# Relationship between Vds and Vgs- MOSFET

I've been experimenting around with a MOSFET and was measuring the $$\V_{gs}\$$ and $$\V_{ds}\$$ for a range of 0-5V. This is the plot I came up with:

Circuit schematic:

Now I'm trying to discern a relationship here but can't figure out what the constant value from 0-3.2V indicates. I've found 3.2 V to be the threshold voltage. Beyond the threshold voltage I see that Vds falls dramatically as Vgs increases until it levels off at about 3.8 V. Now I'm not very familiar with MOSFET's which is why I'm simply trying to get a qualitative view of this $$\V_{ds}\$$ and $$\V_{gs}\$$ plot. What happens before the threshold voltage? What happens after it?

• Your circuit has more than just a MOSFET. It would help answer your question if you included a schematic.
– W5VO
Mar 15 '16 at 23:25
• Oh yes. Forgot that. Please check now. Mar 15 '16 at 23:28
• Is "Cutoff Region" what you are asking about? The wiki page for MOSFETS en.wikipedia.org/wiki/MOSFET Mar 15 '16 at 23:50
• @Tyler I guess. I don't really know too much about MOSFET's. Just did this experiment and just trying to get a big picture view I guess. I understand that when the drain current starts flowing at Vt the drain voltage decreases because the current from the gate driver must go to ground to decrease its voltage. But apart from that I don't much understand the general behavior of the plot. Mar 15 '16 at 23:51

There are three basic regions of operation for a MOSFET. Simplifying a bit, they are:

1. Cutoff (Vgs < Vt) -- No current flows from drain to source.

2. Linear (Vgs > Vt and Vds < Vgs - Vt) -- Current flows from drain to source. The amount of current is roughly proportional to both Vgs and Vds. The MOSFET acts like a voltage-controlled resistor. This region is used for switching.

3. Saturation (Vgs > Vt and Vds > Vgs - Vt) -- current flows from drain to source. The amount of current is proportional to the square of Vgs, and is (almost) independent of Vds. The MOSFET acts like a voltage-controlled current source. This region is used for analog circuits like amplifiers.

In your circuit, R1 limits your drain current to about 1 milliamp, which is pretty small. It looks like it only takes a Vgs about half a volt above Vt to get that much current. If you want to see the relationship more directly, remove R2 and replace R1 with a much smaller resistor, or even just a current meter. This will let you apply a fixed Vds. Be sure to turn up Vgs slowly and carefully to avoid frying the transistor or resistor.

The MOSFET is like a switch. When your gate voltage is low (0V to ~3.2V in your diagram), it is like the MOSFET isn't even there. The switch is open. You have a resistive voltage divider in your circuit which is determining the voltage across the MOSFET as well. If you take the MOSFET out of the circuit, you will see the same voltage.

When you are above Vth, the MOSFET starts conducting and shorts out R2. When the component is shorted, there is no voltage across it (well, a very small voltage across it). If you placed a wire in place of the MOSFET, you would see the same thing.

The nearly vertical line that you see there is the change of the MOSFET between 'off' and 'on'. This is the threshold voltage (Vth).

Overall, the MOSFET works like a switch. Most people either apply 0V to the gate/source or 10V. There are some applications that work near the threshold voltage, but most are using MOSFETS as very fast switches.

The flat line from 0 to 3.2V is the 9.09V your 1K/10K divider provides when your FET is not conducting (Vgs below threshold). The steep dropoff is the region where the transistor is conducting with a resistance comparable to your divider. The flat region above 4V shows where the transistor is largely saturated, and the 1K resistor is doing all the current limiting it can...meaning 10mA off of a 10V supply. You can see this approximate point on the FET curves where Vgs reaches a level where Id>10mA.