Significance of \$V_{GS}-V_{th}\$ for MOSFETs

Question

What is this factor's significance? Please, don't give any long mathematical explanation. Seeing the drain characteristics I can easily place all the regions. But where does this \$V_{GS}-V_{th}\$ lie on the curves? Basically it just dictates how large you have made the voltage at the gate compared to the minimum required to turn it on. Why is it related to \$V_{DS}\$ in bizarre ways in saturation and linear regions?

Answer 1

Well the explanation is not extremely meaningful without math and without being familiar with the physical structure of a MOSFET. Anyway, I'll try to keep math at bay, but don't expect a quantitative answer, though!

Let's focus on an N-channel device (the reasoning is the same for a P-channel one, but all polarities are inverted). \$V_{GS} - V_{th}\$ is called by some the drive voltage or also the overdrive voltage of the FET (this is true also for JFETs). For now let's restrict the analysis to an enhancement-mode device, for which \$V_{th}>0\$.

As you seem to know, \$V_{th}\$ is the threshold voltage: below that the MOSFET is only barely conducting (there is only a tiny current and the MOSFET is in the so-called subthreshold region). When the GS voltage reaches that threshold the behavior of the MOSFET changes, because under the gate a channel running from the drain to the source is formed. That channel is the result of the gate being positive with respect to the bulk semiconductor under the oxide layer. Therefore the gate attracts electrons that end up accumulating under that oxide layer. Think of the gate and the bulk as two plates of a capacitor being charged.

If \$V_{GS}\$ is under the threshold the electrons are too few and the channel is not completely formed. Reaching the threshold enough charge has accumulated under the oxide layer to form a complete channel.

From then on further increase in voltage just increases the charge density in the channel, making it more conductive. Therefore the increase in conductivity depends on the overdrive voltage. In this situation the MOSFET may be either in the ohmic (i.e. linear) region or in the saturation region, depending on the value of \$V_{DS}\$. If \$V_{DS} < V_{overdrive}\$ the MOSFET is in the ohmic region, otherwise it is in saturation.

Why? Well, from an intuitive point of view it goes like this. If \$V_{DS}\$ is exactly zero, the channel may be viewed like a resistor whose conductivity (hence resistance) depends on its charge density, i.e. on the overdrive voltage. If \$V_{DS}\$ is small enough, the current it injects through the channel is small, and this electron flow won't disrupt the situation in the channel, which keeps behaving as a resistance.

As \$V_{DS}\$ rises, the injected charges are more energetic and cause a sensible voltage drop along the channel, that tends to contrast the action of \$V_{GS}\$. At the end this voltage drop becomes so big that the current in the channel ceases to be proportional to \$V_{DS}\$, i.e. the channel doesn't behave as a resistance any longer. Now the MOSFET has entered saturation, i.e. the current in the channel cannot grow further by the action of \$V_{DS}\$ (if we neglect higher order effects like channel length modulation, analogous to Early effect in BJTs).

Why the boundary between ohmic and saturation region is exactly the overdrive voltage? Well, that would need more math to be explained in rigorous way. Intuitively, you may think of the overdrive voltage as the "eccess gate voltage" that prevents the channel from being "saturated" by \$V_{DS}\$. Once \$V_{DS}\$ has "eaten-up" that margin, the channel will "surrender" and saturate.

For depletion devices the principle is the same, but since the channel is already formed for \$V_{GS} = 0\$ then the threshold is negative, i.e. \$V_{th} < 0\$. Thus, even for zero gate voltage you have some overdrive and you have to use negative gate voltages to drive the MOSFET into cut-off.

Let's tackle the last bit of your question: "But where does this \$V_{GS} - V_{th}\$ lie on the curves?". As I said earlier, the overdrive voltage determines the value of \$V_{DS}\$ that separates the ohmic from the saturation region. This point of separation of course depends on \$V_{GS}\$, so it depends on the particular curve you consider on the output characteristics. You can easily determine this point for each curve. Approximately it is the point where the knee that joins the ohmic region with the saturation region ends. But you can draw it precisely by computing the overdrive voltage for each curve and intersecting that curve with a vertical line of equation \$V_{DS} = V_{overdrive}\$ (see the red dots in the image below). It turns out that all those points lie on a parabola, so not so "bizarre" after all!