In an enhancement mode MOSFET, voltage applied to the gate terminal increases the conductivity of the device. The main advantage of a MOSFET is that it requires almost no input current to control the load current, when compared with bipolar transistors (bipolar junction transistors/BJTs). Operating as switches, each of these components can sustain a blocking voltage of 120 V in the off state, and can conduct a continuous current of 30 A in the on state, dissipating up to about 100 W and controlling a load of over 2000 W. This holds true if each enhancement load has more than its threshold across it, and makes current calculation easier as you only need to consider a single nmos with half the supply voltage.Two power MOSFETs in D2PAK surface-mount packages. The circuit you provided with two depletion loads in series acts just like a voltage divider made out of two equal diodes, so the output voltage is half Vdd. A better inverter uses a depletion nmos or CMOS, but that is not relevant here. This makes this circuit an inverter, although it is not a particularly good one, it is better than having a resistor as that takes a large amount of space on a chip. Turning the bottom nmos on will form a voltage divider, getting the output close to zero. 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. First of all, I'm sure you ment Vds >= Vgs - Vth for a MOSFET in saturation.
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