# How would one calculate the transconductance of the FETs in this circuit?

I was given formulas to find the transconductance (gm) of the FETs using IDSS, VP and VGS, however, none of the ways I tried them leads to the correct answer (given as 2,55mS).

The two formulas you show are correct. But you can't compute $$\g_m\$$ without knowing $$\V_{\text{GS}}\$$. And you need the DC operating point to tell you that value.

That means you must work out the DC quiescent operating point for each stage.

In your circuit, you can tentatively consider the gate current as zero. (This assumption is justified by the fact that you aren't given the information that would be needed to compute its actual value.)

So the voltage drop across the $$\3.3\:\text{M}\Omega\$$ resistor can be treated as $$\0\:\text{V}\$$ and therefore the gate voltage is $$\V_\text{G}=0\:\text{V}\$$. From this, you can work out that $$\V_\text{GS}=V_\text{G}-V_\text{S}=0\:\text{V}-I_\text{D}\cdot 680\:\Omega\$$. So $$\I_\text{D}=\frac{-V_\text{GS}}{680\:\Omega}\$$.

Recall that $$\I_\text{D}=I_\text{DSS}\left(1-\frac{V_\text{GS}}{V_P}\right)^2\$$. Substituting, find: $$\10\:\text{mA}\left(1-\frac{V_\text{GS}}{-4\:\text{V}}\right)^2=\frac{-V_\text{GS}}{680\:\Omega}\$$. Solve that for $$\V_\text{GS}\$$.

solve( Eq( 10e-3*(1-vgs/-4)**2, -vgs/680 ), vgs )
[-8.46217412794563, -1.89076704852496]


Obviously, the larger magnitude is wrong. (You should be able to quickly see why.)

Select $$\V_\text{GS}\approx -1.89\:\text{V}\$$.

(2*10e-3/abs(-4))*(1-(-1.89)/(-4))
0.00263750000000000


$$\g_m\approx 2.64\:\text{mS}\$$. That's pretty close to the value you say is the correct answer.

Let's look at your equations, though. Just as a double-check. The definition of $$\g_m\$$ is $$\g_m=\frac{\text{d}}{\text{d}\, V_\text{GS} }I_\text{D}\$$:

diff(idss*(1-vgs/vp)**2,vgs)
-2*idss*(-vgs/vp + 1)/vp


This result validates your equations (recalling that $$\V_P\$$ is negative, so when using $$\\vert V_P\vert\$$ the -2 must become +2.)

Recall: I computed $$\g_m\approx 2.64\:\text{mS}\$$. (With JFETs that's more digits than can be justified.)

The stated answer was $$\g_m\approx 2.55\:\text{mS}\$$.

I'd consider that close enough.

• Gate current is also easily concluded as the gate is pulled to the lowest voltage in the circuit, whereas S and D are same or higher voltage. Commented Jul 5 at 10:04
• @TimWilliams I'm not seeing how it can be concluded from the direct information supplied in the problem. Would you please educate me, given that it is easily concluded? I'm feeling a bit like an idiot, for now. (But I'll happily test out the idea in simulation to learn more.) Commented Jul 5 at 10:07
• At the most basic, a JFET is a complicated diode. D and S are joined by conductive material (channel), and N-channel means G is the P-type joined to it. Hence the symbol's arrow into the channel, indicating a P-N junction. Indeed, JFETs with D+S tied are regularly used as picoampere diodes. The voltage between D and S, or the degree of channel pinchoff, doesn't change this analysis -- or at worst, we can treat G-D and G-S as independent diodes. Cheers! Commented Jul 5 at 10:12
• @TimWilliams I know. But, and this is critical because of the 3.3 MOhm resistor present, how do you quantitatively compute Ig directly from the information provided in the problem. I'm not seeing the computation without more information. Commented Jul 5 at 10:18
• Yeah, not sure about the exact numerical value; also even if there were Ig given, it would only make gm higher. Perhaps there is other background information given in the textbook, but the formulas you used should be pretty standard. Commented Jul 5 at 10:47