Talk:Gyrator/Archive 1

Archive 1

C and L mistake

The "C" shown in the simulated circuit should be an "L" (inductance), the description won't make sense otherwise... --Palapala 22:34, 27 Feb 2004 (UTC)

I fixed it but the picture isn't updating for some reason.

see http://en.wikipedia.org/upload/archive/1/14/20040227215433%21Gyrator.png

if you know how to fix it please do. - Omegatron

I reverted your image to the one you want, I hope.

It is http://en.wikipedia.org/upload/archive/1/14/20040228164914%21Gyrator.png

It has resistors R and R(subscript L) in the top, Gyrator circuit, while the bottom Simulator circuit has one resistor R and the inductor L with the equation L=R(sub L)RC.

Does that match the description here?

"From the diagram, the inductance L of the gyrator is equal to R1R2C. Given the inductance to be simulated and the resistance values, the capacitance can be found by C = L / (R1R2)."

R1 is R(sub L) and R2 is R?

Wikibob 19:17, 2004 Feb 28 (UTC)

Yes, Wikibob, thanks for fixing it. I fixed the equations too. Everyone like it now? R_L is now the resistance a real inductor would have. - Omegatron

Polarity of opamp

The SRL external link has the polarity of the opamp reversed. Does anyone know which is correct? I suppose I could stop being lazy and calculate it myself... - Omegatron

Ok. I did the calculations, and it works. I've also found several other circuits with the same polarity. Only that one example has it reversed. It is vaguely similar to a standard difference amplifier, with a capacitor and a zero ohm feedback resistor. - Omegatron

The circuit in this article (with *negative* feedback) works. I'm almost certain that the other circuit (with *positive* feedback) can't possibly work. But someone should really confirm.

I remember doing calculations with both and finding that they both work, but I could have made dumb assumptions. I will check again some time. - Omegatron 14:03, Aug 3, 2004 (UTC)

Conventional analysys of op-amp circuits relies on the assumption that the voltage difference between the two inputs will become negligable. This is only true if the feedback is negative. If you use postive feedback you end up with a schmitt trigger. Plugwash 02:02, 14 March 2006 (UTC)

Of course.  :-) — Omegatron 03:41, 14 March 2006 (UTC)

I checked in a simulator and it seems to work either way. Pfalstad 06:39, 14 March 2006 (UTC)

Simulators are not to be trusted. Plugwash 11:51, 18 September 2007 (UTC)

An easy way to show it has to be the way arround it is in the article is to consider the steady state case. In this case the capacitor disapears. With the circuit in the article that means the op-amp acts as a voltage follower following zero and so the circuit reduces to simply a resistor to ground. This is the same as the steady state case of a series RL circuit. If the op-amp is turned arround it becomes a schmitt trigger with a zero input and will settle into saturation (which direction it settles in will depend on the internals of the op-amp and possiblly luck as well) Plugwash 12:02, 18 September 2007 (UTC)

As a filter or as just an impedance to ground?

I noticed that the voltage at the output of the op-amp and at the inductor are the same, but I don't know if it's worth mentioning. The same effect can be had with just the R and C, so it is not too important, right? - Omegatron

Perhaps adding another "output node" circle to the schematic diagrams would be sufficient.

I had that at first, but thought it would be too distracting and people would think it was meant to be used as a filter. It is meant to act as an inductor to ground as part of a simulated LC filter. - Omegatron 14:03, Aug 3, 2004 (UTC)

electrical network is smaller than a electrical network

Currently the article claims that "an electronic circuit, which is typically much smaller than a tabletop electrical circuit."

Since *both* electronic circuit *and* electrical circuit are now re-directs to another article, normally I would change them to point directly (rather than indirectly through the redirect) to that other article:

"an electrical network, which is typically much smaller than a tabletop electrical network."

Huh ?

--DavidCary 19:25, 2 Aug 2004 (UTC)

If I were fixing a double redirect, I would probably leave the wording the same, and just change where the link points. You can't always change the wording to wherever it redirects. Some redirects are just conceptual redirects, instead of words that could literally be interchanged. I changed it like this:

Its primary use is to simulate an inductive element in a small electronic circuit or integrated circuit, which is typically much smaller than a tabletop circuit.

- Omegatron 14:03, Aug 3, 2004 (UTC)

Error in description?

It appears that the line "In other words, it can make an inductive circuit behave capacitively," should actually read "In other words, it can make a capacitive circuit behave inductively,"

It wasn't really an error, because the sentence said invert but the main use is to simulate inductors. Thank you for your contribution. Ancheta Wis 13:07, 22 Nov 2004 (UTC)

But it also says ...it can make...a bandpass filter behave like a band-stop filter, and so on.... How it is proposed that this done? - by individually inverting each branch in the circuit? If so, then I'm not convinced it's worthy of mention, and if not then some explanation would help (help me anyway). --catslash (talk) 00:32, 28 April 2010 (UTC)

The gyrator is designed to convert two-terminal elements (one-port circuits) into their inverse duals. So, if one-port band-pass filters exist, they can be converted into band-stop filters by some gyrator circuit. Fortunately, they are such circuits; a series LC circuit is an example of one-port band-pass filter. So that, if you connect a series LC circuit to the input gyrator port, the combination of this circuit and the very gyrator would behave as a two terminal stop-band filter at the output gyrator port. Maybe, a series crystal resonator is another example of a one-port band-pass filter? BTW, I do not know if there are one-port band-stop filters. I think there are not such series configurations; they are built by parallel connected (to the output) band-pass elements. Circuit dreamer (talk, contribs, email) 04:55, 28 April 2010 (UTC)

Negative feedback circuits have the unique property to "invert" two-port circuits connected in the feedback loop (to "reverse" their inputs/outputs). Analog examples are: an op-amp inverting voltage-to-current converter is a reversed passive current-to-voltage converter; an op-amp non-inverting amplifier is a reversed voltage divider and v.v., a negative feedback amplifier with another amplifier connected in the feedback loop is a "voltage divider"; an op-amp integrator is a reversed CR differentiator and v.v., an op-amp differentiator is a reversed RC integrator; an op-amp logarithmic converter is a reversed DR antilogarithmic converter and v.v., an op-amp antilog converter is a reversed RD log converter; the input part of the simple current mirror and the transdiode log converter act as "reversed transistors" whose collector current is the input quantity while the base-emitter voltage is the output quantity; and, finally, an op-amp band-pass can be build by connecting a band-stop filter into the feedback loop and v.v. A "mixed" example is an analog-to-digital converter that is a reversed digital-to-analog converter. A digital example is a PLL circuit that is a reversed frequency divider. There are also interesting non-electrical examples of these unique negative feedback property... but let's continue this interesting discussion about the unique phenomenon on negative feedback, negative feedback amplifier, operational amplifier and operational amplifier applications pages... Circuit dreamer (talk, contribs, email) 05:33, 28 April 2010 (UTC)

If certain governments could discover a way of applying their debts to this mixed example op-amp, their financial problems would be over. Or, how about feeding time into one and making a time machine? Is that just a fantasy? Zen-in (talk) 06:36, 28 April 2010 (UTC)

How can a one-port device act as a filter? It only has one port, and therefore doesn't have both an "input" and an "output". Edit: I suppose it could when terminating a transmission line, but this is almost certainly not what you're referring to above. Oli Filth^{(talk|contribs)} 08:05, 28 April 2010 (UTC)

I mean a series LC circuit acting as a band-pass filter. How do we name it? Maybe, a "two-terminal band-pass filter" or something similar? Circuit dreamer (talk, contribs, email) 10:39, 28 April 2010 (UTC)

A series LC acting as a one-port device isn't a filter, it's just a load. Oli Filth^{(talk|contribs)} 12:12, 28 April 2010 (UTC)

LC circuit says: "...Therefore the series connected circuit, when connected to a circuit in series, will act as a band-pass filter having zero impedance at the resonant frequency of the LC circuits..." Circuit dreamer (talk, contribs, email) 12:44, 28 April 2010 (UTC)

Then I would argue that that description is misleading, at best. A filter (as I understand it) takes a voltage or current as an input, and produces a voltage or current as an output. This is obviously not the case with a one-port device. Oli Filth^{(talk|contribs)} 13:17, 28 April 2010 (UTC)

Let's drop the word filter (because an impedance on it's own isn't a filter) and just say ...makes a series LC circuit behave like a shunt LC circuit....--catslash (talk) 13:27, 28 April 2010 (UTC)

Agreed. BTW, RLC circuit says: "...The RLC circuit may be used as a bandpass or band-stop filter by replacing R with a receiving device with the same input resistance..." Circuit dreamer (talk, contribs, email) 13:34, 28 April 2010 (UTC)

Yes, at which point you have an implicit two-port network! Oli Filth^{(talk|contribs)} 13:45, 28 April 2010 (UTC)

The very load can act as the resistor R. In this case, the series LC circuit is the very filter in a form of a composed two-terminal element. It has a voltage input and a current output (it acts as a frequency-dependent voltage-to-current converter). Circuit dreamer (talk, contribs, email) 13:59, 28 April 2010 (UTC)

(outdent) No, as Catslash above said above, an impedance is not a filter. In the case of the RLC example, the output port is the voltage across the resistor, which just so happens to be the voltage required by the load. The LC on its own is not a filter, as there is no output port.

Incidentally, we're straying off-topic here! Oli Filth^{(talk|contribs)} 14:25, 28 April 2010 (UTC)

Well, we finish... a few sentences only... I agree that "filter" is an inappropriate and misleading word here. But, in a broad sense of the word, a series LC circuit is some kind of "current" band-pass filter that controls the current in the same way as a series connected capacitor (high-pass filter) or inductor (low-pass filter). Circuit dreamer (talk, contribs, email) 14:59, 28 April 2010 (UTC)

Gyrators are 2-port elements. Isn't it?

I think some mistake happened because of different definitions. As I know as a Mechatronic student, Gyrators are two-port elements that have the relations v2= k i1 and v1 = k i2. Most of gyrators use Hall Effect. For more information please refer to:

http://www.paritycomputing.com/jpdfs/ieee/ssc/ireissc3/1959002/01feb/0024grub.pdf

and

"gyrator." McGraw-Hill Encyclopedia of Science and Technology. The McGraw-Hill Companies, Inc., 2005. Answers.com 02 Mar. 2007. http://www.answers.com/topic/gyrator

I am eagerly ready to hear other’s opinions. --Anooshahpour 07:58, 2 March 2007 (UTC)

There is no mistake with definitions, 'gyrator' is the only name I've heard this circuit refered to in the electronics industry. However, there definitely are other devices which are also known as a gyrator. So far we have 3 types:

• Impedence inverter circuit (This article)
• "Hall Effect Gyrator" which uses a slab of semiconductor through which a current (usually DC) flows, these are used in hall effect sensors
• "Microwave gyrator" - which causes phase changes in microwave signals, by using a ferrite device. These are used in phased array radar.

Perhaps a disambiguation page is in order, when there are articles for other types. --Ozhiker 09:47, 2 March 2007 (UTC)

Microwave gyrator

The microwave gyrator mentioned above is the same thing as the gyrator described in this article. In microwave terminology it's a (non-reciprocal) 2-port which 180°-phase-shifts the signal travelling in one direction, while passing unmodified the signal in the other direction. A 180° phase-shift is equivalent to negating (changing the polarity) of the signal, so the gyrator's scattering matrix is

$S={\begin{pmatrix}0&1\\-1&0\\\end{pmatrix}}$

If the characteristic impedance of both ports of the gyrator is $\scriptstyle {Z_{c}}$ and one port is terminated in an impedance $\scriptstyle {Z_{t}}$, then the reflection coefficient at the termination is $\scriptstyle {\rho _{t}={\frac {Z_{t}-Z_{c}}{Z_{t}+Z_{c}}}}$, and the reflection coefficient looking into the other port of the gyrator is just minus this (the round-trip through the gyrator negates the signal). Thus the impedance looking into the free port of the gyrator is

$Z_{in}={\frac {1+\rho _{g}}{1-\rho _{g}}}Z_{c}={\frac {1-\rho _{t}}{1+\rho _{t}}}Z_{c}={\frac {1-{\frac {Z_{t}-Z_{c}}{Z_{t}+Z_{c}}}}{1+{\frac {Z_{t}-Z_{c}}{Z_{t}+Z_{c}}}}}Z_{c}={\frac {Z_{c}^{2}}{Z_{t}}}$

i.e. the load impedance is inverted as described in this article. This sort of gyrator uses non-reciprocal materials (biased ferrites) as stated above, but does not employ any active components. --catslash (talk) 17:58, 22 March 2010 (UTC)

Need to mention that a Gyrator won't simulate a coil for power transmission?

I remember there was a confusion among students at first as to whether it is possible to replace a power inductor, say in a DC/DC inverter. Maybe it would be worth mentioning under ,,things not possible with a gyrator? Marek Lewandowski 15:41, 28 August 2007 (UTC)

It has been recently been proved that switch-mode gyrators could be used to simulate inductors. As a switch-mode circuit, the gyrator L is capable of storing energy and could be used to replace the power inductor in DC-DC converters. MSaad.ego (talk) 22:00, 4 March 2017 (UTC)Mohamed Saad

input impedance mistake

I think that the input impedance is wrong. I think the correct one is: $Z_{i}=R_{L}(sCR+1)/(sC(R_{L}+R)+1)$

—Preceding unsigned comment added by 190.18.25.75 (talk • contribs)

It seems someone has changed this description so that it looks as if the circuit is behaving as a filter. It's not. The circuit's input impedance is supposed to behave as the input impedance of an inductor. The output of the opamp is unused. — Omegatron (talk) 16:23, 12 April 2008 (UTC)

other inductor replacing circuits

Can any inductor-free circuit used to replace an inductor be called a "gyrator"? If "gyrator" only applies to the specific 1 amp circuit shown here, what term should I use for the 2 TZ amp circuit[1]? --68.0.124.33 (talk) 05:36, 21 July 2008 (UTC)

The term applies to any circuit that simulates an inductor - I've seen versions that use a transistor for the amplifier... or two op-amps. --Ozhiker (talk) 09:56, 21 July 2008 (UTC)

Discussion about energy storage in Gyrator circuit

I would like to discuss thoroughly the basic idea, the operation and the properties of this odd circuit. I suggest using the Elliot's example as a base.

The basic idea

We may consider this circuit as consisting of two parts (branches) - an input part (the RC circuit) and an output part (the op-amp and the resistor R_L). We may think of the input part as of a "modeling", "shaping" part that drives the power output part.
Building a CR differentiator. The current flowing through a capacitor is proportional to the derivative of the voltage applied across it; so, a capacitor acts as a simple differentiator with voltage input and current output. But we need a voltage output; for this purpose, we connect in series a resistor acting as a current-to-voltage converter and get the voltage across the resistor as an output.
Building an L differentiator. Dually, the voltage across an inductor is proportional to the derivative of the current flowing through it; so, an inductor acts as a simple differentiator with current input and voltage output. In this way, the voltage across the resistor R of the input CR circuit behaves as the voltage across the inductor; so, we may use this voltage to simulate an inductor. For this purpose, we buffer the weak "shaping" input part by an op-amp follower and apply this voltage back to the input. To limit the maximum current through the "inductor" (the op-amp), we connect a low resistor (R_L); it represents the inductor's internal resistance... Circuit dreamer (talk) 19:04, 3 April 2010 (UTC)

The operation

In the classical voltage supplied RL circuit (voltage source --> resistor --> inductor), the inductor opposes to the input voltage variations by producing a contrary voltage that substracts from the input one (figuratively speaking, when the input source "moves" the left end of the resistor, the inductor "moves" the right end in the same direction). In our circuit, the op-amp does the same by "moving" the right end of the resistor R_L. This "bootstrapping" increases many times the resistance R_L when the input voltage changes; the combination of the resistor R_L and the op-amp acts as a time-dependent dynamic resistor. Let's consider the typical moments of the transition process.

Imagine the input voltage (the voltage at the left side of the resistor R_L) changes suddenly from zero to V_H. The capacitor transfers immediately this change and the voltage drop across R becomes V_H as well. The op-amp follower "copies" this voltage drop and applies the same voltage at the right side of the resistor R_L. As a result, no currents flows through the resistor and the circuit shows infinite large input resistance (impedance). In the course of time, the capacitor charges, the current passing through it decreases and the voltage drop across the resistor R decreases as well. The op-amp follower "lowers" continously the right end of R_L and finally reaches the ground. At the end of the transition, the current is maximum - I_MAX = V_H/R_L. Circuit dreamer (talk) 21:14, 3 April 2010 (UTC)

The mistake you are making is in not looking at the circuit as a whole. By breaking it up into separate parts you are arriving at the wrong conclusion. Op-Amps circuits need to be analyzed mathematically, not figuratively. If you don't like using simple math to analyze electronic circuits, you will always be making the wrong guess. Zen-in (talk) 19:46, 3 April 2010 (UTC)

Energy considerations

Network theory says energy is stored

A network with a complex impedance has reactance(s). A reactance stores energy. These are basic principles of physics and network theory. There are already too many statements in this article that are not backed up by citations. The first two paragraphs make all kinds of claims that are not backed up. So before we start adding incorrect statements, some of the dubious claims need to be researched and backed up with references. Otherwise this article will just become another example of excessive OR and POV, if it isn't already. Zen-in (talk) 19:23, 3 April 2010 (UTC)

The output part dominates over the input one

Well, let's continue... The two parts (branches) of the circuit are connected in parallel. The output part (R_L + op-amp) is the very "inductor" and it has to dominate over the input part (the input part has only "shaping" properties and it has not to disturb the "inductor"). In addition, the capacitor has to have as much as possible small capacitance (dimensions). So, the resistor R has very high resistance. In the Elliot's example, C = 100 nF and R = 100 kΩ whlie R_L is only 100 Ω. The obtained "inductance" is as many as 1 H.

None of the above makes any sense to me, sorry. Over 50 years in electronics and an Engineering degree from a highly rated engineering school has not prepared me to understand what you are trying to say. It can't be any easier for anyone with less electronics experience/education. Zen-in (talk) 21:54, 3 April 2010 (UTC)

Zen-in, my idea is not to place these lengthy explanations on the main article. I would like only to reach a consensus about the basic idea, the circuit operation and the energy problems. Please, join the discussion with more concrete reasoning. Circuit dreamer (talk) 22:39, 3 April 2010 (UTC)

Does the circuit store energy?

Lets' make a final conclusion. The capacitor is the only element that can store energy in this circuit. But it has very, very small capacitance and the resistor R has extremely high resistance; so, the stored energy is negligible. I give a chance to the formal thinking Wikipedians to calculate and compare the energy stored in the capacitor C = 100 nF through a resistor R = 100 kΩ with the energy stored in the equivalent real inductor with inductance 1 H through a resistance R = 100 Ω. So, the final conclusioh is:

The simulated inductor has the same behavior through time as the real inductor but it does not have the same accumulating properties. The simulated inductor does not store energy. Circuit dreamer (talk) 22:47, 3 April 2010 (UTC)

The small energy accumulated in the capacitor is even undesired in this "inductive" circuit as it behaves in the opposite ("capacitive") manner - when the input voltage decreases, the capacitor passes a reverse current through the input source. If we want to eliminate completely this influence, we may connect another op-amp follower before the RC circuit. Circuit dreamer (talk) 10:47, 4 April 2010 (UTC)

The circuit has no need of storing energy - it is an active circuit - it just uses the capacitor as a impedance reference. The energy comes from the power supply of the op-amp or whatever other active element is used. --Ozhiker (talk) 00:01, 4 April 2010 (UTC)

You said it: The capacitor stores energy. I don't know why you are still arguing about this. The circuit either stores energy or it doesn't. I say it does, you say it does, and most people say that it isn't relevant. So why do you want the article to say that the gyrator circuit does not store energy? I think this discussion points out the limits of your way of analyzing circuits. You have come up with two contradictory results. Zen-in (talk) 00:28, 4 April 2010 (UTC)

The capacitor is not part of the gyrator; it is a termination placed on port 2. An ideal gyrator neither stores energy nor requires a power-supply - though this isn't obvious from the article. --catslash (talk) 01:42, 4 April 2010 (UTC)

OK so let's remove the capacitor and if the circuit is still a gyrator (is it an ideal gyrator now?), we have a gyrator that doesn't store energy because it has no reactances. (well except for the junction capacitances of all the transistors in the op-amp, etc.) A gyrator is an impedance invertor. While op-amps don't usually have inductors attached to them you could in theory create a capacitive reactance by using a gyrator circuit and an inductor. Does that gyrator not store energy as well? NO! -> Anytime you have a reactance in a circuit there is energy storage. Not energy storage like a battery (steady state) but in a transient way. That is why a capacitor causes the current to lead the voltage and why an inductor cause the voltage to lead the current. Energy is stored in the reactance during part of the cycle and is removed during another part of the cycle. Excuse me for belaboring the point but this is something any sophomore EE student would know before his/her first problem set. Zen-in (talk) 03:29, 4 April 2010 (UTC)

As I have stated above and several times earlier it is incorrect to say that a gyrator circuit that has a reactance does not store energy. This does not imply that you would use it to replace an inductor in a power switching circuit, like a switching regulator, etc. That would be impractical. Most people reading this article would assume this and anyway it is mentioned near the end of the article. So the article does not need to be further adumbrated upon to the extent that is now present in other articles. Zen-in (talk) 04:49, 5 April 2010 (UTC)

It is self-evident that the capacitor stores a small amount of energy. Although it will not be the same amount of energy as the true inductor, it would simply be incorrect for the article to say "the circuit does not store energy". Oli Filth^{(talk|contribs)} 16:34, 5 April 2010 (UTC)

I think most of you are confusing the inductor example with the strict definition of a gyrator. A gyrator is an impedance transformer; it converts the i–v characteristics at one port to different i–v characteristics at a different port. It is true that in the inductor example, the capacitor "stores" energy. However, if you replaced a DC source at the input of the inductor circuit by a ground short, the energy that would be dissipated across R_L would be different than the energy stored in the capacitor. Thus, some of the energy would come from the amplifier's power supply (which is not shown in the diagram). In fact, in that particular example, all of the energy dissipated across R_L would come from the power supply; the capacitor's energy would be dissipated across R. So it really sounds like the page should reflect that:

• A gyrator is an impedance transformer; it need not store any energy. It just needs to translate one i–v relationship to another. In the case of generating a purely resistive impedance, there would be no storage elements to find anywhere near the circuit.
• The particular inductor example given, which is an active design, the i–v characteristics of a capacitor are being transformed into the i–v characteristics of an inductor. Hence, in the process of impedance transformation, some energy is being stored in the capacitor. However, this "storage" has nothing to do with the gyrator action.

It would be incorrect to say that a transformer dissipated energy just because a generator attached to the primary side of the transformer faces a real load when a resistor is placed on the secondary side of the transformer. Hence, it is also inappropriate to say that a gyrator IN GENERAL must store energy. Having said that, the EXAMPLE CIRCUIT itself does have storage elements, but that's tangential to the topc of gyration. —TedPavlic (talk/contrib/@) 17:35, 5 April 2010 (UTC)

R_L considerations

It is important to explain the role of the load resistance R_L and, especially, to show why it can't be zero. Please, reword the text below but save the meaning. Circuit dreamer (talk) 21:36, 7 April 2010 (UTC)

"There are contradictory requirements to the value of R_L. From one side, it has to have some acceptable resistance to limit the maximum current through the simulated inductor (the op-amp output) and to obtain large inductance. From the other side, it has to have as much as possible low resistance to obtain high Q factor. Note if R_L = 0, the circuit will stop working at all: the op-amp output voltage will be zero, the op-amp will keep (almost) zero voltage across the capacitor and no current will flow through it."

Won't the Q factor be constant? It's going to be (from Inductor#Q factor):

$Q={\frac {\omega R_{L}RC}{R_{L}}}=\omega RC$

Oli Filth^{(talk|contribs)} 22:05, 7 April 2010 (UTC)

I'm not familiar enough with Q factor topic and I need time to realize it. But really it is very important to clarify the role of R_L as someone may try to make an "ideal inductor" (R_L = 0). The results will be unfortunate - zero inductance (as the expression L = R_LRC says) and what is more important, the circuit will not operate at all (here I suppose an ideal op-amp). Circuit dreamer (talk) 06:06, 8 April 2010 (UTC)

Q is a fundamental concept in electronics. So much so that filter design equations are sometimes written in terms of the desired Q. There are even Q meters. I encourage you to learn more about this "mysterious" quantity Q. You may even find that the ubiquitous negative resistance plays a hand in it. Zen-in (talk) 18:59, 10 April 2010 (UTC)

Latest additions

I've reverted these latest additions because:

• As has been mentioned previously, this is your own bespoke analysis. If you can find a reference that supports the idea that this is a bridge topology, then perhaps we'll revisit.
• There have been multiple objections in the past to the hand-drawn diagrams.
• This isn't really a positive-feedback topology.

Oli Filth^{(talk|contribs)} 22:59, 19 April 2010 (UTC)

Thank you for the elucidations. Definitely, talk pages become more interesting than the main pages and I am apprehensive that our readers will begin reading talks instead articles:) Well, let's discuss the section about circuit operation that you have reverted.

Bridge viewpoint. Of course, we may consider the circuit without seeing any bridge idea behind it; we may see only the "trees" (the particular C, R, R_L, op-amp, power supply) without seeing the "forest" (the bridge topology). But the bridge configuration is something familiar to nearly everyone; it is a concept helping understanding. We, human beings, understand new things (systems) by discerning old well-known patterns (subsystems, components, ideas, tricks, etc.) inside them..., by making associations based on what is familiar... Balanced bridge is another powerful concept whose amazing feature is the equality (horizontal symmetry) of the voltage drops. I realized it as far back as 1968 when I, 14-year-old boy, made a simple servo system by two potentiometers, a polarized relay with neutral position, split supply and a motor with reduction gear...

The two pairs C,R and R_L,Vs constitute a balanced bridge

Wikipedia says "...bridge circuit is a type of electrical circuit in which two circuit branches (usually in parallel with each other) are "bridged" by a third branch connected between the first two branches at some intermediate point along them. .." Look at the picture now: R and C constitute the left branch; Vs and R_L constitute the right branch. The two branches are connected in parallel and they are "bridged" by the op-amp differential input. The bridge is balanced since the op-amp follower keeps Vs = V_R... plain, clear and simple... Tell me now, should we look for "a reference that supports the idea that this is a bridge topology"? BTW, an INIC (drawn in such a symmetrical form), is another example of balanced bridge circuit where the op-amp changes the supply voltage to balance the bridge.

Diagram. It is a "snapshot" of the circuit operation where the instantaneous voltages across the elements and the currents flowing through them are shown by proportional voltage bars and current loops. This presentation allows readers to grasp the idea (equal voltages across the opposite elements) only by a glance at the picture. It is a colorful just because our world is colorful. Is your monitor black & white? Do you watch black & white TV? Then, why do you make readers look at this colorful world through black glasses? And what is the problem with this kind of hand-written diagrams in a Forrest Mims style? If you do not like it, feel free to redraw the pictures by an editor.

Positive feedback. There is no feedback in this circuit only if it is driven by a perfect voltage source (with zero internal resistance). Normally, in the case of a real voltage source (with some internal resistance), there is a positive feedback with a closed-loop gain 0 < A.β = 1.β < 1. So, this feedback is "under control" (the Armstrong's regenerative receiver idea). In comparison, in Schmitt triggers, the positive feedback is out of control as A.β >> 1. This is the "trick" here - to "bridle" the positive feedback by using a unity-gain amplifier. Another trick to make the positive feedback stay under control is to "neutralize" it by a negative feedback; an INIC operating in a linear mode is based on this powerful idea. In INIC operating in a bistable mode, the positive feedback dominates and it is out of control.

Circuit dreamer (talk, contribs, email) 22:59, 20 April 2010 (UTC)

I don't know how many times you need to hear it! You can't keep adding your own interpretations and analyses of circuits, unless you can find sources that support them, and unless you can keep their length to a minimum. You can't keep adding these diagrams, because they are cluttered and unprofessional, and cannot be modified by other editors. Oli Filth^{(talk|contribs)} 23:15, 20 April 2010 (UTC)

Agreed, this hypothetical bridge analysis has been reverted before. Electronics analysis is not a "Let's throw everything against the wall and see what sticks" kind of thing. This is too much POV-pushing. Zen-in (talk) 02:00, 21 April 2010 (UTC)

Gyrators and Smith Charts

It would be interesting to include a section comparing Smith Chart operations with the gyrator. With a Smith Chart you can transform an impedance to its conjugate with a simple operation. If I'm not mistaken that is what a gyrator does. Zen-in (talk) 06:45, 28 April 2010 (UTC)

No, because (a) this would be original research, and (b) this doesn't make any sense as far as I'm aware. A Smith chart doesn't do anything; it's simply a graphical tool to facilitate calculation of (amongst other things) complex admittance from complex impedance. This seems like an apples and oranges comparison. Oli Filth^{(talk|contribs)} 08:17, 28 April 2010 (UTC)

Transformer simulation

It seems to me that the gyrator replacement for the transformer will transmit dc current where a transformer wouldn't. Is it still an acurate replacement, and are the other claims that transformers can be replaced correct? 178.83.12.182 (talk) 19:18, 11 September 2011 (UTC)

An ideal transformer will work down to d.c., though a real one will not (see transformer: effect of frequency and electrical element for the concept of ideal components). The section you refer to does talk about ideal gyrators and ideal transformers - perhaps ideal should link to 'Electrical element'? --catslash (talk) 21:43, 11 September 2011 (UTC)

Gyrator as used in electroacoustic circuit analysis

I'll leave it as an exercise to someone else to add a section on how a gyrator is used in electroacoustics, as proposed by Georgia Tech EE professor Prof. W. Marshall Leach in his seminal paper On the Specification of Moving-Coil Drivers for Low-Frequency Horn-Loaded Loudspeakers (Journal of the Audio Engineering Society, vol. 27, no. 12, pp. 950-959, December 1979). I was a student of Leach's at the time, and this as a sidelight introduced a powerful concept in section 2: Using a gyrator model for the voice coil, which "eliminates the confusing parallel-element mobility and admittance circuits from the analysis." Go to figure 2 and you'll see the gyrator with a gyration resistance of Bl between the electrical and mechanical network segments. Although introduced for horn-loaded loudspeakers, in fact it has broad use across electroacoustics, including for conventional piston loudspeakers of all varieties, and also for hearing aids. Ironically, in his final edition of Introduction to Electroacoustics and Audio Amplifier Design, Fourth Edition before his death in November 2010, he did not include his groundbreaking use of the gyrator, because parallel reactive elements are now easily handled in SPICE.

What does a simulated inductor have to do with a gyrator?

Nowhere is this explained. I don't see any two port network here. This section appears to be totally irrelevant to the topic. — Preceding unsigned comment added by 71.214.216.175 (talk) 20:21, 25 July 2012 (UTC)

It does say in at least two places that a gyrator transforms a capacitive load into an inductive load. What may be less clear is that the op-amp circuit minus the capacitor (the capacitor being considered an external load), simulates a gyrator (provided R >> 1/jωC >> R_L). --catslash (talk) 22:23, 27 July 2012 (UTC)

Comment

It is great fun to read the above notes. When I was a student 30y ago, our professor told us "in 10 years all the inductors will be replaced with OPAmps". I laughed and pissed in my pants. Except few kHz frequency band pass filters, the inductors are still in use. Have fun. What I like is the diode model. When you short it, the current flows. That's ready for Nobel prize. — Preceding unsigned comment added by 69.159.36.17 (talk • contribs) 04:07, 25 October 2012

"Applications" caveat

In the Applications bit it says that power conversion and transformers using flyback are cases where gyrators can't be used, but since e.g. parasitically powered voltage doubler systems should theoretically make such applications possible ^{(no, they won't be ideal due to implementation limitations and losses from operation, but good luck finding any truly ideal elements :) )}, shouldn't the wording (I said "caveat" for a reason :) ) be more along the lines of "There are many applications where a gyrator is not enough to replace an inductor"? — Preceding unsigned comment added by 99.23.95.33 (talk) 03:52, 15 May 2014 (UTC)

If that is an entirely theoretical consideration (ie there is no real application) and you don't have a source discussing the possibility then I think we can safely ignore that. However, Now my attention has been drawn to this section, I seriously disagree with the claim gyrators are not used at RF. Impedance inverters (which in black box circuit terms are gyrators) are commonly used in RF circuits. It is true that passive inductors are much easier to make for RF and active gyrators, using an IC for instance, are much harder, but nevertheless passive impedance inverters are very common, see quarter-wave impedance transformer for instance. I'll do something in the article. SpinningSpark 11:32, 15 May 2014 (UTC)

This is unclear

"Simulated elements only imitate actual elements as in fact they are dynamic voltage sources. They cannot replace them in all the possible applications as they do not possess all their unique properties. So, the simulated inductor only mimics some properties of the real inductor." (Comparison with actual inductors[edit])

Please, could someone who thinks that she knows what this means express it in such a way that the pronouns refer unambiguously to particular nouns ? As the passage stands it is almost meaningless. The usual rule in English is that pronouns refer to the nearest noun which agrees in number, gender, etc, so the first sentence seems to say "Simulated elements only imitate real components since some real components are dynamic voltage sources". This would make sense if we read it as "... imperfectly imitate...". It would be helpful also if the concept "dynamic voltage source" were explained; although I am a professional electrical engineer this is not clear to me. Perhaps "source of stored energy" is intended (true of capacitors and inductors, but not of resistors) ?

In the second sentence "they" and "them" are so far from their likely nouns that we are left guessing. My guess is that this is intended:- "Simulated elements cannot replace real components in all applications since the simulations do not possess all of the circuit properties of the real components they represent. For example, the simulated inductor only mimics some properties of a real inductor." If this is intended, it would be helpful if the deficiencies of the simulated inductor were described. Such deficiencies need not be related to the supposed inability of simulations to act as "dynamic voltage sources". For example, a parallel resonant circuit in which the inductor is simulated by a gyrator exhibits perfectly well the expected energy exchange which occurs between capacitor and inductor when the circuit oscillates. This implies that the simulated inductor returns stored energy to the capacitor just as a real inductor would. Of course we know that the simulated inductor is does not conserve energy, and that the "stored" energy is actually derived from a op-amp's power supply, but nevertheless the simulation is faultless in this respect.82.37.54.83 (talk)