Understanding op amp solution

Discussion in 'Homework Help' started by Hitman6267, Jun 2, 2010.

  1. Hitman6267

    Thread Starter Member

    Apr 6, 2010
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    I also don't know V1 V2... are the voltages at what resistors.
     
  2. ELECTRONERD

    Senior Member

    May 26, 2009
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    Nodal analysis is typically used to analyze an op amp with a single voltage source on the input.

    You need to remember the ideal attrributes of an op amp:

    • Infinite Gain
    • Infinite Input Impedance
    • Zero Output Impedance
    • V+ = V-
    • I+ = I- = 0
    Consider the resistor in series with Vs. Since the V+ input has infinite input impedance, the current flow theoretically can't go into your V+ input. By KCL, the current has to go somewhere, so it goes through the feedback network. Based on those hints, see if you can solve it now.
     
  3. Hitman6267

    Thread Starter Member

    Apr 6, 2010
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    Could you tell me the effects of having the grounds the way we do ? Thats what is throwing me off. I have no idea what they do.
     
  4. ELECTRONERD

    Senior Member

    May 26, 2009
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    Take a look at your first attachement. You see how V+ is connected to GND? Therefore, at that point, it is 0V. Also keep in mind that V+ = V-, so V- = 0 as well. I'll show a quick example of solving for the current in the first resistor in series with Vs.

    I_R_1 = \frac{V_S - 0}{R_1}
     
  5. Hitman6267

    Thread Starter Member

    Apr 6, 2010
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    I get that, here's where I'm facing trouble.
    After the current passes the first R in the feedback cable does it divide itself between the R linked to the ground and the other ?

    V2= -Vs -2 isrc R , is relative the the ground linked to the Vs. The V0 they calculated is relative to which ground ?
     
  6. ELECTRONERD

    Senior Member

    May 26, 2009
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    Think of KCL, all your resistors are equal to 0.5kΩ. So after the first feedback resistor, the current splits equally into each of the next resistors.

    Try taking the op amp out of your drawing and just simplify all the resistors. Remember that the first resistor in series with Vs is also in series with the first feedback resistor.
     
  7. Hitman6267

    Thread Starter Member

    Apr 6, 2010
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    Are the first R in the feedback loop and the R connected to the ground in parallel ?
     
  8. ELECTRONERD

    Senior Member

    May 26, 2009
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    Didn't I say to take your op amp out of your drawing and solve for the resistor network?

    Look at the attachement, this can be solved by very simple source transformations.

    Or look at the Vo resistor, it's in parallel with one of the R's. You could simplify the feedback network that way also.
     
  9. Hitman6267

    Thread Starter Member

    Apr 6, 2010
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    First thank you for the effort of drawing it, second I can't do that unless I know if they're considered to be in parallel . In your drawing would the resistor that would result from the transformation of Vs and R1 be considered in parallel with R2 ?
     
  10. ELECTRONERD

    Senior Member

    May 26, 2009
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    It's almost like a puzzle, there's a certain way you have to solve it and you won't be able to solve it otherwise. In this case you'll have to work backwards, because R4 and R5 are in parallel with -8V across each of them. Or, do the source transformations yourself and you'll see which are in parallel or not.
     
    Last edited: Jun 2, 2010
  11. Hitman6267

    Thread Starter Member

    Apr 6, 2010
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    In your drawing between R4 and R5 there is a resistor missing. Did you remove it somehow ? If it still were there would R5 and R4 still be considered in parallel ?
     
  12. Jony130

    AAC Fanatic!

    Feb 17, 2009
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    All you need is to find equivalent resistance of this circuit :
    [​IMG]

    Rz = 1.6R = 800Ω

    Then we need to find the voltage on on this node
    [​IMG]
    So if we find V1 then we know the current that is flow through Vs
     
    Last edited: Jun 2, 2010
  13. Hitman6267

    Thread Starter Member

    Apr 6, 2010
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    That's what I don't know how to do.
     
  14. The Electrician

    AAC Fanatic!

    Oct 9, 2007
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    All the grounds are connected to the same point. Imagine that the circuit was sitting on a copper sheet; all the grounds could be connected to that sheet of copper. Ground is the reference for all the voltage measurements in your circuit.
     
  15. Jony130

    AAC Fanatic!

    Feb 17, 2009
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    We can try solve this circuit without finding equivalent resistance.
    We draw this
    [​IMG]


    Is = I1
    I3 = I1 + I2
    I5 = I3 + I4
    V1 = I2*R2
    V2 = I3*R3 + I2*R2
    Vout = I5*R5 + I4*R4


    Vs/R = (0 - V1) / R1 ----> V1 = -R1/R * Vs

    I2 = V1/R2 = Vs/R2*R1/R

    I3 = I1 + I2 = Vs/R + Vs/R2*R1/R = Vs/R *( (R1+R2) )/R2) =
    Vs/R*(1+R1/R2)

    V2 =
    V2 = I3*R3 + I2*R2 = Vs*(R1/R + R3/R + R3/R*R1/R2)

    I4 =
    (Vs*(R1/R + R3/R + R3/R*R1/R2))/R4 = Vs/R4 * (R1/R + R3/R + R3/R*R1/R2) = Vs/R4 * [ (R3*(R2+R1))/(R*R2) + R1/R ]

    I4*R4 = (R1/R + ((R1 + R2)*R3)/(R*R2)) *Vs

    I5 = I3 + I4 and so on
    We finally find the voltage gain:
    Au = - (R1*R5*( R2 + R3 + R4 ) + R1*R4*( R2 + R3 ) + R2*R5*( R3 + R4 ) + R2*R3*R4 ) / (R*R2*R4)

    It would be much faster if we substitute the nambers in the first place.
     
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