How to classify the voltage transfer curve into various regions, as shown, enter image description here

(In the image, \$V_0\$ refers to \$V_{DS}\$ and \$V_i\$ refers to \$V_{GS}\$)

Since, the various regions of operations are classified based on the output characteristics of the MOSFET, where \$ V_{GS}\$ is constant for each curve.I mean, why is the MOSFET pushed into various regions while we change \$V_{GS}\$.


The circuit, enter image description here

  • \$\begingroup\$ Please add a schematic to show us the circuit in question. This curve is not for a single MOSFET in isolation, it is for some kind of circuit. In particular, we need to know where \$V_I\$ and \$V_O\$ are connected. \$\endgroup\$ – Elliot Alderson Apr 6 at 14:09
  • \$\begingroup\$ @ElliotAlderson I've added it, Please take a look \$\endgroup\$ – Aravindh Vasu Apr 6 at 14:21
  • \$\begingroup\$ This might be probably a transfer curve for an inverter. If VGS is smaller than VT than Vout would be high. This is the region between 0V and 1V in the transfer curve. It is called cutoff region. Mosfet is off. If vGS is greater than VT and vGS - vT>vO than mosfet is operating in triode region. In this situation since mosfet is open. vO would be almost zero. Between triode and cutoff region, there is saturation region. If we want to use mosfet as an amplifier we should operate in saturation region. \$\endgroup\$ – Erdem Apr 6 at 14:26
  • \$\begingroup\$ Yeah, but then, my question is, how does increase in \$V_{GS}\$ pushes the MOSFET into various regions, as the various regions are named based on the output characteristics. \$\endgroup\$ – Aravindh Vasu Apr 6 at 14:29
  • \$\begingroup\$ Are you comfortable with Idrain = K/2 * WL * (Vgs-Vt)^2? The derivative of that provides the GM. For long channel FETs. \$\endgroup\$ – analogsystemsrf Apr 6 at 18:28

As you change \$V_{GS}\$ the transistor's conductivity will change, meaning that \$I_{DS}\$ will change. Now, because you have added a resistor to the circuit, KVL tells us that $$ V_{DS} = V_{SUPPLY} - R \times I_{DS}$$ So, \$V_{DS}\$ must change as you change \$V_{GS}\$, and it changes in a way that satisfies KVL and Ohm's Law.

Try drawing a straight line on your family of curves connecting all of the points that satisfy the equation above. You will see that this is the path that your transistor takes as it moves from cutoff, to saturation, to linear behavior.

  • \$\begingroup\$ Okay cool, that's the load line. I got the direct relation, \$V_{DS}=V_S - \dfrac{k}{2} \,(V_{GS}-V{T})^2 \, R\$ , Thank you. \$\endgroup\$ – Aravindh Vasu Apr 6 at 14:57
  • \$\begingroup\$ Oh, wait, I'm confused again, when we look at the input characteristics ( \$V_{GS} vs i_{DS}\$, the \$I_{DS}\$ keeps on increasing, as we increase \$V_{GS}\$, as the cure is governed by the eqn, \$i_{DS}=\dfrac{k}{2}(V_{GS} -V_{T})^2\$. It never 'saturates'. Please elucidate, I'm missing something evident. \$\endgroup\$ – Aravindh Vasu Apr 7 at 0:43
  • \$\begingroup\$ Look at your family of curves. Look at the point where \$V_{GS} = V_{DS}\$ on a few of the curves. The transistor has just entered saturation at those points. Of course, in reality the situation can be much different depending on the relationship between \$V_T\$ and the supply voltage. \$\endgroup\$ – Elliot Alderson Apr 7 at 12:23

You should better see it in action by trying it in a simulation software or in an oscilloscope.


Consider this inverter circuit. There is a 5V DC voltage source to power the amplifier and a triangle-wave voltage source to generate an input voltage that ramps from 0V to 5V over 500ms.

If we make a transient analysis for 0.5 mili seconds and plot the input voltage at x axis and output voltage at y axis we get transfer curve for the mosfet.

mosfet transfer curve

This is called vo vs. vi curve for the amplifier.


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