Gyrator circuit discussion

[Nitrousoxide]

What I do not understand is why your z-axis coordinates starts from 0 at the righthand side (load itself) ?
In other words, I do not understand why z=-L ? why negative ?

The following skips last part of your proof and also makes thing clearer.

It's an arbitrary decision. Think about it this way. If we define the problem as:

We KNOW where the load is, however, we want to find the input length and/or impedance.

You could have just as easily used the positive z direction. You would have just had to change a few steps in the proof (such as multiplying by e^(-jbl) instead of e^(jbl).

What you have posted is identical to the last section of my proof, however, if it allows you to understand it better in writing it that way then who am I to say. I just thought it might be useful to show that there is an inversely proportional relationship between load and input impedance at a quarter length.

[promach]

How does the concept of quarter-wave transformer helps to transform capacitor to inductor ?

[Nitrousoxide]

As I posted before in my previous proof: The quarter wave transform allows for an input impedance to be inversely proportional to the output impedance. This means that a series inductor will be converted to a parallel capacitor. The whole purpose of the 1/4 wave transformer is to mimic the mathematical inverse.

Heres a good read:
http://home.sandiego.edu/~ekim/e194rfs01/filterek.pdf

[promach]

Why use λ/8 transmission line  instead of quarter-wave (λ/4 transmission line)  ?

[Nitrousoxide]

If we look at some of the previous results. We can make extra conclusions.

Say for example. Taking the derived input impedance to a transmission line and replacing the load with a shunt gives: (i.e. When ZL = 0 in the previous Zin equation)

Zin = j*Z0*tan(beta*l)

Now, If we look at the equation for the impedance of an inductor:

Zind = j*w*L.

Looking into a transmission line, we would want it to look like an element. Therefore, we can equate Zin = Zind so that we can make the transmission line segment behave as a component.

j*w*L = j*Z0*tan(beta*l)

w*L = Z0*tan(betal*l).

To simplify things, let us select l = lambda/8. We know that beta = 2*pi/lambda. Therefore, by multiplying beta*l = 2*pi/8 = pi/4.
And the tan(pi/4) = 1. Which gives us:

w*L = Z0.
OR
L = Z0/w

The same thing can be proved for a transmission line acting as a capacitance. It is of note, that the equation is dependent upon omega (w), and thus the device will only work over narrow bandwidths compared to other solutions.

Note: This is for transmission lines acting as lumped elements. I.e. "A transmission line that appears as X capacitance". Not for an impedance inverter. i.e. "Make capacitance look like inductance"

There are tradeoffs between constructing filters as stubbed lines versus hybrid stub/element filters.

[promach]

equation is dependent upon omega (w), and thus the device will only work over narrow bandwidths compared to other solutions.

Zin(L=λ/8) = Zo * (ZL + j*Zo) / (Zo + j*ZL)

Why narrower bandwidth compared to other solutions ?

There are tradeoffs between constructing filters as stubbed lines versus hybrid stub/element filters.

I am very confused with your terminology used in the above sentence quote.
I do not get what you mean by this sentence.

[Nitrousoxide]

Zin(L=λ/8) = Zo * (ZL + j*Zo) / (Zo + j*ZL)

That's correct so far. Take it a step further. Try placing Zo = 0 (for a shunt TL) or Zo = inf (for an open TL).

Why narrower bandwidth compared to other solutions ?

There are tradeoffs between constructing filters as stubbed lines versus hybrid stub/element filters.

I am very confused with your terminology used in the above sentence quote.
I do not get what you mean by this sentence.

First, I will clarify what I meant by that statement. I meant the difference between constructing a filter out of microstrip line elements (or CPW) and conventional elements (i.e. conventional inductors or capacitors).

Essentially what you're doing is performing an impedance transform, it can also be interpreted as an impedance match. The stripline impedance matcher has a limited bandwidth, this can be alleviated by using stepped transmission line sizes, or other geometries, such as triangular or exponential to increase the bandwidth. i.e. For an impedance inverter layout, the more steps (thus the more poles, and the more stubs) will translate into a larger bandwidth.

For example, a good measure for a single stub capacitor would be the ratio of Zc/Zo, where Zc is the capacitor impedance.

Alternatively, again if we look at a single stub case, we can see that the bandwith (and thus Q) is:

Q = Wo*L/R = 1/(Wo*R*C)

So to increase the bandwidth you would decrease Q. You then would need to either decrease L, or increase the R or C.

Obviously, as you add more poles (either series inductance or extra stubs) the above relationship becomes significantly less trivial.

[nq93]

Could you explain how the circuit works? and what is the maximum inductance we can get it?