Gyrator circuit discussion

[promach]

What do you guys think about the following gyrator circuit ?

Someone told me the following:

    Let's put a voltage on v(in) and see the current drawn by that voltage. The inverter is high impedance, so none goes into that, so it will be just the current out of the diff pair. At DC, We get a current out of the inverter Iinv = Gm1*V(in). That current is multiplied by the output resistance of the inverter Ro,inv to give V(b) = V(in)*Gm1*Ro,inv. Then we multiply by the diff pair Gm2, so Iout = Gm2*V(b) = V(in)*Gm2*Gm1*Ro,inv. Rtest = Vin/Iout = 1/(Gm2*Gm1*Ro,inv).

    The cap on node Vb makes the impedance at that point go down with frequency, making Ro,inv get smaller. But the Ro,inv term is in the denominator of our Zin equation, so it makes the test impedance get bigger as frequency increases. Replace Ro,inv with the impedance of the cap, and you have a rough approximation of the impedance of the gyrator.

    Anyone up for a discussion regarding the mosfet sizing ?

[Nitrousoxide]

Talking generally. A mosfets transconductance parameter and output resistance is determined by the DC bias current (and thus gate voltage), as well as the width/length ratio. So to an extent, you can very either parameter till you get the desired transconductance  (and gain)

Depending on the design constraints there are a few basic rules to follow:

- Usually, you want to minimise channel length for speed and noise performance (for digital designs). However, for analog it is usually chosen to be 2x the minimum manufacturing length of the process.

- Keep in mind when increasing channel width and length, you will increase paracitic capacitive values such as Cox and Cov (which may or may not effect the performance)

- Speaking of the paracitic capacitances, be careful as they may form poles in the RPH and cause oscillation if they exist with high impedance nodes.

- Differential pair should have the same W/L ratio (M3 and M4). The current mirror should have the same ratio (M5 and M6). The current mirror and differential pair do not have to be the same size.

- The inverter should be sized appropriately. For the ideal ratio to minimise propogation delay (and symmetric response). The ratio between Pmos and Nmos should be Wp = Wn * sqrt(Mu,n*Cox,n/Mu,p*Cox,p), this usually come to about ~2.3x. This can be proven by setting the currents to be equal and taking the derivative. However, some people just arbitratily choose a scale of 3-5x the unit width of Nmos as the unit width for Pmos.

- Your ideal current sources would probably be replaced with some form of current mirrors in implementation.

Specifically, Youd only really have to analyze if these may cause instability. Again, a SPICE package and the correct EDA (like -god forbid- cadence virtuoso) will do wonders.

[promach]

Someone told me that I could draw up a parasitic model of the gyrator (active inductor) using only 4 discrete components resembling the following AC plots

Note: the magnitude plot has a peak at 500MHz

Could anyone advise ?

[coppercone2]

from experience playing with these on a breadboard or deadbug, they are kind of hard to use even with good op-amps. I made ones that worked well as LPF. My attempts at higher order BPF using gyrators were met with oscillators.

Don't have much experience but you are on the right track using a simulator to attempt these circuits, it seems like one of those things where you need to use a simulator and a breadboard at the same time to get decent performance.

I had trouble getting it to work in the 500KHz region, let alone many MHz. I think you picked a hard circuit.

[Nitrousoxide]

Someone told me that I could draw up a parasitic model of the gyrator (active inductor) using only 4 discrete components resembling the following AC plots

Do the 4 discrete components include the capacitor? Do they include the current sources? It's possible to form a gyrator out of two FET devices. In fact, there are a lot of different configurations: Single-ended, floating, Lin-Payne, Hara, Wu folded, just to name a few.

What is this for? If you're operating at 500 MHz and this is for a filter. You can always use 1/4 wave transmission lines to rotate the phase by 90 degrees so that a capacitor looks like an inductor.

[promach]

Do the 4 discrete components include the capacitor? Do they include the current sources?

This is a parasitic model of an inductor. Why would an inductor itself contain current sources internally ? The current source is only for testbench.

You can always use 1/4 wave transmission lines to rotate the phase by 90 degrees so that a capacitor looks like an inductor.

I am a bit confused on this statement, I have not studied transmission line (although I do know reflection coefficient) in depth enough. Would you mind writing few more sentences to make it more understandable for laymen ?

[bson]

You can always use 1/4 wave transmission lines to rotate the phase by 90 degrees so that a capacitor looks like an inductor.

You mean 180˚ - that's a very neat idea!  Or, you could take a plain resistor and phase shift it 270˚ into an inductor (at 3/4 wavelength)...  Does this actually work in a circuit, as a general passive 1-port impedance?

[Nitrousoxide]

This is a parasitic model of an inductor. Why would an inductor itself contain current sources internally ? The current source is only for testbench.

I am a bit confused on this statement, I have not studied transmission line (although I do know reflection coefficient) in depth enough. Would you mind writing few more sentences to make it more understandable for laymen ?

To the first question:
You have active devices in your circuit, and the topologies that you use require a current source to set the DC bias (depends on the application, most of the time for a long tail pair you will need a DC bias), the current sources are NOT just for the test bench (This is a good lesson here: You always, ALWAYS separate the device under test from the test bench to remove confusion).

For example, look at your I4 and I5. Think about when you build this circuit, how would you implement that? The answer to that question is to use a current mirror, or some kind of current source circuit which will consist (usually) of at LEAST 2 devices. Remember, one of the active devices must act as a negative resistance source.

In the end, you're modeling an "inductor", it just has to "appear" as an inductor from looking into the input port. That statement has no bearing on wether or not it requires any biasing. (Infact, you have a 1.8v DC source in the circuit too).

For the second question:
This is somewhat related to the reflection coefficient. With the aid of smith charts this is more easily explained (Im going to assume you have no familiarity).

Basically, picture a TEM wave being shifted by 90 degrees. This is the same as converting a series resistance, into a parallel admittance (on the smith chart). This is similar to performing a mathematical inversion i.e. ( jw to 1/(jw) ) Doing this, will cause a capacitor to appear as an inductor. However, you dont nessacerily need a 1/4 wave line to perform this function, it can be done with pi and t networks.

Wiki has not that bad of an article: https://en.wikipedia.org/wiki/Quarter-wave_impedance_transformer

[Nitrousoxide]

You can always use 1/4 wave transmission lines to rotate the phase by 90 degrees so that a capacitor looks like an inductor.

You mean 180˚ - that's a very neat idea!  Or, you could take a plain resistor and phase shift it 270˚ into an inductor (at 3/4 wavelength)...  Does this actually work in a circuit, as a general passive 1-port impedance?

No. Isnt it mathematically 90 degrees? Sure you rotate 180 degrees on a smith chart. But a smith chart is half a wavelength? 180 is half of 360 and a smith chart is 0 to 0.5lambda. Hence 0.5*0.5 = 0.25lambda?

Im not too sure about rotating 270 degrees to form an inductor. I think due to the phase ambiguity and that smith charts only work on 1/2 wavelength basis, it will appear as a capacitor? Im not tooo sure about this one. Would have to give it some more thought.

[promach]

I am trying to model the gyrator (active inductor) AC magnitude and phase plots using only passive devices (pure resistor, pure capacitor, pure inductor) . So, no active sources (voltage source, current source)

Allow me some time to understand the TEM and smith chart discussion above.

[Nitrousoxide]

I am trying to model the gyrator (active inductor) AC magnitude and phase plots using only passive devices (pure resistor, pure capacitor, pure inductor) . So, no active sources (voltage source, current source)

Allow me some time to understand the TEM and smith chart discussion above.

OH! Right. I have misunderstood your request! I thought you were asking to simplify the circuit down to two components (you did say DISCRETE parts, not PASSIVE). Well- In that case you would end up with something along the lines of a parallel RLC tank circuit:



It is obvious to see that the resonant frequency would be w0 = 1/sqrt(LC). As for the Q/damping factor, probably not so trivial. From a rough calc: If R2 becomes small then

There will be considerable mathematical analysis if you want to define the parasitic model in terms of the FET's. Expect (from just a glance): 1/(2*R1*C1)
- C1 would exist prominently of your C1
- L1 to be the capacitance divided by device transconductance.
- R1 and R2 to be functions of both small signal output resistance and transconductance.

[bson]

No. Isnt it mathematically 90 degrees? Sure you rotate 180 degrees on a smith chart. But a smith chart is half a wavelength? 180 is half of 360 and a smith chart is 0 to 0.5lambda. Hence 0.5*0.5 = 0.25lambda?

Im not too sure about rotating 270 degrees to form an inductor. I think due to the phase ambiguity and that smith charts only work on 1/2 wavelength basis, it will appear as a capacitor? Im not tooo sure about this one. Would have to give it some more thought.

For some reason I remembered the chart as being 1 WL, but it's 0.5.  It is however ±180˚... so every 0.25 WLs there is a 180˚ phase shift that conjugates the impedance.  So 0.25 WL would be right for making a capacitor become inductive.

This just repeats as you make the transmission line longer, i.e. [mod 0.5].

So for FR-4 with a VF of say 0.7 that's 299792458/500e6/4*0.7 = 104.9mm.

At 500MHz 33pF is 9.65Ω.
2π*500M*L = 9.65 => L = 3.1nH with 104.9mm of trace.

10pF is 31.8Ω.
=> L = 10nH

So it looks like values from 1-10nH should be very achievable...

Wish I had the right instrumentation to test this, but my 3577A VNA tops out at 200MHz (I rarely do anything RF and measuring low-SPS acquisition filters is far more important to me).

What's also interesting is that if there's filter like a lossless ladder instead of a single cap it will conjugate Z₁₁ and Y₁₁ (the immittance).  This is actually quite interesting in that it would permit synthesizing lossless ladders out of only capacitors...

[Nitrousoxide]

For some reason I remembered the chart as being 1 WL, but it's 0.5.  It is however ±180˚... so every 0.25 WLs there is a 180˚ phase shift that conjugates the impedance.  So 0.25 WL would be right for making a capacitor become inductive.

My god! I wrote that so sleep deprived. You're right. To clarify where I was wrong:

let j = sqrt(-1). Therefore:

1/j = 1/sqrt(-1)
     = (-1)^(-1/2)
     = (-1)^(1/2) * (-1)^(-1)
     = -(-1)^(1/2)
     = -j

Which in layman's terms, is moving from the positive y plane to the negative y plane (inductive to capacitive), or vice versa. This is just an operation of flipping (or moving) in a direction that is 180 degrees from the start.

[promach]

I am not sure how I am going to increase input impedance at low frequencies from 60 Ohm to 10 kiloOhm ?

[promach]

For some reason I remembered the chart as being 1 WL, but it's 0.5.  It is however ±180˚... so every 0.25 WLs there is a 180˚ phase shift that conjugates the impedance.  So 0.25 WL would be right for making a capacitor become inductive.

@bson

I am reading wikipedia article on Quarter-wave impedance transformer

I have googled "WL" but I could not find anything that explains it in terms of smith chart. Could you elaborate more ?

The closest I got on "WL" is from this book: The Electromagnetic Field By Albert Shadowitz but I think your W is not exactly W, but the angle of the reflection coefficient , ψ ?

Please correct me if wrong.

[janoc]

I think he meant WL as wavelength. Aka lambda. 

[Nitrousoxide]

I think he meant WL as wavelength. Aka lambda.

Yes. Its 0.25lambda. Where lambda is the wavelength of the signal in the propagation material.

Regarding low frequencies, how low do you want? Whats your definition of LF? The capacitor will appear as an open circuit and the inductor as a short. So eventually as you reach D.C. the impedance as you look into the network will appear as mostly real, and as approx (1/1 + 1/500)^-1 = 1 ohm.

If you want to increase the impedance at DC, remove the short or add series resistance with the inductor (probably a bad idea). If you want to increase the impedance near resonance, you'd have to decrease the Q of the network.

[promach]

Why 0.25 lambda ?

By the way, I have two parasitic model that gives almost similar frequency response as the gyrator.

[Nitrousoxide]

Ok. I will attempt to do a brief mathematical proof (apologies for the terrible wuality).

Where we want to get to: An equation where we can substitute lengths into, that is also a function of impedances.
Where we start: We know the wave equation.



So. In layman's terms, what does this result mean?

The input impedance looking into a transmission line (Zin) is inversely proportional to the terminating load (Zl). Example:

You have a short, 1/0 which will have Zin tend to infinity, an open circuit.
You have an open, 1/inf which will have Zin tend to zero, a short circuit.

The beauty of the smith chart is that it is a mapping of (essentially) the real and imaginary parts of the above equation. Allowing for the use of graphical methods of solving problems. This is actually quite amazing since back in the day it would have been computationally intensive and bothersome to find roots of trigonometric identities.

[bson]

Are you familiar with the GIC, the general impedance converter?  It can be configured to operate as a gyrator and uses opamps, so works great at low frequencies.

And, yup, WL = wavelength.  No lambda λ on this keyboard... sorry for any confusion.   I wonder if it can be pasted in, or if the forum software will strip it?  λλλ = 3 lambdas

I can see the day when an MMIC has two extra pins and specifies a particular length of trace they should be connected to, the Z0, and how many pF to put at the end.

[promach]

In your derivation ,

Why Z=-L   ?

What is  "unBL" in the Zin expression ?

[Nitrousoxide]

In your derivation ,

Why Z=-L   ?

What is  "unBL" in the Zin expression ?

Apologies. I wanted to cram it all into one page.

The convention I had adopted for the proof was that the transmission line spanned across the "z" axis (it is small z, like the coordinate, not to be confused with impedance). We place a reference point of z=0 at the load and thus moving towards the generator will give us a negative value, this value we call l.

unBL? You mean 2*j*beta*l? Again, apologies for my dog awful handwriting.

j = sqrt(-1).
"l" is the length away from the load towards the generator. (i.e. the length of the TL).
"beta" is the complex component of the propagation constant (also known as the phase propagation constant).

EDIT: OH NO! You mean tan(beta*l)

Now that I think about it.... I tend to accidentally do all my Z's as capitals... damn thats confusing. My bad. I think my logic is to assume that any Z without a subscript is not an impedance. 

[promach]

There is something not right when you directly convert exp() to use tan(beta*l) in your Zin expression.

[Nitrousoxide]

It's correct. It's just a matter of algebraic massaging:



Note:

All these calculations are for losses lines. If it were modeled as lossy, then the beta in the original equation would have been the complex phase propagation constant (lower case gamma, not to be confused with upper case gamma). You'd end up tanh and some scaling.

[promach]

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.