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I tried to do an exercise from my textbook where I have to discover the gain for mid frequencies of the MOSFET below, but I can't solve it. I got stuck at one point.

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I drew the small signal equivalent model of the MOSFET.

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With this model I got that Vout = -gmVGS(RD//RL), so A = -gm*(RD//RL)

I couldn't find a way to calculate gm with the data I have from the exercise description.

Problem description: The circuit below presents a common source amplifier, whose source VDD = 12 V. Considering that the MOS transistor that composes the amplifier has infinite r0 and negligible intrinsic capacitances.

Question Draw the equivalent small-signal model of the amplifier, considering the capacitances indicated in the figure and determine the gain for average frequencies (gain maximum or Vout/Vin)

Data: : ID = 0,8 mA, Vtn = 1,0 V, RG1 = RG2 = 16MΩ, RD = RL = 16kΩ, RSig = 100 kΩ, CC1 = 4nF and CS = 4μF

I would appreciate your help to find the better way to discover the answers.

Andrew Rusell
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  • Gm is a transistor property so I would think the problem either gave you the characteristics of the transistor or information about the input and output. – DKNguyen May 29 '22 at 17:11
  • What data *are* you given? – Hearth May 29 '22 at 17:14
  • I updated the question with the problem description and all the data that I have. – Andrew Rusell May 29 '22 at 17:20
  • What is Vtn? is that the threshold voltage? – Hearth May 29 '22 at 17:25
  • Exactly. Vtn = Vth for NMOS – Andrew Rusell May 29 '22 at 17:28
  • I'm rusty as all hell but I'm pretty sure you need to have calculated the gm from the DC bias version of the circuit, not the small signal (AC) model. Then with the small signal model, it's assumed the input signal is small such that gm can be treated as a constant because the small signal is biased around the DC bias point. – DKNguyen May 29 '22 at 17:28
  • I know two formulas to calculate gm: 1- gm = kn'*(W/L)*(VGS - Vth) 2 - gm = 2Id/(VGS - Vth) But, I don't have VGS neither kn'*(W/L) – Andrew Rusell May 29 '22 at 17:45
  • But you can calculate Vgs without any problem based on the information that was given to you. – G36 May 29 '22 at 17:58
  • I cant see how can I do it, I don't have the Vsig value, so I can't use voltage divider. Could you provide me some tip to understand how can I achieve VGS without that? – Andrew Rusell May 29 '22 at 18:09
  • It sounds like you just don't know the difference between the DC circuit version of the circuit (which is responsible for the biasing the transistor at a quiescent operating point) and the and the AC small signal circuit (that operates about that bias point). That means you are unaware of the first half the analysis process. DC circuit: turn of all AC sources and all capacitors become open circuit and all inductors become short-circuit. Does that ring any bells? You should have already gone over this in class and if you are doing this on your own it means you skipped half the material. – DKNguyen May 29 '22 at 18:24
  • Thank you @DKNguyen, I studied the topic you mentioned and now I understand how to proceed. – Andrew Rusell May 30 '22 at 00:13
  • I have some problems here: 16 MOhm in the source path (RG2)? ID=0.8mA seems to be impossible at 12V supply with this RD and RL – Jens May 30 '22 at 00:35
  • @AndrewRusell Glad to hear it. Good luck. – DKNguyen May 30 '22 at 01:02

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