When a MOSFET is operating in saturation region, i.e. \$V_{GS} > V_{th}\$ and \$V_{DS} ≥ ( V_{GS} – V_{th} )\$, the drain current equation clearly indicates the parabolic relation between the drain current and input voltage:

\[ I_{Dsat}=Kn' \cdot \dfrac{W}{2L} \cdot (V_{GS}−V_{th})^2 \]

hence \$I_{Dsat}\$ is directly proportional to the square of \$V_{GS}\$.

However, when operated in triode region, the drain current equation is given by:

\[ I_D=Kn' \cdot \dfrac{W}{L} \cdot \left[(V_{GS}−V_{th})V_{DS} - \dfrac {V_{DS}^2} {2} \right] \]

Am I right in saying that, for a fixed \$V_{DS}\$, the drain current is linearly dependent on \$V_{GS}\$? Although the equation indicates a linear relationship between drain current and input voltage for a fixed \$V_{DS}\$ such that the device is in linear region, the results of simulation are quite different.

Instead of a straight line, the simulation results in a parabolic curve that saturates for some value of \$V_{GS}\$ (even if there is no resistor connected between the drain and \$V_{DD}\$).


Here are a few questions:

What I understand is that, if the MOSFET is in saturation the ID versus VGS curve would be a parabola as shown:

MOSFET in Saturation

Changing VDS has no effect on this curve (neglecting channel length modulation) right?

But if ID versus VGS is plotted for the MOSFET in triode region, it would be linear(For a fixed VDS such that the device is in triode region) as evident from the equation: \[ I_D=Kn' \cdot \dfrac{W}{L} \cdot \left[(V_{GS}−V_{th})V_{DS} - \dfrac {V_{DS}^2} {2} \right] \]

So in the above equation, if VDS is fixed, ID would vary linearly with VGS.

Why isn't this characteristic of MOSFET exploited? Why do we settle for the "nearly-linear" ID versus VDS relation when we can have a perfectly linear variation of ID with VGS?

  • 1
    \$\begingroup\$ Have you looked at the characteristic shape of a real mosfet? \$\endgroup\$
    – Andy aka
    Jun 16, 2015 at 14:47
  • \$\begingroup\$ Yes. But I did not find the input characteristics of a MOSFET operating in triode region ONLY in any textbook. I hope my question is clear. \$\endgroup\$
    – JGalt
    Jun 16, 2015 at 14:51
  • 1
    \$\begingroup\$ How about this: bass.gmu.edu/~pceperle/WebProjectsSpr03/Johnson-Resistors/… \$\endgroup\$
    – Andy aka
    Jun 16, 2015 at 14:56
  • \$\begingroup\$ Thanks for the link. My question is about the Input Characteristics(ID versus VGS) of a MOSFET operating in linear region. Your link describes the output characteristics. \$\endgroup\$
    – JGalt
    Jun 16, 2015 at 14:58
  • \$\begingroup\$ I used google to search for characteristics of FETs maybe you can do the same - there are plenty out there that show multiple curves with varying Vgs. \$\endgroup\$
    – Andy aka
    Jun 16, 2015 at 15:04

1 Answer 1


What do you mean by "input characteristics"?

Textbooks and datasheets describe the behavior of MOSFETs using two graphs:

  • Output characteristics: \$I_D\$ versus \$V_{DS}\$ with \$V_{GS}\$ as parameter.

  • Transfer characteristic: \$I_{D}\$ versus \$V_{GS}\$ at a given fixed \$V_{DS}\$ value (this latter is chosen so that the MOSFET is in saturation region).

There is no "input characteristic" (such as the \$I_B\$ versus \$V_{BE}\$ curve of a BJT) because the other input quantity besides \$V_{GS}\$, namely \$I_G\$, is virtually zero at DC (and all these curves assume DC operations). Therefore it wouldn't make much sense to plot \$I_G\$ versus \$V_{GS}\$, unless you wanted to analyze leakage gate current, but I assume you are not interested in that.

So it is clear (also by a comment of yours) that by input characteristic you mean the transfer characteristic (TC). Note that the TC is plotted with a fixed drain-source voltage that guarantees that the MOSFET is in saturation for each \$V_{GS}\$ value on the horizontal axis. This is done because the TC is useful when the MOSFET is in saturation, i.e. when the output current depends solely on the input voltage (not considering "Early effect"), for example when you want to use the MOSFET as an amplifier and you need to draw a load line to design its bias circuit.

If you plot the TC for different values of \$V_{DS}\$ you get a family of TC curves. For example consider this circuit simulation with LTspice:

enter image description here

Plotting the TC for different \$V_{DS}\$ values you get:

enter image description here

As you can see, the more you increase \$V_{DS}\$ the more the curve resembles a parabola, as you would expect for the TC in saturation. Notice that this part shows a threshold voltage \$V_{th} \approx 4V\$.

Let's consider what happens if \$V_{DS}\$ is not big enough to drive the MOSFET in saturation for every \$V_{GS}\$ value, like in the lowest blue curve (Note: to present a more revealing plot I selected the curve corresponding to \$V_{DS} = 2V\$, whereas the lowest blue curve above corresponds to \$V_{DS} = 1V\$):

enter image description here

As you can see, in saturation region you get a quadratic curve, whereas in triode region you get a linear curve. Everything as expected, except that real devices don't have an abrupt change between the two regions and that the linearity of the triode region is not perfect because of the device not being ideal (SPICE models usually take into account these effects).

If you see in your simulation an abrupt departure from this behavior it could be that you tried plotting the curves outside the range of the voltages/currents admissible for your device. Notice that I limited the first plot to max 14A/20V which are the absolute maximum ratings for the device I chose. If you don't keep this in mind you will destroy the device (in real life) or get odd results (in simulations).

EDIT (in response to a comment and a question edit)

You ask why the "perfectly" linear curve for \$I_D\$ versus \$V_{GS}\$ in ohmic region is not exploited. Here is some insight:

Why do you need a linear characteristic between input (\$V_{GS}\$) and output (\$I_D\$)? Usually to use the device as a (linear) amplifier. But what are the conditions that allows to have that linearity? \$V_{DS}\$ must be held constant. Therefore to make an amplifier this way you have to insert a load in the output circuit and still keep \$V_{DS}\$ constant. You can understand that such a load cannot be a simple resistor (which is the simplest kind of load). Therefore you need a much more complex circuit (with other active devices).

On the other side, you can use the same MOSFET biased in saturation and get a decent linear amplifier: even if the behavior of the device is not intrinsically linear, but quadratic, there are linearization techniques (e.g. employ simple feedback schemes, like a resistor in series with the source terminal) that allow the overall amplifier to become more linear.

  • \$\begingroup\$ Thank you for the explanation. I'd to make sure that the variation of ID with VGS in triode region is linear since I have not come across this in any of the textbooks. Why isn't this characteristic of MOSFET exploited? Why do we settle for the "nearly-linear" ID versus VD relation when we can have a perfectly linear variation of ID with VGS? Could you please have a look at the edited question? \$\endgroup\$
    – JGalt
    Jun 17, 2015 at 6:17

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