In this video, we will discuss some advanced operation of MOSFET. Before we do that, I want to rewrite the long-channel MOSFET equation, that we derived in the previous video in this form. You can see that this equation has a typical standard drift equation form in that, the charge density, this force to C oxide times W times the quantity in the parenthesis, represents the charge density and then mu times the VD over L is an electric field inside the channel. So, this is a very standard drift current equation and the charge density term represents the charge density at the middle of the channel or the average charge density. Then, VD over L is the electric field produced along the y direction, the channel direction, produced by the drain voltage. So, the long-channel MOSFET equation really represents the current that would flow in the channel containing on average charge density of this guy is here, drift along a constant electric field of VD over L. Now, in reality, of course the inversion layer charge density changes and it actually decreases, charge density actually decreases as it approaches the drain. All the while, your ID should remain constant as a function of y because Kirchhoff law says that as long as there is no branch, your current should be constant everywhere, right? So, our simple model predicts that in saturation, your channel current is determined by the rate at which the carrier arrives at the pinch-off point. Once, the carrier reaches the pinch-off point then it just gets drifted by the large built-in electric field inside the depletion region. So, that was what we assumed. Now, in the long-channel MOSFET, the small movement of the pinch-off point inward is not important. However, as the channel length becomes reduced in modern IC technology, the channel lengths reduction has been one of the main drives in the technological development. So, this pinch-off region with L minus L prime in this figure here, that region width now may not be completely negligible anymore. In that case, we have to somehow take that into account. How do we do that? The best way to do it, is to consider this as a reduced channel length. So, we still assume that the carriers arrived at the pinch-off point gets promptly swept away by the larger built-in electric field. So, the net effect of this pinch-off region is to reduce the channel length and this effect is called the channel length modulation. So, your current actually increases due to the reduction of your effective channel length and if your delta L, which is defined as L minus L prime. If that remains small compared to your total channel length, you can approximate this as this. So, now you can show that your ID, your drain current, increases linearly with the reduction of your channel length. You can calculate the reduced channel length using the Poisson's equation and you can, in much the same way that as we did in the p-n junction and the Schottky contact case and that will give you this equation here. However, this equation is not entirely accurate because we have ignored the free carrier density that's still floating around in the region and also, we have adopted this gradual channel approximation and uses a simply 1D case. So, in order to be more accurate you have to use the full 2D Poisson's equation. Nevertheless, the approach of describing the saturation drain current with the reduced channel length is effective and valid. So, saturation drain current is written in a similar way to the only fact in Bipolar Junction Transistor and is written by this. So, the effect of channel length modulation is represented by this extra term here, which is linearly proportional to the drain voltage. So, this leads to a finite resistance in the saturation region. Then so, if you plot the IV characteristic once again, the saturation region which we use to describe it with a flap constant saturation current, now is linearly increasing with the drain voltage and the slope is determined. Slope is characterized by this parameter V sub A and if you extrapolate this this saturation region, then it will intersect the x-axis at this value negative V sub A, this voltage is called the early voltage. The effect of this, it describes the effect of the channel length modulation leading to a linearly increasing saturation drain current relative to the drain voltage. The next effect I want to discuss is the Body Bias Effect. So far, we assumed the source voltage and the body voltage discharge are the same, but it may not always be the case. What is the effect of the body bias? Well if you look at the source and the body, this is a p-n junction. So, when you apply a positive voltage on the body, then you are applying a reverse bias voltage on this p-n junction. When you apply a reverse bias on the p-n junction, you increase the band bending. So, what it does is, you're effectively changing the potential of your bulk region. So, because of this energy band bending, you are affecting the voltage, gate voltage, that is required to create the inversion layer. So, if you look at the Module one, video four, in this course. You can find that applying a body voltage creates a shift in threshold voltage and the threshold voltage change is related to your VSP, the source to body voltage as this. So, the primary gamma is called the body effect parameter and has a unit of square root of volt and this equation is valid for a long-channel MOSFET but it will lose its validity for a short-channel device and in order to rigorously described this body bias effect, you have to use the full 2D model that describes the actual charge distribution in the short-channel device. But in any case, what is the effect of body bias? Body bias changes the threshold voltage. How does that work? Pictorially, this is the best illustration. So, here is the MOSFET on its side. So, the gate is on the front, substrate is on the back, source on the left and drain on the right. If you draw the three-dimensional energy band diagram, then for zero body bias case, it looks like this. So, this figure B, shows the case of the flat band and then this is the case, this figure C is the case of onset of strong inversion. So, you can see that by applying a voltage on the gate, you create an energy band bending along x direction here. Okay, if the band bending is large enough then you can create strong inversion. If you have a body bias however, there is a larger energy band bending between source and body. So, you start out with a larger band bending here and here. Therefore, if you apply the same voltage on the gate, the energy band bending is not large enough to create the inversion layer you have to apply a larger voltage on the gate to create the inversion layer. That is described by this delta VT shift change in the threshold voltage.
