Electrical – Transconductance of a device


I would like to know what does it mean when someone says "that the transconductance of a FET or MOSFET is less compared to that of a bipolar junction transistor (BJT)".

I even heard someone saying that "This particular JFET has lots of transconductance (about 25mS at its Idss of 6-12mA)" What does this mean? And how does he know this? Is it given in the datasheet of every single JFET?
There is something related to transconductance in a datasheet which is the following:

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I know that the Idss is the maximum current that a JFET can handle.

I also know that the transconductance of a device is useful to calculate the gain of a device, unless we use emitter degeneration.

For a BJT the transconductance is: \$g_m = \frac{I_c}{V_t}\$ where (\$V_t\$ is thermal voltage)

For a JFET the transconductance is: \$g_m = \frac{I_d}{V_{gs}}\$

Best Answer

The transconductance tells you how much the current changes when you increase/decrease the gate/base a very tiny bit. It is a small-signal parameter. So a \$g_m = 25mS\$ at a \$v_{GS} = 0V, v_{DS} = 10V\$ like in your graph will mean that if you increase \$v_{GS}\$ a very tiny bit by \$\Delta v_{GS}\$, that the drain current will also increase a bit by \$\Delta i_d \approx 25mS \cdot \Delta v_{GS}\$.

For BJT's, the transconductance gain can be approximated by

$$g_m \approx \frac{I_c}{n V_T}$$

With \$n\$ the emission coefficient, \$V_T\$ the thermal voltage.

This means that the transconductance is proportional to \$I_c\$, or

$$g_m \sim I_c$$

For MOSFET's (similar to JFET's) the situation is a bit different. The approximation of the transconductance gain is here:

$$g_m \approx \frac{2 I_d}{V_{GS} - V_{TH}}$$

In order to make \$g_m\$ go up, we can just decrease \$V_{GS}-V_{TH}\$, however: the current \$I_d\$ will also decrease when doing that. It turns out that this decrease is approximately:

$$I_d \sim (V_{GS}-V_{TH})^2$$

So you can write that the transconductance is proportional to

$$g_m \sim \sqrt{I_d}$$


$$g_m \sim \frac{1}{\sqrt{V_{GS}-V_{TH}}}$$

And this is a bit annoying. This dependency is much slower! So in order to get the same \$g_m\$ for a FET, you will need a lot of current (limited by power consumption and velocity saturation, where the formula doesn't apply anymore), or almost no \$V_{GS}-V_{TH}\$ voltage (where \$I_d\$ will usually reach impractically low levels \$\sim nA\$). There is one way of solving this, and that is making the FET gigantic, but that is usually impractical as well and it makes other parasitic effects worse.