# 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:

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: $g_m = \frac{I_d}{V_{gs}}$

• Please provide a link to the DS that you got the graph from. – Andy aka Jun 3 '18 at 10:29
• Something extra should be added to that sentence for it to be true: "that the transconductance of a FET or MOSFET is less compared to that of a bipolar junction transistor (BJT) at the same biasing current" Without saying that the biasing current (Ic, Id etc) is the same the statement is not true. – Bimpelrekkie Jun 3 '18 at 10:31
• Here's the datasheet where I got this graph:onsemi.com/pub/Collateral/MMBFU310LT1-D.PDF – Simon Maghiar Jun 3 '18 at 10:32
• Also you should look up what transconductance (gm) actually is and how it relates to a DC transfer curve. You mention Idss which isn't relevant, even with emitter degeneration gm is still relevant. How relevant depends on the value of the emitter deg. resistor and the value of gm. I think you still have some studying to do to fully see the relations between these parameters. Next time you "hear something" ask for an explanation from that person. Remembering statements without understanding them is pointless and not the way to learn electronics. – Bimpelrekkie Jun 3 '18 at 10:36
• gm is a slope of Iout = f(Vin). For a FET's it is Id = f(Vgs). And for BJT's Ic = f(Vbe). And the steeper slope is the "better" amplifier we can build (more gain for a smaller change in the input). electronics.stackexchange.com/questions/302832/… – G36 Jun 3 '18 at 11:19

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

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

or

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

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.