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Gyrator circuit

  1. May 2, 2010 #1
    Hi,

    i have this gyrator circuit that is equivalent to an inductor of 11.11mH and im required to show that if the opamps are ideal and assuming that the circuit is stable that
    Zin = (Z1*Z3*Z5) / (Z2*Z4)
    ive got for an ideal opamp that i+ = i- = 0 and that Aol = infinity

    the ciruit is posted below..
    any assistance would be greatly appreciated in proving Zin
     

    Attached Files:

  2. jcsd
  3. May 2, 2010 #2
    You should try something first, to get some help.

    What's the output from the first op amp given Z1, Z2, and vin?
     
  4. May 2, 2010 #3
    hi,

    well for the first op-amp using just vin,Z1 and Z2 i ended up with 1+Z2/Z1 but im not sure if thats correct...
    i also tried using this circuit attached... to get Zin = (Z1+Z5) / Z2 but again not sure if thats correct or allowed..
     

    Attached Files:

  5. May 2, 2010 #4
    Call the output of the first op amp v1. Call the output of the second op amp v2.

    You're first answer is sort of correct. For a sanity check, see that your units match. You have a voltage equal to a unitless quantity.

    The correct result is v1 = vin(1+Z2/Z1)

    The gain of the first stage is not a direct function of Z5. However vin is a function of Z5, so that eventually Z5 enters into the solution.

    The second stage is not too much harder than the first. What is v2 as a function of vin, v1, Z3 and Z4?
     
  6. May 2, 2010 #5
    It's time I get some sleep, so... I'm gone for a while.
     
  7. May 2, 2010 #6
    ok so for v2 i ended up with v2 = Vin ( 1 + Z4/Z3) + V1

    and because V1 = Vin ( 1 + Z2/Z1)

    which then equates to V2 = Vin ( 1 + Z4/Z3) + ( Vin ( 1 + Z2/Z1))

    should i leave it in terms of V1 or should i evaluate V1...

    Then should the final solution just be the combination of V1 and V2 ie V1+V2 along with Z5 added in ?
     
  8. May 2, 2010 #7
    Try the equation for V2 again. You might notice that V2 shouldn't contribute directly to V1 but depend upon impedance.

    Remember that the current through Z3 is equal to the current through Z4. The voltage at the positive input is equal to V1 and the voltage at the negative input is also V1.
     
  9. May 3, 2010 #8
    so should v2 = Vin ( -V1/Z3 + V1/Z4) ?

    im at a loss as to what v2 should be
     
  10. May 3, 2010 #9
    That's not it. You only need to recall two circuit laws: Kirchhoff's current law (KCL) and I=V/Z.

    First you find the current through Z3. The voltage across Z3 is V1 - Vin. I = (V1 - Vin)/Z3.

    There is no current in or out of the op amp. The current through Z3 all flows through Z4. IZ4= Vin-V2. Eliminate the current from these two equations and solve for V2.

    Remember to check that you have units of voltage on both sides of the equation.
     
  11. May 3, 2010 #10
    well using (Z3) V1-Vin.I and (Z4) IZ4=Vin-V2

    Because the current through Z3 = Current through Z4
    then:
    (V1-Vin)/Z3 = (Vin-V2)/Z4

    V2/Z4 = (-V1+Vin)/Z3 + Vin/Z4

    V2 = Z4 x ( (-V1+Vin)/Z3 ) + Z4 x ( Vin/Z4 )

    V2 = ( Z4(-V1+Vin))/Z3 +Vin

    so
    V2 = Vin + (Z4(-V1+Vin))/Z3 ???
     
  12. May 3, 2010 #11
    Good. That's the answer.

    Now that you have both V1 and V2 (which is Vout) you can solve for Vout as function of Vin.
     
  13. May 3, 2010 #12
    just a quick clarification,

    with v1 = Vin(1+Z2/Z3) AND v2 = Vin + (Z4(-V1+Vin))/Z3

    should i evaluate v2 by substituting in the value of v1 before i solve vout as a function of vin or should i go on and just solve vout as a function of vin
     
  14. May 3, 2010 #13
    I should have warned you I was hitting the sack...

    Substitute V1 into the second equation.
     
  15. May 5, 2010 #14
    ok so i finally got to proving that Zin = Z1Z3Z5/Z2Z4....

    another question in determing Vout/Vin for the attached circuit where Z2 = capacitor and
    Z1,Z3,Z4,Z5 = resistors

    i got the current i1 = current i2
    where i1 = Vin/Z1 and i2=(V1-Vin)/Z2
    so from those two equations v1 = Vin(1+Z2/Z1)

    and the current i3=current i4
    where i3 = (v1-vin)/z3 and i4 = (vin-vout)/Z4
    so again v1= [Z3(Vin-Vout) + Z4Vin] / Z4

    equating both v1
    Vin(1+Z2/Z1) = [Z3(Vin-Vout) + Z4Vin] / Z4

    Z3Vout = Z4x(Z3Vin/Z4) - Z4x(VinZ2/Z1)

    Vout = VinZ3/Z3 - Vin [(Z2Z4)/Z1] / Z3

    Vout = Vin - Vin( Z2Z4/Z1Z3)

    dividing through by Vin

    Vout/Vin = 1 - Z2Z4/Z1Z3

    Vout/Vin = 1 - ( 1/jwC x R4) / (R1R3)

    Vout/Vin = 1 - R4 / R1R3C(jw)

    just wondering whether this is the right expression for Vout/Vin
     

    Attached Files:

  16. May 5, 2010 #15
    Well done. I also solved for Zin and confirmed the result.

    i got the current i1 = current i2
    where i1 = Vin/Z1 and i2=(V1-Vin)/Z2
    so from those two equations v1 = Vin(1+Z2/Z1)

    and the current i3=current i4
    where i3 = (v1-vin)/z3 and i4 = (vin-vout)/Z4
    so again v1= [Z3(Vin-Vout) + Z4Vin] / Z4

    equating both v1
    Vin(1+Z2/Z1) = [Z3(Vin-Vout) + Z4Vin] / Z4

    Z3Vout = Z4x(Z3Vin/Z4) - Z4x(VinZ2/Z1)

    Vout = VinZ3/Z3 - Vin [(Z2Z4)/Z1] / Z3

    Vout = Vin - Vin( Z2Z4/Z1Z3)[/quote]

    Good. This is the result you could have obtained by substituting V1 into Vout=V2 in the previous problem, expressing Vout as a function of Vin.

    That's correct. Usually we want to express the right hand side as the sum of a real part and an imaginary part. This is what you actually have, but we want the imaginary j in the numerator. Multiply the fraction R4/R1R3C(jw) by -j/-j to put it in standard form and you're done.
     
  17. May 5, 2010 #16

    The Electrician

    User Avatar
    Gold Member

    The circuit you've been analyzing is a generalized impedance converter (GIC):

    http://mysite.du.edu/~etuttle/electron/elect66.htm

    There are 24 different arrangements of 2 opamps and 5 impedances to realize a GIC. See the attachment.

    They all have the same impedance expression at the top node, but most are not stable. Your circuit is the one designated B2B.
     

    Attached Files:

    • GIC.gif
      GIC.gif
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  18. May 7, 2010 #17
    Thanks for the nice read, Electrician. Long ago some old duck told that the set of passive components was incomplete without negative reistance, and mumbled something about gyrators and differential equations. I had no idea what what he was talking about at the time. Now I'm slightly better educated.
     
  19. May 11, 2010 #18
    ok so in designing simulating and implementing the circuit with Z2 a capacitor between 10pf and 500pf and the rest being resistors between 0.1K and 10K making the circuit look like an inductor of 11.11mH using a +-12V power supply im having trouble making it look like an inductor of value 11.11mH.

    using L = (Z1Z3Z5) / (Z2Z4)
    L = (R1R3R5) / (1/jwC R4)
    L = C(R1R3R5) / R4 (jw)
    using R1 = R3 = R5 = 1305ohm and R4 = 0.1K and C=500pF

    i get 0.011112238H = 11.112238mH
    but in using these values of resistors its impratical to implement for this purpose so i was wondering whether anyone has any suggestions on what values to use so its simply implemented and tested using readily available resistors and capacitors without the need to use resistors in seriers/parallel to create intended value
     
  20. May 11, 2010 #19

    The Electrician

    User Avatar
    Gold Member

    What is it about those component values that is impractical?
     
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