Gain for CE Amp (Simple)

Discussion in 'Homework Help' started by jegues, Apr 21, 2011.

  1. jegues

    Thread Starter Well-Known Member

    Sep 13, 2010
    See figure attached.

    I was able to successfully apply thevenins theorem and obtain the same circuit as him however I am confused as to how he concludes that,

    v_{I} = V_{T}

    I would agree that,

    v_{o} = V_{C} = -I_{C}*5k

    How does he obtain vi for the gain?

    EDIT: It seems as though hes doing the following,

    A_{vo} = -g_{m} R_{C}

    but isn't that incorrect because,

    V_{o} - 5 = -g_{m}v_{\pi} R_{C} , \quad V_{o} = -g_{m}v_{\pi} R_{C} + 5

    Since, V_{i} = V_{\pi}

    \rightarrow A_{vo} = -g_{m}R_{C} + \frac{5}{v_{\pi}}

    EDIT: I'm gonna give this one another attempt from the start and see what I get.
    Last edited: Apr 21, 2011
  2. jegues

    Thread Starter Well-Known Member

    Sep 13, 2010
    Here's my second shot at it. Is what I'm doing correct?
  3. mik3

    Senior Member

    Feb 4, 2008
    This looks correct. You don't really need to apply Thevenins theorem because when you convert the transistor circuit to its equivalent small signal model the total Rc is 10K//10K=5K.
  4. t_n_k

    AAC Fanatic!

    Mar 6, 2009
    If you recall that the emitter junction small signal equivalent is modeled as a dynamic resistance often approximated by

    re=VT/Ic=26/Ic [Ic in mA]=0.026/Ic [Ic in A].

    Then, in this case with Ic=0.5mA re=VT/Ic = 52Ω

    The CE small signal gain is usually approximated by Av=-(Rc||RL)/re

    So Av=-(10k||10k)/[VT/Ic] which if you leave Ic in mA and the R values in kΩ, leads to the solution provided.