9.3. METAL-OXIDE-SEMICONDUCTOR CAPACITOR 447
The potentialVfbis given by
eVfb=eφm−(eχs+(Ec−EF))
The position of the Fermi level is
EF=EFi+kBT ln
Na
ni
below the conduction band. Also,EFi=Eg/ 2 where for Si,Eg= 1.11 eV. UsingT=
300 K, we get
EF=0.555 + 0.026 ln
(
1014
1. 5 × 1010
)
=0.783 eV
below the conduction band. Thus
Vfb=4. 1 −(4.15 + 0.783) =− 0 .833 V
Example 9.3Consider ap-type silicon doped to 3 × 1016 cm−^3. The SiO 2 has a thickness
of 500A. An ̊ n+polysilicon gate is deposited to form the MOS capacitor. The work
function differenceVfb=− 1. 13 eV for the system; temperature = 300 K. Calculate the
threshold voltage if there is no oxide charge and if there is an oxide charge of 1011 cm−^2.
The position of the Fermi level is given by (measured from the intrinsic Fermi level)
φF=0.026 ln
(
3 × 1016
1. 5 × 1010
)
=0.376 V
Under the assumption that the chargeQsis simpleNaWwhereWis the maximum
depletion width, we get
Qs =(4seNa|φF|)^1 /^2
=
(
4 ×(11.9)×(8. 85 × 10 −^14 F/cm) (1. 6 × 10 −^19 C)(3× 1016 cm−^3 )(0.376 V)
) 1 / 2
=8. 64 × 10 −^8 Ccm−^2
In the absence of any oxide charge, the threshold voltage is
VT = − 1 .13 + 2(0.376) +
(
8. 64 × 10 −^8
)( 500 × 10 −^8
3 .9(8. 85 × 10 −^14 )
)
=0.874 V
In the case where the oxide has trap charges, the threshold voltage is shifted by
ΔVT =
(
1011
)(
1. 6 × 10 −^19
)( 500 × 10 −^8
3. 9 × 8. 85 × 10 −^14
)
= − 0 .23 V
It can be seen from this example that oxide charge can cause a significant shift in the
threshold voltage of an MOS device.
