3.4.3: Threshold Voltage

Our task now is to figure out how much voltage we need to get \(V_{g}\) up to \(V_{T}\), and then to figure out how much negative charge there is under the gate, once \(V_{T}\) has been exceeded. The first part is actually pretty easy. It is a lot like the problem we looked at, with the one-sided diode, but with just a little added complication. To start out, let's make a sketch of the charge density distribution under the conditions of this image, just when we get to threshold. We'll include the sketch of the structure too, so it will be clear what charge we are talking about. This is shown in Figure \(\PageIndex{1}\). Now, we just use the equation we developed before for the electric field, which came from integrating the differential form of Gauss' Law. \[E(x) = \int \frac{\rho (x)}{\varepsilon} \ dx\]

Figure \(\PageIndex{1}\): Charge distribution at threshold

As before, we will do the integral graphically, starting at the left hand side of the picture. The field outside the structure must be zero, so we have no electric field until we get to the delta function of charge on the gate, at which time it jumps up to some value we will call \(E_{\text{ox}}\). There is no charge inside the oxide, so \(\frac{dE}{dx}\) is zero and thus \(E(x)\) must remain constant at \(E_{\text{ox}}\) until we reach the oxide/silicon interface.

Figure \(\PageIndex{2}\): Electric field in the oxide

If we were to put our little "pill box" on the oxide-silicon interface, the integral of \(D\) over the face in the silicon would be \(\varepsilon_{\text{Si}} E_{\text{Si}} \Delta (S)\) where \(E_{\text{Si}}\) is the strength of the electric field inside the silicon. On the face inside the oxide it would be \(-\left( \varepsilon_{\text{Si}} E_{\text{Si}} \Delta (S) \right)\) where \(E_{\text{ox}}\) is the strength of the electric field in the oxide. The minus sign comes from the fact that the field on the oxide side is going into the pill box instead of out of it. There is no net charge contained within the pill box, so the sum of these two integrals must be zero. (The integral over the entire surface equals the enclosed charge, which is zero. \[\varepsilon_{\text{Si}} E_{\text{max}} \Delta (S) - \varepsilon_{\text{ox}} E_{\text{ox}} \Delta (S) = 0\]

or \[\varepsilon_{\text{Si}} E_{\text{max}} = \varepsilon_{\text{ox}} E_{\text{max}}\]

Figure \(\PageIndex{3}\): Using Gauss's Law at the silicon/oxide interface

This is just a statement that it is the normal component of displacement vector, \(D\), which must be continuous across a dielectric interface, not the electric field \(E\). Solving Equation \(\PageIndex{3}\) for the electric field in the silicon: \[E_{\text{Si}} = \frac{\varepsilon_{\text{ox}}}{\varepsilon_{\text{Si}}} E_{\text{ox}}\]

The dielectric constant of oxides about one third that of the dielectric constant of silicon dioxide, so we see a "jump" down in the magnitude of the electric field as we go from oxide to silicon. The charge density in the depletion region of the silicon is just \(- \left(q N_{a}\right)\) and so the electric field now starts decreasing at a rate \(\frac{- \left(q N_{a}\right)}{\varepsilon_{\text{Si}}}\) and reaches zero at the end of the depletion region, \(x_{p}\).

Figure \(\PageIndex{4}\): Electric field and voltage drops across the entire structure

Clearly, we have two different regions, each with their own voltage drop. (Remember the integral of electric field is voltage, so the area under each region of \(E(x)\) represents a voltage drop.) The drop in the little triangular region we will call \(\Delta \left(V_{\text{Si}} \right)\) and it represents the potential drop in going from the bulk, down to the bottom of the drooping conduction band at the silicon-oxide interface. Looking back at the earlier figure on threshold, you should be able to see that this is nearly one whole band-gap's worth of potential, and so we can safely say that \(\left( \Delta \left(V_{\text{Si}}\right) \simeq 0.8 \right) \rightarrow 1.0 \mathrm{~V}\).

Just as with the single-sided diode, the width of the depletion region \(x_{p}\), is (which we saw in a previous equation): \[x_{p} = \sqrt{ \frac{2 \varepsilon_{\text{Si}} \Delta \left(V_{\text{Si}}\right)}{q N_{a}} }\]

from which we can get an expression for \(E_{\text{Si}}\) \[\begin{array}{l} E_{\text{Si}} &= \frac{q N_{a}}{\varepsilon_{\text{Si}}} x_{p} \\ &= \sqrt{ \frac{2q N_{a} \Delta \left(V_{\text{Si}}\right)}{\varepsilon_{\text{Si}}} } \end{array}\]

by multiplying the slope of the \(E(x)\) line by the width of the depletion region, \(x_{p}\).

We can now use Equation \(\PageIndex{4}\) to find the electric field in the oxide: \[\begin{array}{l} E_{\text{ox}} &= \frac{\varepsilon_{\text{Si}}}{\varepsilon_{\text{ox}}} E_{\text{Si}} \\ &= \frac{1}{\varepsilon_{\text{ox}}} \sqrt{ 2q \varepsilon_{\text{Si}} N_{a} \Delta \left(V_{\text{Si}}\right) } \end{array}\]

Finally, \(\Delta \left(V_{\text{ox}}\right)\) is simply the product of \(E_{\text{ox}}\) and the oxide thickness \(x_{\text{ox}}\): \[\begin{array}{l} \Delta \left(V_{\text{ox}}\right) &= x_{\text{ox}} E_{\text{ox}} \\ &= \frac{x_{\text{ox}}}{\varepsilon_{\text{ox}}} \sqrt{ 2q \varepsilon_{\text{Si}} N_{a} \Delta \left(V_{\text{Si}}\right) } \end{array}\]

Note that \(\varepsilon_{\text{ox}}\) is simply one over \(c_{\text{ox}}\) the oxide capacitance, which we described earlier. Thus \[\Delta \left(V_{\text{ox}}\right) = \frac{1}{c_{\text{ox}}} \sqrt{2q \varepsilon_{\text{Si}} N_{a} \Delta \left(V_{\text{Si}}\right) }\]

And the threshold voltage \(V_{t}\) is then given as \[\begin{array}{l} V_{t} &= \Delta \left(V_{\text{Si}}\right) + \Delta \left(V_{\text{ox}}\right) \\ &= \Delta \left(V_{\text{Si}}\right) + \frac{1}{c_{\text{ox}}} \sqrt{ 2q \varepsilon_{\text{Si}} N_{a} \Delta \left(V_{\text{Si}}\right) } \end{array}\]

which is not that hard to calculate! Equation \(\PageIndex{10}\) is one of the most important equations in this discussion of field effect transistors, as it tells us when the MOS device is turned on.

Equation \(\PageIndex{10}\) has several "handles" available to the device engineer to build a device with a given threshold voltage. We know that as we increase the acceptor density \(N_{a}\) the Fermi level gets closer to the valance band, and hence \(\Delta \left(V_{\text{Si}}\right)\) will change somewhat. But as we said, it will always be around \(0.8\) to \(1 \mathrm{~V}\), so it will not be the driving term which dominates \(V_{T}\). Let's see what we get with an acceptor concentration of \(10^{17}\). Just for completeness, let's calculate \(E_{f} - E_{v}\). \[\begin{array}{l} p &= N_{a} \\ &= N_{v} e^{\frac{E_{f}-E_{v}}{kT}} \end{array}\]

Thus, \(E_{f} - E_{v} = kT \ \ln \left(\frac{N_{v}}{N_{a}}\right)\).

In silicon, \(N_{v}\) is \(1.08 \times 10^{19}\) and this makes \(E_{f} - E_{v} = 0.117 \mathrm{~eV}\), which we will call \(\Delta (E)\). It is conventional to say that a surface is inverted if, at the silicon surface, \(E_{c} - E_{f}\), the distance between the conduction band and the Fermi level is the same as the distance between the Fermi level and the valance band in the bulk. With a little time spent looking at Equation \(\PageIndex{4}\), you should be able to convince yourself that the total energy change in going from the bulk to the surface in this case would be \[\begin{array}{l} q \Delta \left(V_{\text{Si}}\right) &= E_{g} - 2 \Delta (E) \\ &= 1.1 \mathrm{~eV} - 2 \times \left(0.117 \mathrm{~eV}\right) \\ &= 0.866 \mathrm{~eV} \end{array}\]

Figure \(\PageIndex{5}\): Example of finding \(\Delta \left(V_{\text{Si}}\right)\)

Using \(N_{A}=10^{17}\), \(\varepsilon_{\text{Si}} = 1.1 \times 10^{-12} \ \frac{\mathrm{F}}{\mathrm{cm}}\) and \(q = 1.6 \times 10^{-19} \mathrm{~C}\), we find that \[\sqrt{2 q \varepsilon_{\text{Si}} N_{a} \Delta \left(V_{\text{Si}}\right)} = 1.74 \times 10^{-7}\]

We saw earlier that if we have an oxide thickness of \(250 \ \AA\), we get a value for \(c_{\text{ox}}\) of \(1.3 \times 10^{-7} \ \frac{\mathrm{F}}{\mathrm{cm}^2}\), or \(\frac{\mathrm{Coulombs}}{\mathrm{V} \cdot \mathrm{cm}^2}\), and so \[\begin{array}{l} \Delta \left(V_{\text{ox}}\right) &= \frac{1}{c_{\text{ox}}} \sqrt{ 2q \varepsilon_{\text{Si}} N_{a} \Delta \left(V_{\text{Si}}\right) } \\ &= \frac{1}{1.3 \times 10^{-7}} 1.74 \times 10^{-7} \\ &= 1.32 \mathrm{~V} \end{array}\]

and \[\begin{array}{l} V_{t} &= \Delta \left(V_{\text{Si}}\right) + \Delta \left(V_{\text{ox}}\right) \\ &= 0.866 + 1.32 \\ &= 2.18 \mathrm{~V} \end{array}\]