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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

E_{c} - E_{f} at
Si surface= E_{f} - E_{v}
in bulk.

The following figure illustrates this condition.

Figure

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

N_{a} = 10^{15} cm^{-3},

t_{ox} = 34 nm = 340 Angstrom,

M = Al.

Q_{ox} = 6E10 q/cm^{2}.

Solution:

There are three contributions to the threshold voltage:

V_{T} = V_{i} + V_{ox} + V_{ms},
----- (1)

where,

V_{i} = potential drop in bulk Si at threshold,

V_{ox} = potential drop over oxide layer due to charges in oxide
and in semiconductor (space charge), and

V_{ms} = potential needed to compensate for the work function difference
between M and Si.

(1) Potential drop in Si

V_{i} = 2 kT/q ln(N_{a}/n_{i}) = 2 * 0.026 *
ln(1E15/1E10) = 0.6 V

because,

n = N_{c} exp[ -q(Ec-Ef)/kT) = ni exp[ (Ef - Ei)/kT ], Maxwell
approx.

thus, Nc = n_{i} exp(Eg/2kT), Maxwell approx.

At the inversion threshold, n (channel) = N_{a}(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 e_{Si} Vi/q Na]1/2 = [ 2
x 10^{-12} x 0.6 / 1.6E-19x10^{15}]^{1/2} = 0.87
mm.

Thus, Q_{d} = - q 1E15 cm^{-3} x 0.87E-4 cm = - 8.7
E10 q.

Qox + Qd = ( 6E10 q - 8.7E10 q) = -2.7E10 q/cm^{2} = - 4E-9
Coul/cm^{2}.

Capacitance:

Cox = e_{ox}/t_{ox} = (3.4E-13
F/cm) / 3.4E-6 cm = 1E-7 F/cm^{2}

Vox = - (Qox + Qd)/C_{ox} = ( 4E-9 Coul/cm^{2} ) / (1E-7
F/cm^{2}) = 0.04 V.

(3) M-Si workfunction difference

V_{ms} = -0.9 for Al on p-Si. [For Al on n-Si, V_{ms}
= -0.3]

(4) The Threshold voltage

V_{T }= 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
> V_{T}. (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 = V_{FB}).

- 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 = V
_{FB}. 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 Qb^{2}/2e_{s}qN. 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, V_{FB}= -Qox/Cox + (Qox + Qb)/Cox - Qb^{2}/2e_{s}qN= Qb/Cox - Qb^{2}/2e_{s}qN. - Inversion at equilibrium:

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

2. From Flat-Band (Vgb = V_{FB}) to Threshold (Vgb = V_{T}).

3. Beyond Threshold.