11.5.2. Limitations and Continuity Conditions

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 th. So, the first assumption is that this is higher than V th; that means, the channel is existing. And then we apply the other potential, so we do have the V DS.

This section discusses how the channel for the MOSFET is determined by the applied voltages. The expression of current I depends on the dimensions of the channel (width W and length L) and the applied voltages (V_GS and V_DS). Assuming V_GS is greater than the threshold voltage V_th indicates that an inversion layer (the conducting channel) has formed, allowing current to flow. The main aspect discussed here is that the current I is influenced by the vertical and lateral electric fields created by the voltages.

Think of a water hose. If you increase the width of the hose (analogous to increasing W in the MOSFET), more water can flow through. If the hose is too long (like having a longer L), less water will flow, as there is more resistance. Similarly, the applied voltages dictate how much water can flow through.

So, you may say directly that I is proportional to (W/L). Now, how about the other parameters? So, this will be proportional to the conductivity in the channel regions which is controlled by this V_GS - V_th, which means that whatever the excess voltage you do have beyond the threshold voltage that is effectively contributing to the conductivity.

This part indicates that the current I is not only dependent on the dimensions of the channel but also on the conductivity, which is influenced by the excess voltage (V_GS - V_th). When the voltage exceeds the threshold, it enhances the channel's conductivity, allowing more current to flow. The relationship can be summarized by stating that the current increases as the width increases and the length decreases, while it also rises with greater excess voltage.

Imagine a river. The width of the river represents the width of the channel (W). If the river expands, more water can flow. However, if you have a long narrow channel (like having a longer L), the water flows slower due to resistance. If you increase the water pressure (akin to increasing V_GS), more water can flow through, which enhances the current.

Whenever we say that V_DS is higher than V_th and whatever the excess amount we have it is contributing towards the conductivity of the channel, but this is valid probably in this portion. This equation assumes that the V_DS is very small compared to V_GS - V_th.

At this point, the section highlights an important limitation to the previous equations. The equation derived assumes that V_DS is relatively small compared to the excess voltage (V_GS - V_th). If V_DS becomes significant, the conductivity will vary along the channel length, requiring a correction to the current expression. This means that under certain circumstances, the assumptions made earlier for simplicity may break down, necessitating adjustments in calculations.

Imagine you are trying to fill a cup with water. If the water level in the cup rises too high (akin to having a significant V_DS), it may spill over the edges, changing how you would fill it. In the same way, if V_DS is too high compared to the excess voltage, the expected behavior of the MOSFET changes.

So, we are keeping the device parameters same as µ then epsilon ε, oxide t and those things it is remaining as is, but then the conductivity we need to change... So, now the new equation it is like this, and this is valid as long as V_GS - V_DS is higher than V_th.

In this chunk, the discussion turns to how changes in V_DS imply that we need to adjust the expression for current again. The approach is to take an average of the effective voltages at both ends of the channel, considering the variations due to the applied voltage. The equation remains valid under the condition that the channel persists, allowing the current to flow despite the changes in voltage across it, and highlighting the varying conductivity along the channel.

If you think about a long highway with some traffic lights, the average speed of cars will fluctuate based on their starting point. In the same way, different portions of the channel have varying conditions of voltage affecting how current flows.

If V_GD is equal to V_th, we can say that this is equal to V_GS - V_DS. However, beyond this point, the channel starts to disappear, hence this equation assumes that the channel is about to vanish.

This section introduces the concept of 'pinch-off,' where the channel is effectively becoming incapacitated at a critical voltage. When V_GD approaches V_th, the condition of the channel deteriorates, making it unable to conduct current effectively. The implications of pinch-off are that while the channel weakens, the device behavior transitions into a saturation region where the current flow is significantly reduced, thus requiring awareness of this threshold voltage.

Imagine a sponge that is drying out. As you continue to apply pressure (like increasing V_DS), the sponge loses its ability to hold water, and it becomes less effective at soaking any additional water. Similarly, when V_GD reaches V_th, the MOSFET's ability to maintain a conducting channel diminishes.

In this condition, if I consider the voltage V higher than (V_GS - V_th), then of course in this portion the channel is disappearing. So, then practically, from here to here there is no potential drop.

Here, we see the behavior of the MOSFET in saturation. When the drain voltage (V_DS) is high enough to reach saturation, the current does not significantly change despite further increases in V_DS. In this condition, the effective channel length is shortened, and it generally results in a constant current output regardless of increases in V_DS, indicating a transitional behavior from linear to saturation modes.

Think of a water slide where the slope becomes shallow at the end. No matter how much you try to push the rider (akin to increasing V_DS), their speed does not significantly change as they exit the slide. They reach an endpoint where additional push doesn’t lead to more speed, similar to how the current stabilizes in saturation.

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