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Mathematical Analysis

Calculation of threshold voltage

Threshold will be defined as when the carrier concentration in the channel is equal to the bulk doping. This is equivalent, approximately, to saying that

Ec - Ef at Si surface= Ef - Ev in bulk.

The following figure illustrates this condition.

Figure

Question: Find the threshold voltage for n-channel MOS when

Na = 1015 cm-3,
tox = 34 nm = 340 Angstrom,
M = Al.
Qox = 6E10 q/cm2.

Solution:

There are three contributions to the threshold voltage:

VT = Vi + Vox + Vms, ----- (1)

where,

Vi = potential drop in bulk Si at threshold,
Vox = potential drop over oxide layer due to charges in oxide and in semiconductor (space charge), and
Vms = potential needed to compensate for the work function difference between M and Si.

(1) Potential drop in Si

Vi = 2 kT/q ln(Na/ni) = 2 * 0.026 * ln(1E15/1E10) = 0.6 V

because,

n = Nc exp[ -q(Ec-Ef)/kT) = ni exp[ (Ef - Ei)/kT ], Maxwell approx.
thus, Nc = ni exp(Eg/2kT), Maxwell approx.

At the inversion threshold, n (channel) = Na(Si bulk), and
qVi = Eg - 2 (Ec - Ef). [see the figure]
Thus, n = Nc exp[ -(Ec - Ef)/kT ] = ni exp[ {Eg - 2(Ec-Ef)}/2kT] = ni exp(qVi/2kT) = Na, or
Vi = 2 kT/q ln(Na/ni).

(2) Voltage drop in the oxide capacitor

Charge comes from the positive fixed oxide charge at the oxide-Si interface and the areal density of depletion charge in Si. Charge per unit area = Qox + Qd where Qox = 6E10 q, and Qd = - q Na dp.

dp = [ 2 eSi Vi/q Na]1/2 = [ 2 x 10-12 x 0.6 / 1.6E-19x1015]1/2 = 0.87 mm.

Thus, Qd = - q 1E15 cm-3 x 0.87E-4 cm = - 8.7 E10 q.

Qox + Qd = ( 6E10 q - 8.7E10 q) = -2.7E10 q/cm2 = - 4E-9 Coul/cm2.

Capacitance:

Cox = eox/tox = (3.4E-13 F/cm) / 3.4E-6 cm = 1E-7 F/cm2

Vox = - (Qox + Qd)/Cox = ( 4E-9 Coul/cm2 ) / (1E-7 F/cm2) = 0.04 V.

(3) M-Si workfunction difference

Vms = -0.9 for Al on p-Si. [For Al on n-Si, Vms = -0.3]

(4) The Threshold voltage

VT = Vi + Vox + Vms = 0.6 + 0.04 - 0.9 = - 0.26.

This means that this nMOSFET is ON even at V = 0 Volts because V = 0 > VT. (Normally ON)

(5) In the applet, you can check this out by setting

M = Al, tox=300A, Qox=3E10, nMOS, N=1E15.

Click on "Equilibrate" button (i.e., V = 0Volts). You see that there is an inversion channel already formed !

Let us now calculate the bias voltage needed to go from one condition to another. This will help you understand the physical reasons for tht threshold voltage, etc.

1. From Equilibrium (Vgb = 0 V) to Flat-Band (Vgb = VFB).

The equilibrium MOS can be accumulation, depletion or inversion. The flat-band MOS is zero charge and zero potential drop in Si, and thus -Qox charge in the gate metal.

• Accumulation at equilibrium: For example, p+poly gate and n-channel MOS. Potential drop in the Si is negligible. The potential drop across oxide goes from - (Qox + Qch)/Cox at equilibrium (accumulation) to - Qox/Cox at flat-band, where Qch is the accumulation charge at equilibrium (Vgb = 0). Thus the external bias is -Qch/Cox. Note that -Qch/Cox = VFB. In this case, removing the accumulation charge Qch from Si established the flat-band condition.
• Depletion at equlibrium: For example, n+poly gate, 150A oxide, n-channel (p-Si) with Na = 1E16 cm-3. To achive flat-band, one just needs to remove the band bending due to depletion in Si. At equilibrium, suppose the depletion charge is Qb per unit area. The potential drop in Si (due to band bending) is Qb2/2esqN. The potential drop in oxide is - (Qox + Qb)/Cox. At flat-band condition, there is no potential drop in Si and a potential drop of -Qox/Cox in oxide. Therefore, VFB = -Qox/Cox + (Qox + Qb)/Cox - Qb2/2esqN= Qb/Cox - Qb2/2esqN.
• Inversion at equilibrium:

2. From Flat-Band (Vgb = VFB) to Threshold (Vgb = VT).

3. Beyond Threshold.