VCCS Analysis

Discussion in 'Homework Help' started by tAllann, Oct 27, 2013.

  1. tAllann

    Thread Starter New Member

    Oct 26, 2013
    11
    0
    [​IMG]


    I solved for Vo in terms of Vi and I get the negative of the correct answer according to the book. Someone told me that the TA set the node to the left of Vo = to -Vo, and this explains why my answer is the negative, but I don't understand why the node adjacent to Vo is -Vo instead of just Vo.

    In this problem, gm = 2m / 1 ohm, and I get that Vo = -2 Vi, but the book gets that V0 = 2 Vi.
     
  2. WBahn

    Moderator

    Mar 31, 2012
    17,777
    4,805
    gm = 2m/Ω = 2mA/V

    Vo = -(gm*Vx)*2kΩ

    Vx = Vi/2

    Vo = -((2m/Ω)*(Vi/2))*2kΩ

    Vo = -2Vi

    You're right, they're wrong.

    I have NO idea why they are putting Vo on the top-right node of the left hand circuit (that's what you are talking about, right?).
     
    anhnha and tAllann like this.
  3. LvW

    Active Member

    Jun 13, 2013
    674
    100
    tAllan,
    the circuit shown equals the classical equivalent small-signal ac circuit diagram for a common emitter transistor amplifier. As you will know, this amplifier has inverting gain characteristics.
    Thus, for a positive input voltage, the output is negative, indeed.
     
    tAllann likes this.
  4. tAllann

    Thread Starter New Member

    Oct 26, 2013
    11
    0
    But can't you solve the circuit without knowing this with simple analysis? Is my analysis wrong?
     
  5. WBahn

    Moderator

    Mar 31, 2012
    17,777
    4,805
    Of course you can solve it without relying on this. You did. I did.

    He is just pointing out that this is a circuit that happens to match a common circuit that you will run into and, if you DID happen to know that, that it confirms that the output voltage is inverted relative to the input voltage.
     
Loading...