Gyrator circuit discussion - Page 1
Forum section: electronics / beginners

author: promach
author_rank: Frequent Contributor
timestamp: February 15, 2019, 02:25:37 pm
content: What do you guys think about the following gyrator circuit ?Someone told me the following:QuoteAnyone up for a discussion regarding the mosfet sizing ?

author: Nitrousoxide
author_rank: Regular Contributor
timestamp: February 15, 2019, 03:03:21 pm
content: 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.

author: promach
author_rank: Frequent Contributor
timestamp: February 16, 2019, 03:34:05 pm
content: 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 plotsNote: the magnitude plot has a peak at 500MHzCould anyone advise ?

author: coppercone2
author_rank: Super Contributor
timestamp: February 16, 2019, 05:45:55 pm
content: 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.

author: Nitrousoxide
author_rank: Regular Contributor
timestamp: February 16, 2019, 11:59:45 pm
content: Quote from: promach on February 16, 2019, 03:34:05 pmDo 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.

author: promach
author_rank: Frequent Contributor
timestamp: February 17, 2019, 01:58:03 am
content: QuoteThis is a parasitic model of an inductor. Why would an inductor itself contain current sources internally ? The current source is only for testbench.QuoteI 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 ?

author: bson
author_rank: Unknown
timestamp: February 17, 2019, 03:51:07 am
content: Quote from: Nitrousoxide on February 16, 2019, 11:59:45 pmYou 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?

author: Nitrousoxide
author_rank: Regular Contributor
timestamp: February 17, 2019, 04:10:37 am
content: Quote from: promach on February 17, 2019, 01:58:03 amTo 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

author: Nitrousoxide
author_rank: Regular Contributor
timestamp: February 17, 2019, 04:13:44 am
content: Quote from: bson on February 17, 2019, 03:51:07 amNo. 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.

author: promach
author_rank: Frequent Contributor
timestamp: February 17, 2019, 05:00:25 am
content: 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.

author: Nitrousoxide
author_rank: Regular Contributor
timestamp: February 17, 2019, 05:42:47 am
content: Quote from: promach on February 17, 2019, 05:00:25 amOH! 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 thenThere 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.

author: bson
author_rank: Unknown
timestamp: February 17, 2019, 11:52:03 pm
content: Quote from: Nitrousoxide on February 17, 2019, 04:13:44 amFor 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 = 10nHSo 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 Z11and Y11(the immittance). This is actually quite interesting in that it would permit synthesizing lossless ladders out of only capacitors...

author: Nitrousoxide
author_rank: Regular Contributor
timestamp: February 18, 2019, 01:52:57 am
content: Quote from: bson on February 17, 2019, 11:52:03 pmMy 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)= -jWhich 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.

author: promach
author_rank: Frequent Contributor
timestamp: February 23, 2019, 04:13:17 pm
content: I am not sure how I am going to increase input impedance at low frequencies from 60 Ohm to 10 kiloOhm ?

author: promach
author_rank: Frequent Contributor
timestamp: February 23, 2019, 04:34:26 pm
content: Quote@bsonI am readingwikipedia article on Quarter-wave impedance transformerI 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 thisbook: The Electromagnetic Field By Albert Shadowitzbut I think your W is not exactly W, but the angle of the reflection coefficient , ψ ?Please correct me if wrong.

author: janoc
author_rank: Super Contributor
timestamp: February 23, 2019, 07:52:31 pm
content: I think he meant WL as wavelength. Aka lambda.

author: Nitrousoxide
author_rank: Regular Contributor
timestamp: February 26, 2019, 06:39:09 am
content: Quote from: janoc on February 23, 2019, 07:52:31 pmYes. 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.

author: promach
author_rank: Frequent Contributor
timestamp: February 28, 2019, 02:49:16 am
content: Why 0.25 lambda ?By the way, I have two parasitic model that gives almost similar frequency response as the gyrator.

author: Nitrousoxide
author_rank: Regular Contributor
timestamp: February 28, 2019, 05:34:13 am
content: 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.

author: bson
author_rank: Unknown
timestamp: February 28, 2019, 06:06:08 am
content: 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 lambdasI 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.

author: promach
author_rank: Frequent Contributor
timestamp: March 01, 2019, 04:21:56 pm
content: In your derivation ,Why Z=-L ?What is "unBL" in the Zin expression ?

author: Nitrousoxide
author_rank: Regular Contributor
timestamp: March 02, 2019, 05:45:15 am
content: Quote from: promach on March 01, 2019, 04:21:56 pmApologies. 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.

author: promach
author_rank: Frequent Contributor
timestamp: March 02, 2019, 12:57:26 pm
content: There is something not right when you directly convert exp() to use tan(beta*l) in your Zin expression.

author: Nitrousoxide
author_rank: Regular Contributor
timestamp: March 02, 2019, 02:33:08 pm
content: 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.

author: promach
author_rank: Frequent Contributor
timestamp: March 03, 2019, 06:42:13 am
content: 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.

author: Nitrousoxide
author_rank: Regular Contributor
timestamp: March 03, 2019, 11:19:03 am
content: Quote from: promach on March 03, 2019, 06:42:13 amIt'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.

author: promach
author_rank: Frequent Contributor
timestamp: March 04, 2019, 01:42:39 pm
content: How does the concept ofquarter-wave transformerhelps to transform capacitor to inductor ?

author: Nitrousoxide
author_rank: Regular Contributor
timestamp: March 05, 2019, 07:15:47 am
content: 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

author: promach
author_rank: Frequent Contributor
timestamp: March 05, 2019, 01:47:03 pm
content: Why use λ/8 transmission line instead of quarter-wave (λ/4 transmission line) ?

author: Nitrousoxide
author_rank: Regular Contributor
timestamp: March 06, 2019, 01:53:08 am
content: 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.ORL = Z0/wThe 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.

author: promach
author_rank: Frequent Contributor
timestamp: March 06, 2019, 01:27:22 pm
content: QuoteZin(L=λ/8) = Zo * (ZL + j*Zo) / (Zo + j*ZL)Why narrower bandwidth compared to other solutions ?QuoteI am very confused with your terminology used in the above sentence quote.I do not get what you mean by this sentence.

author: Nitrousoxide
author_rank: Regular Contributor
timestamp: March 08, 2019, 10:01:36 am
content: Quote from: promach on March 06, 2019, 01:27:22 pmThat's correct so far. Take it a step further. Try placing Zo = 0 (for a shunt TL) or Zo = inf (for an open TL).Quote from: promach on March 06, 2019, 01:27:22 pmFirst, 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.

author: nq93
author_rank: Newbie
timestamp: February 08, 2023, 07:34:38 am
content: Could you explain how the circuit works? and what is the maximum inductance we can get it?