0 (x = 0) = N C e
EV
qVG′ > 0 ~ eqψ S kB T
∞ ∞

x QS = − ∫ q ⎡⎣ n0 (x) + qN A− ⎤⎦ dx ≈ − ∫ qn0 (x)dx

0 0
W
QS ~ eqψ S kB T
(not quite, but close)
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 MOS electrostatics
accumulation flat band depletion/
inversion

qψ S < 0 qψ S > 0
EC EC EC
VG’
EF VG’ EF EF
VG’
EV EV EV

ψS < 0 ψS = 0 ψS > 0
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 MOS electrostatics

log10 QS (ψ S )
C/cm2
EC
~ e−qψ S / 2kB T ~ e qψ S / 2kB T
VG’=0 EF

EV

~ ψS

ψS ψS = 0
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 next steps

1) Understand how to do the problem ‘exactly.’

2) Use simpler arguments to understand QS(ψS).

3) Briefly discuss ultra-thin body electrostatics.

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