Transistors analysis

Discussion in 'Homework Help' started by ___, Apr 19, 2011.

  1. ___

    Thread Starter New Member

    Apr 19, 2011
    Hello all. I've been working on this question and it's killing me. I have solved the entire question (there are more parts :p), but I could only do it once I was told that \small V_{GS} = - V_S + V_G = - V_S.
    I was just wondering why \small V_{G} is at 0. I am having a hard time getting my head around that.
    Shouldn't \small V_{G} simply be \small V_{i}?


    dI = \frac{\partial I}{\partial V_{GS}} dV_{GS} + \frac{\partial I}{\partial V_{DS}}dV_{DS}
  2. Adjuster

    Late Member

    Dec 26, 2010
    You are asked to calculate the operating point. This is usually understood to mean the quiescent condition, where vi = 0.

    Depending on whether the signal source is assumed to be able to pass DC, you might also want to consider whether the gate leakage current would develop a significant voltage in the gate resistor. Unless you can determine the gate leakage current, you might assume it to be negligible.

    Edit: you are told the transistor is ideal, so definitely ignore gate current.
  3. ___

    Thread Starter New Member

    Apr 19, 2011
    Aha! So the operating point isn't calculated for a small signal input model?
    Thanks a lot :D
    I have a couple of more questions regarding small signal model (Don't want to start a new thread just for that):

    How do you decide whether a component is going to be a short or open circuit. For example, for midband frequencies a capacitor is a short. What about an inductor? How do you decide what is ground and what signals to keep?

    And here:


    My first response was that it behaves as a high pass filter, but could you please shed some light into this too?

    Any help would be greatly appreciated.