When the bottom nmos is turned off, the output will be Vdd - Vth. One circuit where this is useful is one in which an enhancement load is placed on top and an nmos is on the bottom, the output is taken from between them, as follows: In this configuration, the nmos is called an enhancement load, this makes the nmos act like a diode but the curve is follows a square expression instead of exponential, thus making it a non-ohmic resistor, as follows: Once you managed that, it becomes an increadibly useful tool while designing and analysing circuits. At least for me, it was the hardest part to wrap my mind around the fact that three quantities are plotted in the same graph. then have a deep look at the output characteristics graph.make sure you understand what the voltages Vds and Vgs mean in general and for your circuit.However, keep in mind that it is not a hard boundary and the whole model of underlying theory of how a MOSFET operates just an approximation. To be honest, I'm not sure if there is a deeper physical explanation for this formula or if it is just a convenient coincidence. If Vds is smaller, we are on the left, in the linear region. If Vds is larger, we are on the right of the red line in the saturation region. Check it yourself!Īt the point where it crosses the blue Vgs - Vth = 4V, Vds is also 4V. This line follows the equation Vds = Vgs - Vth. Take a closer look at the red line seperating the regions (in reality, this is not a hard transistion but rather a soft change). In the linear region, the drain current is dependent on Vds, and the MOSFET behaves roughly like an ohmic resistor. The saturation region is the region in the plot, where the drain current is independent of Vds and therefore is just a horizontal line. That is why plotting Vgs - Vth is more usefull to us than plotting Vgs right now. Once Vgs is larger than Vth, all MOSFETs more or less share the shown behavior. If Vgsis smaller than Vth, the MOSFET is basically completely blocking. You can see how the drain current increases with increasing Vgs (or rather Vgs - Vth). Now have a look at the output characteristics of a standard MOSFET below (graphic taken from this answer). Simulate this circuit – Schematic created using CircuitLabīy shorting gate and drain, they share the same potential. Vds is defined as the potential difference between drain and source, Vgs as the potential difference between gate and source. From figure 3 you can also have an idea of Kn, considering that in that case the MOS is saturated and taking one or two points and substituting Vgs and Ids in the formula.First of all, I'm sure you ment Vds >= Vgs - Vth for a MOSFET in saturation. Then, if you take the value of \$R_ \$ = 5 Ohm.Īn approximated threshold voltage can be argued from the figures at page 4, respectively figure 3 and 6 counting from left to right it appears to be about 1.5 V in that curve. You can find a first approximation of Kn if you fix Vth at an arbitrary value (possibly obtained from a measurement of your device). You could obtain approximate values of Vth and Kn taking some values of IDsat for different Vgs and trying to find the relationship (one is responsible for the linear increase while the other for the quadratic one) but since you don't have exact values, and for the considerations specified, you will hardly find reasonably accurate values. In your case, it's unlikely that Vth is 2.5V: as you can see from the curve for Vgs = 2V, Id would be near 0 with Vgs below threshold. In that datasheet it's specified to be between 0.8 and 2.5V. First observation: Vth is not accurately specified as you can see in the datasheet this because the threshold voltage is depending heavily from temperature, and there is poor interest in making it precise.
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