# 3.4 Mos regimes

Introducing the Sah equation, and discussing some properties of this equation.

This equation looks a lot like the I-V characteristics of a resistor! ${I}_{d}$ is simply proportional to the drain voltage ${V}_{\mathrm{ds}}$ . The proportionality constant depends on the dimensions of the device, W and L as they intuitivelyshould. The current increases as the transistor gets wider, it decreases as it gets longer. It also depends on ${c}_{\mathrm{ox}}$ and ${\mu }_{s}$ , and on the difference between the gate voltage and the threshold voltage ${V}_{T}$ . Note that if we adjust ${V}_{\mathrm{gs}}$ we can change the slope of the I-V curve. We have made a voltage-controlled resistor!

Caution is advised with this result however, because we have overlooked something quite important. Lets go back to our picture ofthe gate and the batteries involved in the operation of the MOS transistor. Here we have explicitly shown the channel as a black bandand we have introduced a new quantity, ${V}_{c}(x)$ , the voltage along the channel, and a coordinate $x$ , which tells us where we are on the channel relative to the source and drain. Note that once weapply a drain source potential, ${V}_{\mathrm{ds}}$ , the potential in the channel ${V}_{c}(x)$ changes with distance along the channel. At the source end, ${V}_{c}(0)=0$ , as the source is grounded. At the drain end, ${V}_{c}(L)={V}_{\mathrm{ds}}$ . We will define a voltage ${V}_{\mathrm{gc}}$ which is the potential difference between the gate voltage and the voltage in the channel.

${V}_{\mathrm{gc}}(x)\equiv {V}_{\mathrm{gs}}-{V}_{c}(x)$
Thus, ${V}_{\mathrm{gc}}$ goes from ${V}_{\mathrm{gs}}$ at the source end to ${V}_{\mathrm{gs}}-{V}_{\mathrm{ds}}$ at the drain end.

The net charge density in the channel depends upon the potential difference between the gate and the channel at each point along the channel , not just ${V}_{\mathrm{gs}}-{V}_{T}$ . Thus we have to modify the equation of another module to take this into account

${Q}_{\mathrm{chan}}={c}_{\mathrm{ox}}({V}_{\mathrm{gc}}(x)-{V}_{T})={c}_{\mathrm{ox}}({V}_{\mathrm{gs}}-{V}_{c}(x)-{V}_{T})$

This, in turn, modifies the integral relation between ${I}_{d}$ and ${V}_{\mathrm{gs}}$ .

$\int_{0}^{{V}_{\mathrm{ds}}} {\mu }_{s}{c}_{\mathrm{ox}}({V}_{\mathrm{gs}}-{V}_{T}-{V}_{c}(x))W\,d {V}_{c}(x)=\int_{0}^{L} {I}_{d}\,d x$

is only slightly harder to integrate than the one before (Now what is the integral of xdx), and so we get for the drain current

${I}_{d}=\frac{{\mu }_{s}{c}_{\mathrm{ox}}W}{L}(({V}_{\mathrm{gs}}-{V}_{T}){V}_{\mathrm{ds}}-\frac{{V}_{\mathrm{ds}}^{2}}{2})$

This equation is called the Sah Equation after C.T. Sah, who first described the MOS transistor operation thisway back in 1964. It is very important because it describes the basic behavior of the MOS transistor.

Note that for small values of ${V}_{\mathrm{ds}}$ , a previous equation and will give us the same ${I}_{d}-{V}_{\mathrm{ds}}$ behavior, because we can ignore the ${V}_{\mathrm{ds}}^{2}$ term in . This is called the linear regime because we have a straight-line relationship between the drain current and the drain-source voltage. As ${V}_{\mathrm{ds}}$ starts to get larger however, the squared term will begin to kick in and the plot will start tocurve over. Obviously, something is causing the current to drop off as ${V}_{\mathrm{ds}}$ gets larger. This is because the voltage difference between the gate and the channel is becomingless, which means there is less charge in the channel to provide conduction. We can graphically show this by making the channellayer look thinner as we move from the source to the drain. , and in fact, would make us think that if ${V}_{\mathrm{ds}}$ gets large enough, that the drain current ${I}_{d}$ should actually start decreasing again, and maybe even become negative!. This does not seem veryintuitive, so lets take a look in more detail at the place where ${I}_{d}$ becomes a maximum. We can define ${V}_{\mathrm{dsat}}$ as the source-drain voltage where ${I}_{d}$ becomes a maximum. We can find this by taking the derivative of ${I}_{d}$ with respect to ${V}_{\mathrm{ds}}$ and setting the derivative to 0.

$\frac{d {I}_{d}}{d {V}_{\mathrm{ds}}}}=0=\frac{{\mu }_{s}{c}_{\mathrm{ox}}W}{L}({V}_{\mathrm{gs}}-{V}_{T}-{V}_{\mathrm{dsat}})$
On dropping constants:
${V}_{\mathrm{dsat}}={V}_{\mathrm{gs}}-{V}_{T}$
Rearranging this equation gives us a little more insight into what is going on.
${V}_{\mathrm{gs}}-{V}_{\mathrm{dsat}}={V}_{T}={V}_{\mathrm{gc}}(L)$

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