Proving Zin Equation for Ideal Gyrator Circuit

[Law91]

Hi,

i have this gyrator circuit that is equivalent to an inductor of 11.11mH and I am 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

 

[Phrak]

You should try something first, to get some help.

What's the output from the first op amp given Z₁, Z₂, and v_in?

 

[Law91]

hi,

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

 

[Phrak]

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 = v_in(1+Z₂/Z₁)

The gain of the first stage is not a direct function of Z₅. However v_in 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 v_in, v₁, Z₃ and Z₄?

 

[Phrak]

It's time I get some sleep, so... I'm gone for a while.

 

[Law91]

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 ?

 

[Phrak]

Try the equation for V₂ again. You might notice that V<sub>2 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.</sub>

 

[Law91]

so should v2 = Vin ( -V1/Z3 + V1/Z4) ?

im at a loss as to what v2 should be

 

[Phrak]

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 - V_in. I = (V₁ - V_in)/Z₃.

There is no current in or out of the op amp. The current through Z₃ all flows through Z₄. IZ₄= V_in-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.

 

[Law91]

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 ?

 

[Phrak]

Good. That's the answer.

Now that you have both V1 and V2 (which is V_out) you can solve for V_out as function of V_in.

 

[Law91]

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

 

[Phrak]

I should have warned you I was hitting the sack...

Substitute V1 into the second equation.

 

[Law91]

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

 

[Phrak]

ok so i finally got to proving that Zin = Z1Z3Z5/Z2Z4...

Well done. I also solved for Zin and confirmed the result.

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)[/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.

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

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.

 

[The Electrician]

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.

 

[Phrak]

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.

 

[Law91]

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 I am 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

 

[The Electrician]

What is it about those component values that is impractical?