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Compute the effective Ron in the triode region of the v-i characteristic when Vgs = 3V, i.e., measure the current through the MOSFET switch when, say, Vds = 1V and report Ids/Vds as your answer below. Remember this should be for a device whose W/L is 1.

So I realize you can't just give me the answer, I just wanted to know if there was an equation to find Ron. I haven't learnt it yet, sorry. I only started MOSFETs a week or two ago.

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    \$\begingroup\$ When in doubt, try Ohm's Law... \$\endgroup\$ – Adam Lawrence Mar 1 '19 at 16:39
  • \$\begingroup\$ You're saying use the MOSFET as a resistor? \$\endgroup\$ – user213230 Mar 1 '19 at 16:40
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    \$\begingroup\$ Not so much use as suffer the fact that a MOSFET has resistance. You are being asked to calculate that resistance at a particular point of operation. Looking up the meaning of "triode region" would probably help your understanding as well. \$\endgroup\$ – Chris Stratton Mar 1 '19 at 16:41
  • \$\begingroup\$ something like K *w/l * Ve, perhaps from differentiating the transconductance \$\endgroup\$ – analogsystemsrf Mar 1 '19 at 18:03
  • \$\begingroup\$ Yeah, but I can't know K without know ids \$\endgroup\$ – user213230 Mar 1 '19 at 20:39
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I would say that a MOSFET can be modeled as a resistor for small changes in voltage or current. I think it is misleading to say that a MOSFET has resistance.

Pick some point on one of the curves you show. Find the slope of that line (actually the tangent of the line) at that point. The slope is \$\Delta I / \Delta V\$, which is \$1/R\$ and \$R\$ is the effective resistance of the MOSFET at the chosen voltage/current point. The resistance will obviously be different at different points on the curves.

Of course, this only makes sense for the regions of operation where the current changes as a function of voltage (the linear or triode region). When the transistor becomes saturated it no longer acts like a resistor, more like a constant current source.

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  • \$\begingroup\$ @WhipStreak23 concepts are infinitely more valuable than equations. If you have a solid understanding of the concepts, equations become rather easy to derive as needed. This answer provides all the needed concepts in this case. \$\endgroup\$ – Edgar Brown Mar 1 '19 at 20:43
  • \$\begingroup\$ Not everything in real life (most things actually) will be completely characterized into a into a single nice neat equation. In particular non-linear things, like the curves you have posted above. You could approximate by breaking up the curves into a workable number of sections that are approximately linear with the same slope and assigning a constant R all voltages within that section and swap out as required. The more segments you have, the more accurate but cumbersome it becomes until you reach what a simulation does where it's basically a lookup table. \$\endgroup\$ – DKNguyen Mar 1 '19 at 20:49
  • \$\begingroup\$ @WhipStreak23 The simple equation you are looking for is Ohm's Law. \$\endgroup\$ – Elliot Alderson Mar 1 '19 at 21:00
  • \$\begingroup\$ 😑 Should have realized that earlier. Thanks Elliot! \$\endgroup\$ – user213230 Mar 1 '19 at 21:03

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