Analyzing an op amp

Discussion in 'General Electronics Chat' started by JasonMcG, Apr 28, 2013.

  1. JasonMcG

    Thread Starter New Member

    Apr 27, 2013
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    See image:
    http://i1362.photobucket.com/albums/r687/JasonMcG1/Circuit_zps98e971ed.jpg

    I have tried to decipher what i think is correct. Using the gain on each amplifier and multiplying it by the voltages i acquired Vout to be +1.9 volts. Need to know if i am correct. Also need to know what to do for part ii. Also where do you connect the battery in this circuit? Any and all help is much appreciated. Thanks.

    Jason M
     
  2. ScottWang

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    Aug 23, 2012
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    The 0.9V output voltage is right.

    How to connecting the batteries to the 741, where is the +9V of battery that it will be connecting to which is the +9V on the circuit, and where is the -9V of battery that it will be connecting to which is the -9V on the circuit.
     
  3. JasonMcG

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    Apr 27, 2013
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    0.9V or 1.9V ?
     
  4. ScottWang

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    Aug 23, 2012
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    The gain of inverting amplifier = -(9V*(Rf/Ri)) = -(9V * 0.1) = -0.9V
    The gain of noninverting amplifier = 9V*((Ri+Rf)/Ri) = 9V*(11/10) = 9.9V.

    The +V only 9V, so the output could not reach to 9.9V.
     
  5. Ron H

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    Apr 14, 2005
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    You are both wrong.

    Jason, show your work.
     
  6. ScottWang

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    Aug 23, 2012
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    When I got the answer, I just felt the values is little strange, but that is according to the formular, so what's wrong with the formular or those values?
     
  7. JasonMcG

    Thread Starter New Member

    Apr 27, 2013
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    I have renamed the components for easier referencing.
    See image:
    http://i1362.photobucket.com/albums/r687/JasonMcG1/Ciruit_3_zpsac535a8d.jpg

    Amp A: inverting
    Gain A = R2/R1 = -0.1
    -0.1x9V= -0.9V

    Amp B: non-inverting
    Gain B = 1+(R4/R3) = 11V
    11x0.1 = 1.1V

    Amp C: non-inverting 'buffer'
    Gain C = 1
    1x1.1 = 1.1V

    Amp D: 'Summing amp'
    G1=-(R7/R5) = +0.9V
    G2=-(R7/R6) = -11V
    R5//R6
    G3=1+(100/9.091) =1+11=12
    12x1 = 12V

    Therefore Vout = 0.9-11+12 = 1.9V
     
  8. screen1988

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    Mar 7, 2013
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    I think in your image Vin = -0.1V => output of B: 11x(-0.1)= -1.1V
    Gain C = 1
    1x(-1.1) = -1.1V
    Vout = 0.9+11+12 = 23.9V
     
  9. Ron H

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  10. JasonMcG

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    Apr 27, 2013
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    I have gone through it again and calculated Vout to be 23.9V. Is this the correct value?
     
  11. Jony130

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    Feb 17, 2009
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    No you are wrong. For Vcc equal to +9V LM741 won't give you more then 9V at the output.
     
    Last edited: Apr 29, 2013
  12. Ron H

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    Apr 14, 2005
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    If you had an ideal op amp with no limits on output voltage, that would be correct. As jony130 says, the output can never exceed the rail voltages. With a 741, and ±9V supplies, your output will never exceed ≈±8V.

    That is either a trick problem, or a very poorly designed one.
     
  13. JasonMcG

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    Apr 27, 2013
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  14. Ron H

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    For now, just imagine that the RC network is not there, and the sine wave appears unattenuated at the output of op amp C. What would the voltage be at that node? What would the output of op amp D be?
     
  15. JasonMcG

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    Apr 27, 2013
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    The output at the node would be 1.1V and at op amp D it would be 23.9V
     
  16. Ron H

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    You might want to think a little harder. A sine wave is not a DC voltage.
     
  17. JasonMcG

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    Apr 27, 2013
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    It would have a max peak at 55mV and a minimum at -55mV
     
  18. Ron H

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    How did you come up with that? :confused:
     
  19. JasonMcG

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    Apr 27, 2013
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    the output voltage being 1.1V =110mV peak to peak
    meaning 50mV up and 50mv down from the origin
     
  20. Ron H

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    I regret to inform you that 1.1V=1100mV, NOT 110 mV.:D
    Do you really think carefully before replying? I'm starting to feel like a babysitter.:rolleyes:
     
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