3.2 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-sideddiode, but with just a little added complication. To start out, lets make a sketch of the charge density distribution under theconditions of this image , just when we get to threshold. Well include the sketch of the structure too, so it will be clear what charge weare talking about. This is shown in . Now, we just use the equation we developed before for the electric field , which came from integrating the differential form of Gauss' Law.

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

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 ${\epsilon }_{\mathrm{Si}}{E}_{\mathrm{Si}}\Delta (S)$ , where $E_{\mathrm{Si}}$ is the strength of the electric field inside the silicon. On the face inside the oxide it would be $-({\epsilon }_{\mathrm{ox}}{E}_{\mathrm{ox}}\Delta (S))$ , where $E_{\mathrm{ox}}$ is the strength of the electric field in the oxide. The minus sign comes from the fact that the field on the oxide side isgoing into the pill box instead of out of it. There is no net charge contained within the pill box, so the sum of these twointegrals must be zero. (The integral over the entire surface equals the enclosed charge, which is zero.)

This is just a statement that it is the normal component ofdisplacement vector, $D$ , which must be continuous across a dielectric interface, not the electricfield, $E$ . Solving for the electric field in the silicon:

The dielectric constant of oxides about one third that of the dielectric constant of silicon dioxide, so we see a "jump" downin the magnitude of the electric field as we go from oxide to silicon. The charge density in the depletion region of thesilicon is just $-(q{N}_a)$ and so the electric field now starts decreasing at a rate $\frac{-(q{N}_a)}{{\epsilon }_{\mathrm{Si}}}$ and reaches zero at the end of the depletion region, $x_p$ .

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

Just as with the single-sided diode, the width of the depletion region $x_p$ , is (which we saw in a previous equation ):

We can now use to find the electric field in the oxide

Finally, $\Delta (V_{\mathrm{ox}})$ is simply the product of $E_{\mathrm{ox}}$ and the oxide thickness, $x_{\mathrm{ox}}$

Note that $x_{\mathrm{ox}}$ divided by ${\epsilon }_{\mathrm{ox}}$ is simply one over $c_{\mathrm{ox}}$ , the oxide capacitance, which we described earlier. Thus

has several "handles" available to the device engineer to build a device with a given thresholdvoltage. We know that as we increase $N_a$ , the acceptor density, that the Fermi level gets closer to the valance band, and hence $\Delta (V_{\mathrm{si}})$ will change some. But as we said, it will always be around 0.8 to 1 Volt, so it will not be the driving term whichdominates $V_T$ . Let's see what we get with an acceptor concentration of $10^{17}$ . Just for completeness, let's calculate $E_f-E_v$ .

In silicon, $N_v$ is $1.08E19$ 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 leveland the valance band in the bulk. With a little time spent looking at , you should be able to convince yourself that the total energy change in goingfrom the bulk to the surface in this case would be

Using $N_A=10^{17}$ , ${\epsilon }_{\mathrm{Si}}=1.1E-12\frac{F}{\mathrm{cm}}$ and $q=1.6E-19C$ , we find that

We saw earlier that if we have an oxide thickness of 250Å, we get a value for $c_{\mathrm{ox}}$ of $1.3E-7\frac{F}{\mathrm{cm}^2}$ ( $\frac{\mathrm{Coulombs}}{V\mathrm{cm}^2}$ ), and so