11.3.2. Average Effect on Conductivity

So, what will be the expression of this I ? So, we do have, so this is the big question. First of all let me quickly put the biases. For vertical field we do have V here, so that creates vertical field. And let me assume that this V it is higher than V . So, the first assumption is that this is higher than V ; that means, the channel is existing. And then we apply the other potential, so we do have the V which is providing the lateral field. Then, if you see here I think it is let me go with intuitive way that I it is proportional to what? It is proportional to W. In fact, it will be proportional to because if you see here this is the L and this is the orthogonal dimension it is the W. So, if you are having higher length for everything is remaining same it is expected that the resistance here it will increase. So, as a result the corresponding current it will decrease. So, on the other hand if the W is increasing the corresponding resistance it will decrease. So, you may say directly that I it is proportional to or you can say that aspect ratio of the channel.

This chunk discusses how the current flowing through a MOSFET is influenced by the physical dimensions of the device, specifically width (W) and length (L). The volume of the channel must be considered, where a wider channel allows for more electrons to flow, thereby increasing the current (I), while increasing the length (L) will increase resistance, decreasing the current. Therefore, the aspect ratio (W/L) determines the amount of current that can flow through the device.

Imagine a water pipe: if the width of the pipe increases (akin to an increase in W), more water can flow through at once, representing an increase in current. Conversely, if the length of the pipe increases, it becomes harder for the water to flow, akin to increased resistance which reduces the current.

Now, how about the other parameters? So, this will be proportional to the conductivity in the channel regions which is controlled by this V ‒ V which means that whatever the excess voltage you do have beyond the threshold voltage that is effectively contributing to the conductivity or it is helping to increase the conductivity in the channel. So, we can say V is, V ‒ V it is directly increasing this current.

The current (I) is also influenced by the difference between the gate-source voltage (V_GS) and the threshold voltage (V_th). The higher this difference, the more significant the conductivity in the channel becomes. This excess voltage enhances the number of charge carriers in the channel, thus increasing the overall current, as it effectively drives more electrons into conduction.

Think of a hill: the height of the hill is like the threshold voltage; if you have to throw a ball over the hill, the further you can throw it (the excess voltage above the threshold), the more room (conductivity) there is for the ball to roll down once it crosses the hill, representing the current.

So, if I combine all of them, so what we can say here it is I it is say proportionality constant say × (V ‒ V ) × V , ok. And this K, this K it encapsulates whatever the device parameter is there in fact, this K if you see, if the mobility of the electrons is flowing in this way. So, if the mobility of the electron is higher of course, the current it will be better. So, mobility of the electrons is there. And also the dielectric constant of the oxide . So, intuitively you may say that this represents the capacitance of this structure which is per unit area of course.

In this section, the relationship between current (I) and fundamental parameters such as voltage and channel width is mathematically expressed. K serves as a proportionality constant that encapsulates vital device parameters like electron mobility (how easily electrons can move) and the dielectric constant (which affects capacitance). This relationship helps us understand how these device parameters affect the overall behavior of the current under varying conditions.

Imagine a race car: the speed of the car depends not only on the engine power (similar to voltage) but also on how easily the car can move through the air (mobility) and the friction with the ground (dielectric constant). A car that is built with better aerodynamics (higher K) will perform better in races, just like a better MOSFET setup will conduct more current.

So, whenever we say that V is higher than V and whatever the excess amount we have it is contributing for towards the conductivity of the channel, but this is valid probably in this portion. Then if I consider we are applying a voltage V which is say +ve, that means, the voltage across this structure it is not same as the V here. In fact, that supposed to be V which is V ‒ V . So, necessarily the conductivity of the channel towards this drain side and source side they are different. So, this equation it assumes that the V is very small compared to V ‒ V.

The assumptions made regarding voltage differences emphasize the complexity within the MOSFET's operations. The drain-source voltage (V_DS) affects conductivity differently across the channel, indicating that V_DS should be considerably smaller than the difference between the gate-source voltage (V_GS) and the threshold voltage (V_th). If these conditions are not met, the derived equations for current and conductivity become invalid.

Consider a stretch of highway with vehicles (electrons) flowing. If heavy traffic (high V_DS) builds up on one end, that impacts the flow of cars on the entire highway, much like how a high V_DS could disrupt the expected current flow through the MOSFET.

--