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Differential amplifier

  1. Sep 29, 2006 #1
    Hello!
    I'm trying to determine Vout on this amplifier and I'm having some problems with the equations,because I'm not sure if they're correct!
    This's the link to the amplifier:

    http://i75.photobucket.com/albums/i281/esmeco/Amplifier-1.jpg

    The equations I've got so far are these:

    IB=0
    IA=0
    V2-V3=R3xI3
    V3-0=I4xR4
    V4-V1=R1xI1
    Vo-V4=R2xI2

    I have some questions:how do I know from which way does the current flow(On the wire that is connected to Vo I've determined that current flows from right to left,and,on the top,the current flows from left to right)?Is the voltage on node V3 and V4 the same as VA and VB respectively(As Va=Vb the voltage in both points are 0 volts)?
    Any help on this is really appreciated!
    Thanks in advance!:smile:
     
  2. jcsd
  3. Sep 29, 2006 #2
    When you are using the ideal op amp assumptions (for quick circuit analysis), you assume:
    [tex] i_{+} = i_{-} = 0 [/tex]
    [tex] v_{+} = v_{-} [/tex]

    Thus the node you labeled V4 is at the same potential as V3.

    Write the KCL equations for the current flowing into the nodes.
     
  4. Sep 29, 2006 #3
    Oh, and VA and VB are not zero volts. You have not grounded either of them.
     
  5. Sep 30, 2006 #4
    So,since there is negative feedback,V4=V3,but would not they be equal to 0?JUSt another question:if there wasn't a resistor connected to V4,would V4 be equal to V1 and thus V3 equal to V1 and the current that would flow from V4 to V1 equal to 0?
     
    Last edited: Sep 30, 2006
  6. Sep 30, 2006 #5
    So,since there is negative feedback,V4=V3,but would not they be equal to 0?

    No.

    V4 = V3, because that is an assumption you make. When you are working with an ideal op-amp model.

    I pulled this from wikipedia,
    "For any input voltages the ideal op-amp has infinite open-loop gain, infinite bandwidth, infinite input impedances resulting in zero input currents, infinite slew rate, zero output impedance and zero noise."

    So you assume infinite open-loop gain.
    A differential amplifier the output is equal to [itex] A(v_+-v_-) [/itex].

    Ideally, the voltage between the terminals is zero, thus:
    [tex] A(v_+-v_-)=v_0 [/tex]
    [tex] A(0) = v_0 [/tex]
    [tex] 0 = \frac{v_0}{A} [/tex]

    Since A is infinite, v- must equal v+.

    Is the fact that there is zero current flowing into the op-amp messing with you? Why do you think V3 = V4 = 0?

    Zero current into the op-amp essentially means you can take the op amp out of there, since it acts as an open circuit. For example on the bottom you would have,

    V1-----R1------V4-------R2-------V0

    and on the top you would have,

    V2-----R3------V3-------R4------GROUND

    Now since you are using ideal op amp assumption, V3 = V4, so you get a system of equations:
    V1-----R1------VN-------R2-------V0
    V2-----R3------VN-------R4------GROUND


    For your second question,
    if there wasn't a resistor connected to V4,would V4 be equal to V1 and thus V3 equal to V1 and the current that would flow from V4 to V1 equal to 0?

    What do you mean if there wasn't a resistor connected to V4. You have R1, and R2 connected to V4. Which one are you talking about?
     
  7. Oct 1, 2006 #6
    Thanks for the explanations!I'm now understanding this...Well,when I said about a resistor not connected to V4 I meant R1,and in that situation would V4=V1?
     
  8. Oct 1, 2006 #7
    If you remove node 1 (short it for example), yup, V4 would equal V1.

    And no problem at all man. I actually have a test on Wednesday about this stuff (it's for the first class after basic circuit analysis though).
     
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