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This plot represents drain current versus gate-source voltage (with different ranges of temperature):

enter image description here

Often the transconductance of MOSFET is defined as enter image description here.

  • What is the difference between upper equation and gm = Id/Ugs (without changes)?
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3 Answers 3

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The transconductance is the ratio of the relative change of current to the relative change of gate-source voltage.

Looking at the black line of your graph, the current at VGS=4.25 is 0.6A and at VGS=4.5V it is about 1.1A.

The transconductance is given by gm=(1.1A-0.6A)/(4.5V-4.25V)=2A/V.

So, it can be used to have a linear expression to calculate the change of current for a change of voltage around the point where the transconductance was calculated.

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  • \$\begingroup\$ But if you calculate transconductance at certain point (same equation but without relative changes), you get the value of transconductance for that point only, right? Which isn't really useful in practice, right? \$\endgroup\$
    – lucenzo97
    Commented Jul 9, 2017 at 18:46
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    \$\begingroup\$ Without relative changes it is not the transconductance, because we need to calculate the derivative (or an approximation) of the curve. Calculating the ratio Id/Vgs is useless. The whole point is to make a linear approximation of the curve. \$\endgroup\$
    – Mario
    Commented Jul 9, 2017 at 18:52
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Id / Vgs is certainly a ratio appears to have no significance, just as a diode V/I has no significance as fixed ratio where we call the derivative ESR or Rs.

But let me show you my insight.

We would expect the 1st derivative of this function is quite different than the function. 1st yr calculus right? This derivative can be measured as \$\frac{\Delta Id}{\Delta V_{gs}}=g_m\$

When you choose to operate in the linear mode at a set bias Id, then you choose the Rd , compute gm and compute the Drain voltage gain.

But since this curve is exponential and we know the derivative of d(e^x)/dx is the same e^x dx, and derivative of \$e^{kx}~~ \text{is just }~~ke^{kx} dx\$

Thus k can be seen in the graph I made below from the blue curve by a constant gap or ratio.

What is the difference between upper equation and gm = Id/Ugs (without changes)?

The difference is a ratio k which appears to be constant (30+/-1) up to the Vgs threshold of 4V.

At this point both gm are flattening out as RdsOn rapidly reduces to its rated
ON resistance and gm has very low gain.

Is it relevant? not really. Unless you find some use for it.

Ideally one should use 3x Vth minimum or Vgs =12V to get near rated RdsOn and use Vgs < Vth=4V for higher gm values. However the resistance is higher so there are other factors if one wants to choose a linear operating point for max power gain.

enter image description here

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Running the FET as a switch, use Id/Vgs.

Running the FET as small-signal amplifier, use deltaId/deltaVgs.

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