MOS Regimes

Module by: Bill Wilson

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

This equation looks a lot like the I-V characteristics of a resistor! Id Id is simply proportional to the drain voltage Vds Vds . The proportionality constant depends on the dimensions of the device, W and L as they intuitively should. The current increases as the transistor gets wider, it decreases as it gets longer. It also depends on cox cox and μs μs , and on the difference between the gate voltage and the threshold voltage VT VT . Note that if we adjust Vgs Vgs 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 of the gate and the batteries involved in the operation of the MOS transistor. Here we have explicitly shown the channel as a black band and we have introduced a new quantity, V c x V c x , the voltage along the channel, and a coordinate x x, which tells us where we are on the channel relative to the source and drain. Note that once we apply a drain source potential, Vds Vds , the potential in the channel V c x V c x changes with distance along the channel. At the source end, V c 0=0 V c 0 0 , as the source is grounded. At the drain end, V c L= V ds V c L V ds . We will define a voltage Vgc Vgc which is the potential difference between the gate voltage and the voltage in the channel.

Thus, Vgc Vgc goes from Vgs Vgs at the source end to V gs - V ds V gs V 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 gs - V T V gs V T . Thus we have to modify the equation of another module to take this into account

This, in turn, modifies the integral relation between Id Id and Vgs Vgs .

Equation 3 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

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

Note that for small values of Vds Vds , a previous equation and Equation 4 will give us the same I d - V ds I d V ds behavior, because we can ignore the V ds 2 V ds 2 term in Equation 4. This is called the linear regime because we have a straight-line relationship between the drain current and the drain-source voltage. As Vds Vds starts to get larger however, the squared term will begin to kick in and the plot will start to curve over. Obviously, something is causing the current to drop off as Vds Vds gets larger. This is because the voltage difference between the gate and the channel is becoming less, which means there is less charge in the channel to provide conduction. We can graphically show this by making the channel layer look thinner as we move from the source to the drain. Equation 4, and in fact, Figure 3 would make us think that if Vds Vds gets large enough, that the drain current Id Id should actually start decreasing again, and maybe even become negative!. This does not seem very intuitive, so lets take a look in more detail at the place where Id Id becomes a maximum. We can define Vdsat Vdsat as the source-drain voltage where Id Id becomes a maximum. We can find this by taking the derivative of Id Id with respect to Vds Vds and setting the derivative to 0.

On dropping constants: Rearranging this equation gives us a little more insight into what is going on.

At the drain end of the channel, when Vds Vds just equals Vdsat Vdsat , the difference between the gate voltage and the channel voltage, V gc L V gc L is just equal to VT VT , the threshold voltage. Any further increase in Vds Vds and the difference between the gate and the channel (in the channel region just near the drain) will drop below the threshold voltage. This means that when Vds Vds gets bigger than Vdsat Vdsat , the channel just near the drain region disappears! We no longer have sufficient voltage between the gate and the channel region to maintain an inversion layer, so we simply revert to a depletion condition. This is called pinch off, as seen in Figure 5.

What happens to the drain current when we hit pinch off? It looks like it might go to zero, but that is not the right answer! Although there is no active channel in the pinch-off region, there is still silicon - it just happens to be depleted of all free carriers. There is an electric field, going from the drain to the channel, and any electrons which move along the channel to the pinch-off region are sucked across by the field, and enter the drain. This is just like the current that flows in the reverse saturation condition of a diode. There are no free carriers in the depletion region of the diode, yet Isat Isat does flow across the junction region.

Under pinch-off conditions, further increases in Vds Vds , does not result in more drain current. You can think of the pinched-off channel as a resistor, with a voltage of Vdsat Vdsat across it. When Vds Vds gets bigger than Vdsat Vdsat , the excess voltage appears across the pinch-off region, and the voltage across the channel remains fixed at Vdsat Vdsat . If the channel keeps the same charge, and has the same voltage across it, then the current through the channel (and into the drain) will remain fixed, at a value we will call Idsat Idsat .

There is one other figure which sometimes helps in seeing what is going on. We will plot potential energy for an electron, as it traverses across the channel. Since the source is at zero potential and the drain is at + V ds V ds , an electron will loose potential energy as it flows from the source to the drain. Figure 6 shows some examples for various values of Vds Vds :

For the first two drain voltages, Vds1 Vds1 and Vds2 Vds2 , we are below pinch-off, and so the voltage drop across Rchannel Rchannel increases as Vds Vds increases, and hence, so does Id Id . At Vdsat Vdsat , we have just reached pinch-off, and we are starting to see the "high field" depletion region begin to develop. Since electric field is just the derivative of the potential, the slope of curves in Figure 6 gives you an idea of how big the electric field will be. For further increases in Vds Vds , Vds4 Vds4 and Vds5 Vds5 all of the additional voltage just shows up as a high field drop at the end of the channel. The voltage drop across the conducting part of the channel stays fixed (more or less) at Vdsat Vdsat and so the drain current stays more or less fixed at Idsat Idsat .

Substituting the expression for Vdsat Vdsat into the expression for Id Id we can get an expression for Idsat Idsat

We can define a new constant, kk, where So that

What this means for Figure 3 is that when Vds Vds gets to Vdsat Vdsat , we simply hold Id Id fixed from then on, with a value of Idsat Idsat . For different values of Vg Vg , the gate voltage, we are going to have a different I d - V ds I d V ds curve, and so once again, we end up with a family of "characteristic curves" for the MOSFET. These are shown in Figure 8.

This also gives us a fairly easy way in which to "sketch" a set of characteristic curves for a given device. Suppose we have a MOS field effect transistor which has a threshold voltage of 2 volts, a width of 10 microns, and a channel length of 1 micron, an oxide thickness of 150 angstroms, and a surface mobility of 400cVsec 400 c V sec . using ε ox =3.3×10-13Fcm ε ox 3.3 10 -13 F cm , we get a value of 2.2×10-7Fc 2.2 10 -7 F c for cox cox . This then makes k have a value of

Comments, questions, feedback, criticisms?

Send feedback

• E-mail the author of the module, MOS Regimes

Footer

More about this content: Metadata | Version History

• How to reuse and attribute this content
• How to Cite and attribute this content

This work is licensed by Bill Wilson under a Creative Commons Attribution License (CC-BY 1.0).

Last edited by Raymond Wagner on Aug 14, 2007 11:37 am GMT-5.