Can anyone explain me how I can find gm from this schematic below (for a nmos transistor)when VGS=6V and VDS=6V?
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1\$\begingroup\$ At Vgs = 8v, Iout is 5.4 amps; at Vgs = 4v, Iout is 2 amps; Difference is 3.4 amps, with 4 volts change in Vgs. Result? approximately 0.82 amps/volt. \$\endgroup\$– analogsystemsrfCommented May 13, 2017 at 12:48
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\$\begingroup\$ Thanks analogsystemsrf.Are you sure about this answer?Because I have read that the only way to find gm from a graph, is the slope of a curve from a graph ID-VGS. \$\endgroup\$– gr1Commented May 13, 2017 at 12:55
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\$\begingroup\$ Also, what is your opinion on how to find Vth.We can say Vth=2V where the current is very smallo or you can think a different way to find it? \$\endgroup\$– gr1Commented May 13, 2017 at 12:59
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\$\begingroup\$ @gr1 - Look at the change of Id vs Vgs at 6 volts. Stepping from 2 to 4 to 6 to 8 gives close to constant increases in current. So you can assume the transconductance is close to constant over that interval and you can calculate it as was done. It's not exact, but it was not represented as such. And without knowing the threshold current criterion, there is no way to figure the threshold. \$\endgroup\$– WhatRoughBeastCommented May 13, 2017 at 13:04
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1\$\begingroup\$ FETs, at least on silicon, have a V^2 region (if gate length > 1micron) and a subthreshold region which is exponential and is useful for taking LOG(V) in some applications. Thus "Vt" does not exist. \$\endgroup\$– analogsystemsrfCommented May 13, 2017 at 18:00
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Well \$g_m = \frac{\Delta I_D }{ \Delta V_{GS}} \approx \frac{3.7A - 2A}{6V - 4V} \approx \frac{1.7A}{2V} \approx 0.85\; S \$
For \$V_{GS2} = 6V, I_{D2} = 3.7A\$ and \$V_{GS1}=4V , I_{D1} = 2A\$
But this way was already shown.
But from the plot, we can also find \$V_{TH}\$ using this equation:
$$V_{TH} = \frac {V_{GS1} \sqrt{I_{D2}} -V_{GS2} \sqrt{I_{D1}}}{\sqrt{I_{D2}} - \sqrt{I_{D1}}} $$
And \$K_P\$ factor
$$K_P=\left ( \frac{\sqrt{2I_{D1}}-\sqrt{2I_{D2}}}{V_{GS1} - V_{GS2}} \right )^2$$
But in your case, these equations do not give any sensible results. So there is something wrong with your plot.
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1\$\begingroup\$ those devices look short so between that and the improbable square law compliance, it is what it is \$\endgroup\$– b degnanCommented Nov 18, 2017 at 13:49