MOS Threshold Voltage

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MOS Threshold Voltage
• This is the gate voltage required to strongly invert the surface of the substrate :
N-Type substrate  we need to Bend the Bands Upward by 2q F.
P-Type substrate  we need to Bend the Bands Downward by 2q F.
G
O
S
2q F
Depletion
Region
Vth = Flat Band voltage (VFB) – 2q F + Voltage to sustain the Depletion
region charge (-QB/Cox) + Voltage to overcome the Oxide trapped charge (-Qox/Cox)
• Flat Band Voltage VFB: The Voltage required to flatten the energy Bands, at
the surface of the substrate VFB = EFG – EFS = ФG – ФS
Initially Before Contact:
Oxide
Gate
Semiconductor
Ec
EFG
EFS
Ev
After
G
O
S
qVFB
EF
ФG = Gate Work function
ФS = Substrate Work function (Depends on the Doping)
= s + Eg/2q –  F
s : Electron Affinity (Eo – Ec)
The Work function is the Energy required to remove an electron from the
Fermi level to outside the Solid.
Vacuum Level
Eo
q s
Ec
Eg/2
Ei
EF
Ev
q F
Vacuum Level: Energy level that electrons outside Semiconductor
 +ve VFB means that Bands are Bent Upward at zero Bias (since We need
a +ve voltage to push them down to make them flat).
 -ve VFB means that Bands are Bent Downward at zero Bias (since We
need a -ve voltage to pull them up to make them flat).
 F = VT ln(ND/ni)
For N-Type substrate.
 F = - VT ln(Na/ni)
For P-Type substrate.
• QB, the Depletion charge = +
(2q NSub sub |2 F |)
ε
For N-Type substrate
=–
(2q NSub sub |2 F |)
ε
For P-Type substrate
NSub = net substrate Doping.
εsub = Dielectric constant of substrate.
e.g) For Si
εsi = εo εsir : εo = 8.85x10-14; εsir = 12
• Qox = this is a charge trapped in the Oxide during manufacturing.
Example 1: A MOS device is made with an AL Gate (ФAL = 4.15 ev) and PType substrate (Na = 1016 cmˉ³). The Oxide capacitance =2 fF/μm².
f = 1exp-15. Assume Qox = 0.
Calculate Vth
si = 4.1 ev Egsi = 1.1 ev = 1.1 * q J
Sol
Vth = VFB – 2 F – QB/ Cox – Qox/ Cox
VFB = ФG – ФS = 4.15 – [Xs +Eg/2 q –  F] = 4.15 – [4.1 + 0.55 –  F]
 F = -0.025 ln (1exp16/1exp10) = -0.345 V
VFB = 4.15 – 4.1 – 0.55 – 0.345 = – 0.89 V
at zero Bias
G
O
S
EF
QB = – √ (2 * 1.6*10-19 * 1016 *8.85*10-14 * 12 * 0.69) = – 4.8*10-8 C/cm²
Cox = 2 fF/μm² = 2 * 1*10-15 * 1 * 108 = 2 * 10-7 F/cm²
Vth = – 0.89 + 0.69 + [4.8*10-8 / 2 * 10-7] = 0.04V
at VG = Vth
G
O
S
0.69 ev
EF
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