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Another non-linear device question

  1. Jun 26, 2010 #1
    1. The problem statement, all variables and given/known data
    See figure.

    2. Relevant equations

    3. The attempt at a solution

    I tried KVL (see figure), but still got stuck with only 1 equation and two unknowns. Any ideas/tips on what I should try next?

    Attached Files:

  2. jcsd
  3. Jun 26, 2010 #2
    First, I don't like the wording of this question. It is a little confusing. I also don't like the notation. It is also a little confusing. I don't normally give full answers when no good attempt has been made, but I can see why you were confused.

    My interpretation of this confusing question is that they are asking you to linearize the response for small signals. The capital V would represent a large DC signal and the small v (with a small s subscript) would be a small AC signal. The solution is to simply take the derivative of the nonlinear equation dv_Q/dv_S and this should approximately equal v_q/v_s. I get dv_q/dv_s=3v_S-3=12 when V_S=5.
    Last edited: Jun 26, 2010
  4. Jun 26, 2010 #3
    How did you get this?

    [tex]\frac{3}{2}(v_{S} - 1)^{2}

    \Rightarrow 3v_{S} - 3

    \Rightarrow 3(5 + v_{s}) -3


    What is this equal to anyways, we don't know what vQ is?
  5. Jun 27, 2010 #4
    Sorry, I just noticed that you asked a question about my response.

    First, let me say that my answer is my best guess on how to solve the problem. When I look at the wording of the problem I find it vague and misleading. It's possible that I'm missing an obvious thing, but all I can do is give you my opinion.

    That said, I can answer your question.

    The way I interpret the problem is that you need to linearize the nonlinear equation and find the response of small changes in v_Q to small changes in v_S. The interpretation I make is that v_q and v_s are small signals that take place around the operating point V_S=5 V and V_Q=24 V;

    Hence, v_q/v_s can be approximated by the derivative dv_Q/dv_S evaluated at the DC operating point of V_S=5 V and V_Q=24 V. Hence the following

    [tex] {{v_q}\over{v_s}}={{\Delta v_Q}\over{\Delta v_S}}\approx{{d v_Q}\over{dv_S}}\bigg|_{v_S=5V}=12 [/tex]

    The fact that I arrive at one of the possible answers gives me some confidence that this is correct, but I'm not totally confident that another interpretation isn't possible.
  6. Jun 27, 2010 #5
    How did you know that V_Q = 24 V?

    [tex] {{v_q}\over{v_s}}={{\Delta v_Q}\over{\Delta v_S}}\approx{{d v_Q}\over{dv_S}}\bigg|_{v_S=5V}=12 [/tex]

    When you evaluate this you find that [tex]v_{S} = 5\cdots v_{s} = 0 [/tex]

    And if [tex] v_{s} = 0 \cdots \frac{v_{q}}{v_{s}} = undefined [/tex]
  7. Jun 28, 2010 #6
    Bump, still looking for some help!
  8. Jun 28, 2010 #7
    Sorry, somehow I keep missing this thread.

    I found V_Q simply by plugging V_S=5 in for v_S in the nonlinear equation.

    I interpret v_s to mean a small signal that is not zero, but just small compared to V_S. The small response changes of v_q in response to small changes in v_s can be approximated by the derivative. Since the derivative changes at different points on a curve, if the function is nonlinear, we need to evaluate the derivative at the operating point.

    This probably looks very confusing, the first time you see it, but it's a very common procedure for linearizing nonlinear circuits. If you can find a good analog circuit design book that covers transistors and diodes, you'll find this concept explained in a much better way than I've done here.

    Somewhere I have a help-note I wrote on linearization. I'll try to find it and I'll post it if I find it. [EDIT: I can't seem to find it]
    Last edited: Jun 28, 2010
  9. Jun 28, 2010 #8
    Thanks for your responses stevenb, things have become MUCH more clear.
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