Talk:Capacitor

TODO

I would suggest doing the following:

- Move capacitor non-idealities before the section describing types of capacitors. The former is needed to understand the latter.

- Remove the bit about replacing old electrolytics in audio equipment. Or possibly move this into the section on capacitor types

- Are the ESR numbers good? I've seen caps with much higher ESR. Someone should verify this.

- I'd clean up/organize the section on non-idealities. Parts of it are a bit rambling. I'd have one paragraph on each type of non-ideality, I'd also move the parts on accuracy/precision from the top into this section.

At some point, it'd be very nice to have a table of type of capacitor vs. accuracy, tempco, ESR, ESL, resonant frequency, leakage, etc. This would be a valuable resource.

I don't think it's bad to have more information in a Wikipedia article (as someone mentioned about keeping it basic/for beginners, rather than an engineering reference). But I do agree that the additional information should come in seperate sections at the end. A beginner should be able to read the first bit. An expert should be able to use the rest for reference. The reference should be in a reference format. We could add a section 'engineering practice' at the end for this. Organize it as:

Overview

Physics

History

Non-idealities

Types of capacitors

Applications/Engineering practics

Trimpots

A previous version of this page referred to small variable capacitors "trimpots". AFAIK, trimpot is short for "trimming potentiometer", a type of small variable resistor you turn with a screwdriver. A quick Google search showed that my usage is certainly common, so I deleted that reference. -- Tim

Yes, trimpots are resistors. There are trimmer capacitors, though. - Omegatron

Biological capacitors

We need to add the technical use of "capacitor" in biology. -- Anon.

And what is this technical use? - Omegatron

Charge on plates or insulator?

It isn't always recognised that the charge stored in a capcitor is not stored on the plates but on the surface of the insulator or dielectric.

Is this correct? Since you can have air capacitors or vacuum capacitors I can't see how this is the case. I always understood that the charge was the electrons pulled towards the other plate, and therefore on the surface of the plate. I only studied Electronics and Physics to A-level so I will wait for an expert to verify rather than changing the article myself. -- Chris Q 14:14 Feb 18, 2003 (UTC)

I changed it. It is not really a contrast: The charge is at the boundary. - Patrick 14:31 Feb 18, 2003 (UTC)

It is possible to experimentally prove that the charge is not on the plates. It can easily be done in a system where you create the charged capacitor, remove the plates, short circuit them to demonstrate no electrical energy and then replace them on the insulator. Yes you can have air capacitors but if the air was stirred the charge would dissipate. Vacuum capacitors store the charge on the glass insulator with the vacuum providing a high resistance to electrical discharge between the two opposing chargesRjstott

??? How do you remove the plates from the system without removing the dielectric? Stirring the air does not discharge an air cap, except maybe the charge that is transferred through moisture from one plate to the other. Air capacitors are used commonly in radio. What glass insulator? Wouldn't that make it a glass-dielectric capacitor? - Omegatron

I've always thought that the energy was stored in the dielectric (which actually can be true for a vacuum, the energy density is 0.5*epsilon*E^2), but not the charge - the existence of vacuum capacitors should illustrate that clearly. What does happen for a dielectric is that the charge in the dielectric shifts slightly, still giving a net neutral charge for the bulk of the dielectric, but partially cancelling the free charge at the surface so that one can fit more free charge for the same voltage.

Might not be entirely relevant to the article, as the overview is correct as it stands, but it may be important to keep in mind. StuartH 05:36, 9 Jun 2004 (UTC)

http://www.amasci.com/emotor/cap1.html

I just read Bill Beaty's article at the above URL. It is excellent, and should be read by all contributors to this Wikipedia article. In fact, I wish we could steal the whole thing word-for-word. Can we hire Bill to write for us? I think our article has a slight touch of the 'physicist's capacitor' versus the 'engineer's capacitor' problem.

By the way, Bill's article explains why the statement "It is possible to experimentally prove that the charge is not on the plates." above is wrong. -- Heron 14:36, 10 Jul 2004 (UTC)

seriously. has anyone ever tried inviting him to wikipedia? - Omegatron 20:20, Jul 10, 2004 (UTC)

I've been here for a couple of years, mostly wreaking havoc on entries such as Electricity, Conduction, Charge, etc. --Wjbeaty 01:25, Dec 28, 2004 (UTC)

This would be pretty easily proved by building two identical jars, charging one, and swapping components. Which one has the charge now? "The charge is on the dielectric", according to the Leyden jar article, which I have moved to the talk page. - Omegatron 00:33, May 31, 2005 (UTC)

Oh. I see that it is not so easy because of the corona effect. - Omegatron 00:37, May 31, 2005 (UTC)

Interesting, I was not aware that this experiment had been done in the way described and perhaps the explanation has some merit. However, the way I saw it done involved two large flat plates and a large flat dialectric where the separation was done by moving the plates perpendicularly away from the dielectric rather than horizontally. No corona effect?--Rjstott 03:46, 12 September 2005 (UTC)

Stored charge or energy?

Is it true that "a capacitor stores electric charge", as the article claims? I think it's more accurate to say that a capacitor stores energy by redistributing its internal electric charges to create an internal electric field. The electrons and protons are there regardless of the capacitor's state of charge, but in the case of a discharged capacitor the opposite charges are mixed in equal concentrations on both plates. In fact, the word "charged" in this context means "full of energy", not "full of electric charge". If a capacitor really stored electric charge, the whole component would become either positive or negative, and the charge would soon leak away to the environment.

Perhaps I'm just splitting hairs. I'll wait for comments before I make any changes to the article. -- Heron

I rephrased it. Indeed the plates get "full of electric charge". The energy is a consequence, not the primary thing. And you make it sound as if the charge moves internally, but it moves externally through the circuit from one plate to the other. - Patrick 09:53, 3 Sep 2003 (UTC)

Patrick, I dont know if youre still here, but what you have said in the above post is most important. Charge separation occurs via the external circuit not across the plates. So you dont need to postulate any thing passing between the plates. We know charge cannot pass, but does any form of current really need to pass to charge the capacitor?? --Light current 01:26, 12 September 2005 (UTC)

Triboelectricity

I don't see what triboelectricity has to do with capacitors, besides being a possible source of charge to charge one up. Yet there is this paragraph in the article:

Triboelectricity is an important associated phenemonon. This is the generation of electricity by rubbing an insulator. The ancient Greeks experimented with this. Not all capacitors are intended by man; as lightning will attest. Taking the carbon black out of automobile tires has improved their functioning as a dielectric, as any (prob. elder) one who has been shocked at the gas pump can attest. Electrodynamics is even of interest to astronomers. Triboelectricity and magnetics (associated with inductors) are the two possible ways to directly convert mechanial energy into electrical energy.

Are they trying to say that the two separated surfaces are a capacitor? Then any two completely irregular, non conducting surfaces would be a capacitor, as long as there was a difference in charge between them. - Omegatron 18:38, Jun 8, 2004 (UTC)

Recent edits by 24.12.159.248

248, I've been watching your edits (as 128.12.178.70) and I want you to encourage you to add information. It's your organization and style that I think could use some work. Some points:

• Stay on topic in the articles and sections that you modify. If you wander into a side topic, and then a side topic of that topic, that's not good. Try to work your parenthetical remarks into the main text; if you can't do this, perhaps the remark is too far off-topic.
• Avoid replicating information that is in other articles.
• Use complete sentences and avoid informal language.
• When you use pronouns, make sure that the antecedents are clear.
• Don't use first-person or second-person references. Wikipedia is a collective effort with open access.
• Remember that the audience includes people who know nothing about the article subject, and that WP cannot hope to be a substitute for formal engineering training. Therefore focus on fundamentals and hard facts before engineering practice.
• Please, no jokes. Since the audience does include people who know nothing about the subject at hand, your well-intended, light-hearted comments are just as likely to confuse as to entertain.

Again, I want to encourage you to add relevant information. Just make sure that it's added in such a way as to maximize its usefulness to readers. Gazpacho 02:43, 9 Jun 2004 (UTC)

Some recent edits

In 'information theoretic' terms capacitors are used both to "store" and "destroy" information. They can be used to store binary or voltage information, in digital memories or switched-capacitor circuits. They are also used to destroy information, when used to shunt away high frequencies in filters. Even when it is used to store information, it is destroying other information, for example, when it is used as the storage element in a DRAM, one duty of the capacitor is to 'forget' when the bit was written. When they were used in 'bucket brigade' delay lines, they also functioned as an Low Pass Filter.

What is "information theoretic"? How does a capacitor destroy information? A capacitor can store information either in the form of an analog voltage (switched-capacitor filters) or a digital voltage (DRAM). In the DRAM example, the capacitor is not destroying information, it is simply storing a new value. You could say perhaps that the resistor shorting the capacitor charge to ground is destroying the information in the capacitor, or that the memory chip itself is destroying the information by writing over it with something else, but the capacitor isn't doing it, and I don't understand the relevance anyway.

Capacitors exploit electrostatics, inductors, electrodynamics. This leads to a 'symmetry breakdown.' Yes, inductors can be thought of as being 'inverse capacitors,' but there is no 'transformer (i.e. proximity) effect' associated with capacitors. There is no transformer-like effect associated with capacitors.

Capacitors exploit electrostatics, and have parasitic, unwanted inductance, and I suppose are electrodynamic. Can you please explain in more detail why inductors and capacitors are not duals? What is the "transformer effect"? Capacitors can be used as coupling devices in a way roughly similar to transformers.

Intercell (capacitive) nerve connections limit athletic reaction time.

What does this have to do with parasitic capacitance?

a property of the ideal current amp) for speed, given the same Qdiss. Quie

What does Qdiss stand for? please elaborate. - Omegatron 07:09, Jul 10, 2004 (UTC)

----

I fixed most of the points that you questioned above, Omegatron. (They weren't my points, by the way.) Next, I consulted The Penguin Dictionary of Electronics (ISBN 0140511873), which gave me a much clearer picture of how capacitors work, and helped explain the "capacitor/inductor dualism" thing.

The dictionary first defines capacitance as the ability to store charge, which for an isolated conductor is C = Q/V, where V is the voltage change after the addition of a charge Q. It then mentions that two such conductors make a capacitor, and that the capacitance of this thing is the ratio of the charge on either conductor to the voltage difference between them. Note that the isolated conductor has capacitance, but is not a capacitor. I interpret all this to mean that the charge (and now I am confident enough to call it charge, and not energy) stored in a capacitor is held on the plates: +Q on one plate and -Q on the other. The "dismantling a Leyden jar" experiment, which would appear to contradict this, can be explained by corona discharge, as Bill Beaty does in his article referred to above. If this is true, we can stop beating about the bush and proudly state that a capacitor stores charge (or, more accurately, that it stores two equal but opposite charges). The stuff about polarisation of the atoms in the dielectric explains dielectric losses, but has nothing to do with why a capacitor is a capacitor.

I now also understand a bit better the symmetry between capacitors and inductors, and it is neater than our article presently admits. Both devices allow some of their field to leak out, although the leakage is more noticeable with inductors than with capacitors. If two inductors become magnetically coupled, then they make a transformer. The degree of coupling is called mutual inductance, which is defined as the voltage in one inductor caused by a changing current in the other. Similarly, two capacitors can be electrostatically coupled, a fact which is sometimes overlooked, although you (Omegatron) alluded to it. The degree of coupling is called - guess what - mutual capacitance, which is defined (thanks to my Penguin dictionary) as the current in one capacitor caused by a changing voltage in the other. A quick Google for "electrostatic transformer" finds lots of hits related to particle detectors, so it seems that this effect has a practical use, but not one that many electrical engineers would be familiar with.

The symmetry argument breaks down when you try to imagine the dual of the 'isolated conductor'. I think this is related to the fact that magnetic monopoles do not exist. -- Heron 21:21, 10 Jul 2004 (UTC)

Thanks. I didn't know that an object by itself in empty space could have a capacitance. Here is another link explaining this: http://www.qprox.com/background/capacitance.php - Omegatron 23:17, Jul 10, 2004 (UTC)

Mutual capacitance

Oh dear. That article contradicts my Penguin Dictionary's definition of mutual capacitance. So do many other web pages. My dictionary said that mutual capacitance was the interaction of two capacitors, but it seems that it is really the interaction of any two charged conductors. A capacitor is a special case of mutual capacitance in which the two objects are brought close together, so that the electric field is confined mostly to the small volume of space between them and does not affect other nearby objects. -- Heron 11:12, 11 Jul 2004 (UTC)

Oh dear. I don't know the answer.

I would still like an explanation of how capacitors destroy information. "Information theoretic" was probably supposed to link to information theory, and I have some familiarity with it, but i still don't understand what the author was trying to say. - Omegatron 19:34, Jul 16, 2004 (UTC)

It's incorrect anyway. Lossless filters, involving only capacitors and inductors, either reflect energy or transmit it. A lossless low-pass filter doesn't destroy high frequency information, it reflects it back to where it came from. Resistors on the other hand dissipate signals as heat, increasing the disorder of the system, thereby creating information. Nothing destroys information. -- Tim Starling 01:18, Nov 30, 2004 (UTC)

Capacitor vs Capacitance

Although I find Capacitor to be a very good article, it deals with both the physical objects (capacitors) and the physical quantity (capacitance). This causes issues for the categorisation of the article. For clarity, I would prefer to split the current article into separate (but linked) articles on each aspect. Does anyone have any objection to this, please? Ian Cairns 00:32, 14 Nov 2004 (UTC)

RF stubs

I quote from "types of capacitor":

stubs: In RF circuits, a length of transmission line less than a quarter-wave, that is shorted at the end, or a length greater than a quarter-wave left open, has the electrical properties of a capacitor.

I'm not an RF engineer, but I'm pretty sure this is the wrong way around (ie both examples would behave inductively). Could someone with more relevent knowledge make the edit if it is required?

Field-theoretic items

Since a capacitor is inherently a "lumped" circuit element, perhaps the lengthy Maxwellian discussion could be moved elsewhere? All that grad div and curl stuff is intimidating and everything after "forcing function" could be removed with a useful shortening effect to this article. An observation that practical capacitors have all three elements of capacitance, inductance and resistance, and a note that at high frequencies it's no longer useful to use the lumped circuit approximation, should be all that's required in this article. --Wtshymanski 17:53, 1 Jan 2005 (UTC)

Seen on electronics-forum.info

Date: 15 Nov 2004 23:57:00
From: hoo-dair hoo-dat
Subject: Capassitor in series?
Sum 1 tellin me to put passitor in series, whada ye meen series, world
series.

Edits similar to inductors

It would be nice if inductors, capacitors, and resistors followed a similar structure. Any of the previous suggestions are OK with me as long as there is consistency. I like the idea of physics first, then electrical, practical etc. I tried to add some of the same content found in inductor article, e.g. energy. I haven't looked, but this could all be tied into KVL and KCL circuit analysis at some point. Just my $0.02 - Madhu

I said the same thing when I first saw the articles, but never finished it. Be bold and do it. - Omegatron 05:12, Feb 17, 2005 (UTC)

Oh. For some dumb reason capacitance is a separate article from capacitor. So some of what you are adding might be in the other already. - Omegatron 05:15, Feb 17, 2005 (UTC)

Flux capacitor

Watch out for this device, I hear Marty McFly has one of those and he ain't afraid to use it! — Rickyrab | Talk 04:23, 15 Apr 2005 (UTC)

Metal plates

Someone says not all capacitor plates are metal. I'd appreciate a citation? News to me. I'm taking it out for now. --Wtshymanski 04:28, 15 Apr 2005 (UTC)

It may be news to you, but capacitors using conductive plastic plates have been available for quite a few years. There are those that use aluminized mylar (sort of semi-metallic), but there are also those that use conductive polymers without coatings.

Cool. I've never heard of those. Can you show an example datasheet? Please explain this in the article. - Omegatron 12:47, Apr 16, 2005 (UTC)

Agreed, I'd like to know more; someone has changed it again but not listed a reference. Like I said, news to me. I can't see the utility - conductive plastics are very poor conductors. Who makes these? --Wtshymanski 14:45, 16 Apr 2005 (UTC)

Um... Supercapacitors? ultracapacitors? Last I heard, carbon was not a metal. Also, I think I recall seeing "Plastic capacitors" in the digikey catalog. Search google for solid polymer aluminum capacitors, or just "conductive polymer" +capacitors --Wjbeaty 21:57, Apr 17, 2005 (UTC)

I'd forgotten about supercapacitors. And the Kemet paper at http://www.kemet.com/kemet/web/homepage/kfbk3.nsf/vaFeedbackFAQ/134ECF096F57820885256F72006669BA/$file/1999%20CARTS_Replacing_MnO2.PDF

shows that even tantalum capacitors have a non-metallic anode plate. The ignition hazard of tantalum capacitors is also something I'd never read about before. --Wtshymanski 22:23, 17 Apr 2005 (UTC)

condenser

called a condenser because it "condenses" more charge into the same space? for instance, applying a voltage to two metal plates a meter apart would only imbalance the charge a tiny bit. placing them very close together and then applying the voltage allows much more charge imbalance in the same volume. as far as electrons are concerned, they are now "condensed" tighter together. more are able to be "stored" (and vice versa for holes, of course). - Omegatron 02:14, May 21, 2005 (UTC)

I think you are right. This article on Ben Franklin gives the same explanation. Also, here are some citations from the OED:

I had rather call it a condenser of electricity..using a word which expresses at once the reason and cause of the phenomenon. -- Volta in Phil. Trans. LXXII. App. 8, 1782

and

Accumulators are sometimes called Condensers, but I prefer to restrict the term ‘condenser’ to an instrument which is used not to hold electricity but to increase its superficial density. -- Maxwell Electr. & Magn. I. 50, 1881

and Abraham Bennet wrote to the Royal Society in 1787 thus:

The labours of M. VOLTA have been very successful on this subject by the application of his condenser (as he terms it), which, by means of a thin coated electric, is capable of receiving a greater quantity of electrical fluid than a common insulated conductor... (Philosophical Transactions V77 (1787) pages 288-296, found at [1])

which corroborates Volta's claim to have invented the term (which he probably translated from the Italian condensatore). --Heron 11:09, 21 May 2005 (UTC)

Excellent work! Thanks. We should include this in the article, of course. - Omegatron 16:08, May 21, 2005 (UTC)

I have done so. Now I want to know more about those "Ancient Greeks". Which ones, exactly? --Heron 20:10, 21 May 2005 (UTC)

I can't find an etymology of the word capacitor. Most non-English languages use a variant of condenser or condensatore. I wonder who the first was to use capacitor. This is the best I can find and I don't trust it:

The name "capacitor" was given in the US due to its capacity for charging electricity. When capacitors were introduced to Japan, the English word "capacitor" was translated as "chikudenki," which means a component that can condense and store electricity. Later, people in Japan thought it was called condenser in the US when they retranslated it into English. That is the reason the electric component called capacitor in the US is still called condenser in Japan. Moreover, in another theory it used to be called the "condenser" in the world, but recently there is also an opinion of having came to be called "capacitor". - Omegatron 13:38, May 22, 2005 (UTC)

Hmm. That sounds like an answer to the question "Why do the Japanese still call them condensers?"

The OED doesn't record the exact moment of conception, but it has a quote from the 1922 BSI Glossary of Terms in Electrical Engineering that describes 'capacitor' as a 'new term' and recommends its use to avoid confusion with the steam 'condenser'. The OED gives the etymology as from 'capacity', which meant 'capacitance' as early as 1777 (in Tiberius Cavallo, A complete treatise of electricity in theory and practice). --Heron 17:17, 22 May 2005 (UTC)

Definition of capacitance in words

Capacitance is a measure of the amount of charge stored (separated?) in a capacitor for a given voltage across the capacitor. That is, for a given V, Q is proportional to C. The original wording implied that for a given Q, V is proportional to C when in fact V is inversely proportional to C. Alfred Centauri 01:21, 26 May 2005 (UTC)

Transfer Function

From the "In electric circuits" section:

"The transfer function for an ideal capacitor can be written as a differential equation in time domain"

A transfer function lives in the frequency domain and so is not the appropriate term here. Also, the use of the term differential equation is a stretch. Although the equation does involve the derivative of v(t), it does not involve v(t) itself or any other derivatives of v(t) thus placing zero constraints on the function v(t) other than that its derivative must exist. I've changed the wording to a more standard form:

"For an ideal capacitor, the capacitor current is proportional to the time rate of change of the voltage across the capacitor where the constant of proportionality is the capacitance, C:"

Alfred Centauri 02:39, 26 May 2005 (UTC)

not exactly

"Because each plate stores an equal but opposite charge, the total charge in the device is always zero."

Not exactly.  :-) Scuff your feet across the floor and then touch one side of a capacitor which is otherwise insulated from everything else. That side of the capacitor will now have a net charge. Whether we should actually include this or not, I don't know. It probably confuses understanding more than anything... Electricity can flow in things that are not circuits, though, which is probably an important thought process that is being lost in our age... Would learning about the fundamentals of charged objects help or hinder understanding of circuits? I don't know. - Omegatron 00:04, May 28, 2005 (UTC)

I wouldn't include it since a capacitor is, by definition, a two terminal circuit element used in electric circuits where, by definition, charge is made to flow along a closed path (circuit). Let's face it, if you scuff your feet and touch the chassis of an isolated, battery powered circuit, the entire circuit has net charge. In this case, we can talk about the capacitance of the entire circuit. Perhaps this kind of stuff would be better in the Capacitance article, which has, IMHO, a lot of material related to Capacitors that doesn't belong there. Alfred Centauri 16:53, 28 May 2005 (UTC)

Well, technically a capacitor is any two conductive objects separated by a dielectric. A coax cable is a capacitor, for instance. Two pins of an IC form a parasitic capacitor, and so on.

But, that's exactly my point - a capacitor is a two-terminal device that is charged by moving electrons from one plate (conductor) to the other via an external circuit. On another note, would you agree that the term capacitor refers to, in practice, an intentional capacitor and not the inescapable and ubiquitous parasitic capacitors found in all electronic components (including resistors, inductors, p-n junctions, etc. etc.)? Alfred Centauri 17:58, 29 May 2005 (UTC)

Yes, "capacitor" usually refers to intentional components. It also refers to parasitic components as part of a circuit quite often. Rarely does it refer to the abstract capacitance between two completely isolated conductive objects. That is usually just referred to as "capacitance", if you're talking about the two words themselves. Obviously they're both talking about the same thing. - Omegatron 20:50, May 29, 2005 (UTC)

Hmmm... OK. However, I personally do prefer to use the term capacitor to refer to the device and parasitic capacitance to refer to those uninvited guests and capacitance to the physical phenomena due to the electric field. Alfred Centauri 01:55, 30 May 2005 (UTC)

Yeah. That's probably how we should keep the articles, too. - Omegatron 02:30, May 30, 2005 (UTC)

Yes, you could touch and charge an entire circuit, but any capacitors in the circuit would isolate the charge, inductors would resist the sudden charge imbalance, etc. The kind of stuff that causes noise in signal processing gear from triboelectric effect in cables and so on. If you did scuff your feet and touch one side of an otherwise floating cap, you could then discharge the cap and do (a small amount of) work, since there is a charge imbalance across its plates. And, if my understanding is correct, you can fit more charge on one side of a capacitor than you could on the same piece of metal if the capacitor were disassembled. It's not completely trivial.

Yeah, it probably should be in the capacitance article instead, maybe with a little "(see capacitance for more details)". - Omegatron 20:51, May 28, 2005 (UTC)

I don't quite understand what you are getting at in your first sentence above. Can you elaborate? I'm not sure about being able to store more charge on one plate of an assembled capacitor versus a plate in isolation. Do you have any further info on this? Finally, when you say 'discharge the cap', do you mean to discharge through an external circuit connected across the cap or to a discharge to ground the lead that you touched? Alfred Centauri 17:58, 29 May 2005 (UTC)

Sure. I was just saying that you can charge a floating circuit with a static charge, but the entire circuit doesn't charge at once. Your injection of electrons or holes into the circuit causes currents to flow, etc. Any section of the circuit that is completely isolated with capacitors will not become charged, too.

Consider a metallic chassis upon which an electric circuit is constructed in some form or fashion. The chassis is used as the circuit 'common' (a term I much prefer over 'ground').

Good point! - Omegatron

Assume that the circuit is isolated from Earth ('ground') and is battery powered. Now, place charge on the metallic chassis. Because the chassis is conductive, the charge placed on the chassis distributes itself such that the electrical energy is minimized. The chassis is the zero volt reference point for the circuit. But, since the chassis has net charge, the chassis is at some potential w.r.t ground. Thus, we can define a capacitance for the circuit that is the total charge place on the chassis divided by the potential of the chassis (the Earth is considered to be our zero reference). Now, if a particular node in the circuit is 5V w.r.t. the chassis and the chassis is, say, 1000V w.r.t. the Earth, the node, by KVL, must be at a potential of 1005V w.r.t Earth, right? Perhaps we're saying the same thing differently so let's just leave it at that. Alfred Centauri 01:55, 30 May 2005 (UTC)

I was just saying that the chassis/common node is the only thing that gets that charge, at first. The charge "distributing itself such that the electrical energy is minimized" will take time to go through inductors, will never go through the dielectric of capacitors, etc. So sections of the circuit that were capacitively coupled would not become charged. That is all I'm saying. - Omegatron 02:30, May 30, 2005 (UTC)

I'm not sure about being able to store more charge either, but I think you can. Hence the term "condenser". I am learning about this on other sites as well.  :-)

I'm going to venture that the term condenser is due to the notion that a capacitor is, in effect, a way of separating charge versus a way to store charge. Think about it... Each plate of the capacitor has plenty of charge but there is an equal amount of positive and negative charge so there is no 'net' charge. When we move electrons from one plate to the other, we are more or less separating or organizing the charge that already exists in the plates into separate regions. Alfred Centauri 01:55, 30 May 2005 (UTC)

Yes, we are separating charge when we use a capacitor in a circuit. I wonder if the original (I've been reading some old patents lately) formulation of the capacitor was more general, and the term "condenser" came about because you could store more charge on one plate if it was near another plate that was connected to a charge reservoir than you could if the plate were floating by itself. And when I say "more charge" I am talking for the same voltage or same initial charged object or something. I am asking about this elsewhere as well... - Omegatron 02:30, May 30, 2005 (UTC)

When I say "discharge the cap", I mean discharge the cap by connecting its two terminals together. Say you have a cap floating through the air, lifted by invisible magical hands (or just held up by some piece of plastic). It has two terminals, A and B. If you scuff your feet along the floor and touch terminal A, the A side of the capacitor now has a net charge, while the B side is neutral. If you then connect a little lamp between A and B, the charge imbalance will settle out by sending a current through the lamp. Now both sides of the cap will have a static charge, but less so than side A was. - Omegatron 20:50, May 29, 2005 (UTC)

But, even though the charge is now distributed equally on the two plates, the capacitor has a net charge an thus a potential w.r.t. to Earth. Thus, by connecting both leads together on one side of a litte lamp with the other side of the lamp connected to Earth ('ground'), the charge you orginally placed on the terminal A can move from the capacitor through the lamp to ground, right? Somehow, this seems like double dipping. I've got to think about this a little longer... Alfred Centauri 01:55, 30 May 2005 (UTC)

Well, if you charge up any object and then connect it to another object that is differently-charged, a current will flow as the charge redistributes itself. In my example plate A is charged and plate B is neutral, so the charge redistributes itself when you connect them together, causing a current to flow, even though you didn't pull the charge off one plate and put it on the other. You just added extra to one plate from another object. Yes, the entire cap will always have net charge now unless you connect (both plates!) to something else. There is a certain amount of work involved in moving a certain amount of charge to plate A of the capacitor. If you do work redistributing that charge equally to plate A and plate B, does that reduce the work you can do in draining that charge back to ground? It seems like it would in order to conserve energy. So, something doesn't smell right to me so I'll have to get back with you on this one... Alfred Centauri

It takes work input to add charge to plate A. This creates a charge imbalance. You get work back out by letting A and B settle out, same as any other discharging of a cap. I am guessing half the work back out. Whoa I just thought of a way to try this with real capacitors.  :-) - Omegatron 03:19, May 30, 2005 (UTC)

OK, let's look at this another way. This problem is essentially the same as the 'missing energy paradox' when an uncharged capacitor is connected directly across an identical but charged capacitor. Plates A and B represent two identical conductors in free space (for simplicity here, assume the plates are in a vacuum). There is some capacitance C associated with each plate (this is the capacitance associated with this configuration of conductors and 'ground' and is NOT the same as the capacitance of the capacitor formed by the two conductors). Now, place Q amount of charge on plate A so that plate A has some voltage $V={\frac {Q}{C}}$ w.r.t. 'ground' . The energy stored in this configuration is ${\frac {Q^{2}}{2C}}$. Now, connect plate B to A via some resistance. Eventually, plate A and plate B will each have a net charge of Q/2. The total stored energy of this configuration works out to one-half the original configuration - half of the original electric energy was converted to thermal energy in the resistor as the charge redistributed. Alfred Centauri 14:25, 30 May 2005 (UTC)

Yeah I understand that all fine. - Omegatron 15:04, May 30, 2005 (UTC) Hmm... not what I expected. There's more to this than I think. - Omegatron 03:33, May 30, 2005 (UTC)

are you missing energy? Let the resistance mentioned above tend to zero and there seems to be paradox. If the resistance is zero, no electrical energy can be converted to thermal energy. So, where did the missing energy go? The answer is suprising! Alfred Centauri 14:25, 30 May 2005 (UTC)

No, different problem. I understand the resistance of wires just fine.

I know that you do. But that's not the source of the paradox. With near zero resistance, the flow of charge from one plate to the other occurs so quickly that we can no longer ignore the effects of EM radiation. Some of the electrostatic energy is converted to EM waves. Remember spark-gap radios?

Take four neutral conductive spheres, A, B, C, and D. Now connect a source between A and B. Some charge will move from A to B, and now A will have charge Q while B has charge -Q. Now touch A-->C and B-->D. A = Q/2, B = -Q/2, C = Q/2, D = -Q/2. Right? V_AB = V_CD and you should get an equal current when you discharge A to B and C to D.

Although there may be some configuration of spheres for which this is true, it is not true in general. When you charge the capacitor formed by spheres A and B, you do get Q on A and -Q on B. However, it is not necessarily the case that you get the Q/2 charge distribution when you connect A to C and B to D. Remember that the lowest energy configuration for equal and opposite charge concentrations is when the distance is minimized. The amount of charge that moves to sphere C depends entirely on where sphere C is w.r.t. spheres A and B! IMHO, this 4 sphere scenario introduces to much complexity. There is a simple setup that will answer your original question. Alfred Centauri 16:44, 30 May 2005 (UTC)

But now try the same thing with two identical capacitors, one with plates A and B, and the other with plates C and D. No (noticeable) charge is transferred and V_AB= the original source voltage, while V_CD=~0. If any charge is transferred, it is negligible compared to the charge imbalance on AB. I am clearly missing some fundamental capacitor concept. - Omegatron 15:04, May 30, 2005 (UTC)

I'm not sure if you should be thinking of the Earth as the universal reference point. The earth is just a big conductive sphere. It's not necessarily neutral. In my opinion, anyway. - Omegatron 02:30, May 30, 2005 (UTC)

What we are looking for is something that has, in effect, an infinite capacitance - a infinite 'sink' (or source) for charge. The Earth is, for all practical purposes, that object. Thus, it is our best zero reference point. Alfred Centauri 03:08, 30 May 2005 (UTC)

---

Here is a better description of my "condensed charge" question.

In case A, a metal sphere is charged up and brought in contact with another, neutral, identical sphere. The excess charge redistributes itself so that each sphere now has 1/2 the excess charge, right?

In B, the neutral sphere is now 1/2 the mass/volume/whichever is important here. This time only 1/3 of the charge is distributed to the smaller sphere, because there is "less room" for it to fit.

In C, the exact same thing as B, except the smaller sphere has been hammered out into a thin plate. The physical shape of the second object has no effect on how much charge will be transferred, does it?

Hmmm... The lowest energy configuration is for the individual charges to be infinitely far apart, right? Now, take the hammering of the smaller sphere into a thin plate to the limit - create an infinitely thin disk that is infinite in extent. My guess is that all of the charge from the first sphere moves onto the disk. What think? Alfred Centauri 03:34, 30 May 2005 (UTC)

Oh right! The shape does matter! Because they are trying to get as far away from each other as possible. Obviously the sheet can only be spread out to one atom thick, but most of the charge would be on that very wide disk, since it can be farther apart that way.

But what does this revelation do to the difference between C and D, if they have identical plates? - Omegatron 03:51, May 30, 2005 (UTC)

In D, the same exact plate as B, except now a thin insulating plate has been stuck to the metal plate, and another conductive plate placed very close. How much charge gets transferred to the plate now?? - Omegatron 03:14, May 30, 2005 (UTC)

Perhaps this will help. Consider a single metal plate isolated above an infinite conductive plane (a ground plane). The plate and ground plane form a capacitor. Connect the positive terminal of a 1V battery to the plate and the negative terminal to the ground plane. There will be charge Q deposited on the plate. Thus, the capacitance of the plate is Q farads. Now, remove the battery and discharge the plate to the ground plane. Stick the insulator to the plate and place a second, identical plate on the insulator. If we connect the positive terminal of the 1V battery to plate 1, will the amount of charge deposited by the same as before? If so, then the answer to part D of your problem is 'the same amount as in part C'. In effect, we are asking if the presence of the insulator and conductor changes the capacitance of the capacitor formed by plate 1 and the ground plane. It seems to me that the capacitance must change. The presence of the insulator and plate 2 change the electric field around the charged plate 1 which must change the capacitance. Further, it would seem likely that the orientation of the assembly w.r.t. to the ground plane will affect the capacitance. Thus, I think the answer to part D is that the amount of charge transferred is different from the amount transferred in part C. Alfred Centauri 16:44, 30 May 2005 (UTC)

Here is some added detail to my comment about orientation above. Assume plate 1 is initially above and parallel with the ground plane. If you place the insulator and plate2 beneath plate 1, you have inserted two additional dielectric layers between plate 1 and the ground plane (recall that a conductor has in infinitely large dielectric constant). This must act to increase the capacitance so more charge would be stored on plate 1 for the same potential. On the other hand, if the insulator and plate 2 are placed on top of plate 1, I don't see how the capacitance would be effected (ignoring fringing effects) so the amount of charge stored would be the same as without the insulator and plate 2.

Regarding the term 'condensor', I have a hunch that it might be based on the idea that the electric field is 'condensed' into the region between the plates instead of all space as in the case of an isolated object with a potential referenced to the zero at infinity. This has been a fun discussion. Thanks! Alfred Centauri 23:28, 30 May 2005 (UTC)

charge pump

So everyone knows that if you connect a voltage source to a capacitor, the voltage across the cap logarithmically ramps up until it reaches the voltage of the source, and then stops. What happens if you connect a current source? I guess this is equivalent to have an open-circuited ideal current source or a short-circuited ideal voltage source; an impossibility. (What's the difference between a charge pump and a current source?) - Omegatron 00:07, May 28, 2005 (UTC)

Right; you'd have to have an ideal current cource if you wanted to see the capacitor voltage go to infinity. But just like real-world voltage sources don't have a zero-ohms output impedance and can't deliver infinite current, real-world current sources don't have infinite impedance and can't deliver infinite voltage.

Charge pump usually refers to a circuit that places charge on one or more capacitors in parallel, then re-arranges the circuit to place the capacitors in series or reverses their polarity. It's sort of a "transformer for DC". They're used a lot in electronic circuits where you need a small amount of current at a higher-voltage and/or opposite polarity than the ordinary supply voltage. For example, the supply voltage may be +5 volts but the communications circuitry may need a couple of milliamps of +/- 12 volts. A charge pump can be used to produce this power. Charge pumps are related to voltage doublers/voltage multiplier and Cockroft-Walton voltage multipliers.

As you probably realize now, a current source is a different concept.

Atlant 00:18, 28 May 2005 (UTC)

Ok. I knew some of that. Somehow had the idea that charge pumps are the same as current sources, though. I've also seen "charge amplifiers" with capacitive negative feedback. So it seemed to me like voltage, current, and charge circuitry are different things. Maybe that charge is just another word for one of the others... - Omegatron

To be precise, an ideal voltage source is that two-terminal circuit element where the voltage across the element is independent of the current through it. Thus, if you connect an ideal voltage source across an ideal capacitor, the voltage across the capacitor instantaneously becomes the voltage across the voltage source which implies that the source supplied an impulse of current (an infinitely large and infinitely short 'pulse' of current). Obviously, there is no such thing as an ideal voltage source. All real voltage sources are modeled as an ideal voltage source in series with a resistor. So, when a real voltage source with open-circuit voltage V is connected to a capacitor, an RC circuit is formed so that the time for the capacitor to charge to about 0.99V is about $5RC$ seconds. The voltage across the capacitor is an inverse exponential:

$v(t)=V(1-e^{\frac {-t}{RC}})$

On the other hand, an ideal current source is that two-terminal circuit element where the current through the element is independent of the voltage across it. Thus, if you connect an ideal current source across an ideal capacitor, the current through the capacitor is constant so, by $i=C{\frac {dv}{dt}}$, the slope of the voltage across the capacitor is constant. That is, the voltage across the capacitor is a linear function of time which says that the voltage across the capacitor can become arbitrarily large. Once again, there is no such thing as an ideal current source. A real current source is modeled as an ideal current source in parallel with a resistor. So, when a real current source with short-circuit current I is connected to a capacitor, an RC circuit is formed just as is the case with a real voltage source. The voltage across the capacitor is given by:

$v(t)=IR(1-e^{\frac {-t}{RC}})$

Alfred Centauri 14:42, 28 May 2005 (UTC)

That's significant. Didn't know that. Apparently Van de Graaff generators are "current sources" though both sources can be made equivalent to each other through source conversion. What happens if you connect one to a cap? I guess source transformation is the key here. Both are really the same thing. - Omegatron 15:37, May 28, 2005 (UTC)

That's correct. Consider an ideal voltage source with a series resistance equal to (V / x) ohms where V is the source voltage. Let V tend to infinity and, in the limit, you get an ideal current source of x amps. That is, an ideal current source is, loosely speaking, equivalent to an 'infinite' voltage source in series with an 'infinite' resistance. The dual of this is an ideal current source with a parallel conductance equal to (I / x) siemens (or mhos) where I is the source current. Let I tend to infinity and, in the limit, you get an ideal voltage source of x volts. A Van de Graaff generator is a very high voltage source with a very high internal resistance so it can approximate a (weak) ideal current source. If you connected this to, for example, a 1uF capacitor, I imagine that the capacitor would charge rather slowly and quite linearly up to its dielectric breakdown voltage... Alfred Centauri 16:31, 28 May 2005 (UTC)

When you think about how a Van de Graaff generator works, it makes sense that it can be modeled pretty closely as an ideal current source connected to a capacitor. Think about it: We have the electrode at the base injecting a pretty-steady stream of electrons onto the belt. A steady stream of electrons is, of course, the exact definition of a constant current. These electrons are then transported on the belt from which they are dumped onto the metal sphere. The sphere accumulates the electrons and, in doing so, acts as one plate of a capacitor with the surroundings acting as the other plate and the air acting as the dielectric. Electron transport on the belt continues and the voltage on the sphere keeps rising until limited by electron transport off the sphere via corona, spark discharges, or leakage along the column and belt. Atlant 20:04, 28 May 2005 (UTC)

I've just taken a look at the article Van de Graaff generator and there are some problems. The first problem is related to this discussion. From the introductory paragraph:

"The generator can be thought of as a constant-current source connected in series with a very large electrical resistance."

This is just silly. Connect a 1A constant-current source across a resistor of R ohms. The voltage across the resistor is R volts, right? Now, put a 'very large electrical resistance' between the constant-current source and the resistor. The voltage across the resistor doesn't change - it is still R volts. Alfred Centauri 18:25, 29 May 2005 (UTC)

You're right. I changed series to parallel. --Heron 20:04, 29 May 2005 (UTC)

Re-factor to "Electrolytic capacitor"?

DavidCary asks: "Should I move all but a brief summary (of the Electrolytic capacitor section) to Electrolytic capacitor ?"

I'm not a big fan of re-factoring into a lot of smallish articles, but I suspect there's enough meat here to make a reasonable sub-article. So I guess it's okay by me. Let's see what some others think as well...

Atlant 23:58, 1 Jun 2005 (UTC)

I agree. - Omegatron 01:53, Jun 2, 2005 (UTC)

Assuming this gets consensus approval, please be sure to re-write the Electrolytic capacitor so that it includes more than just ordinary aluminum electrolytics; right now it's written as if they were the only electrolytic caps in the world; our own summary here provides better coverage of the varieties.

Atlant 11:36, 2 Jun 2005 (UTC)

Applications

A friend and I are both guitarists and we've been having a discussion about the tone controls on a guitar. It turns out that a bass tone control is just a high-pass filter made of a capacitor and a pot in series while a treble tone control is a low-pass filter with a pot and capacitor in parallel. We understand the what, but don't understand the how. How/why does the capacitor have this effect on a signal? Maybe I'm missing the information in the article, but I don't see how the capacitor has this function. Might be a good addition to the applications or some other section. - MordredKLB 21:40, July 14, 2005 (UTC)

First, you should probably be looking at lowpass filter and highpass filter. A simplistic way to think of it is that a resistor resists all frequencies the same, but a capacitor has more resistance at low frequencies than high. (The lowest frequency; 0 Hz, or DC, can't pass through a capacitor at all.) But why does that happen, you ask? Hmmm... - Omegatron 02:00, July 15, 2005 (UTC)

I understand that and the low/high pass articles were where I got my information originally. Okay if a capacitor has more resistence at low frequencies then why does altering the resistence in series with the capacitor vary the amplitude of the low frequency? When the bass knob (pot) on my guitar is at ten, why does the capacitor have no effect (yes there is some effect just from being in the circuit, but it's minor) on the tone? - MordredKLB 15:51, July 15, 2005 (UTC)

Because it's a voltage divider. The output voltage is the input voltage times the ratio of the two resistances. So if the capacitor is a smaller resistance for high frequencies, the output voltage will be different at those frequencies.

Can you describe your guitar circuit in more detail? Does it look like the images in the filter articles? - Omegatron 16:15, July 15, 2005 (UTC)

Yes the wiring diagrams do look like the images in the filter articles. For the treble control I've got a standard 3-prong pot. The input of the pot is coming from the input of the volume pot. The capacitor is hooked up to variable prong on the pot and the other end is connected to ground. I'm wishing I had paid more attention in my E&M class. Maybe I should just stop worrying about why it works and be content that it does. - MordredKLB 17:21, July 15, 2005 (UTC)

Capacitor- Capacitance Distinction

I agree with Ian Cairns (Dec 2004) that capacitance and capictors should be treated as separate but linked articles. A capacitor is a manufactured component-- capacitance is a physical phenomenon. Does anyone want to do the split?? Al I removed the capacitor networks bit from capacitance anp ut it in to capacitor wher Ithought most people had agreed it should go. But my deletion was reverted by Func who said it was 'blanking'. Can anyone tell me how to legitimately move material from one article to another?? Thanks Light current 05:33, 1 August 2005 (UTC)

It's quite common for vandals to just arbitrarily delete text from articles, so if a bunch of text disappears without coment, many folks will automatically revert it back in.

To avoid this, just make sure you add an appropriate "comment" to the audit trail in the article from which you deleted the text. And if you can't explain your change using an audit-trail "one-liner", use the audit trail comment to refer people to the article's talk page and explain yourself there.

(And if you did explain yourself, then I have no explanation for why you were reverted.)

Atlant 11:47, 1 August 2005 (UTC)

Size of Page

The capacitor article is now reaching its upper size limit. Any suggestions on how to split it??Light current 00:52, 2 August 2005 (UTC)

The limit is arbitrary, and imposed solely for the benefit of ancient browsers that didn't want to edit "text" input fields of greater than 32KB size. Ignore it until the article gets a lot bigger.

Atlant 12:07, 2 August 2005 (UTC)

electrosplitic

Yes, split all of the electrolytic capacitor section except a summary into its own article. - Omegatron 23:26, August 18, 2005 (UTC)

Huh?

C0G and NP0 (negative-positive-zero, i.e. ±0) dielectrics have the lowest losses

What does that mean? - Omegatron 23:26, August 18, 2005 (UTC)

I dont think the statement is necessarily true. It may be true that some C0G/NP0 have low loss but I think for instace an air spaced cap or a vacuum cap woul show even lower loss! Light current 23:42, 18 August 2005 (UTC)

Omegatron, which part of that didn't you understand? (And I'm not being snippy or sarcastic; I'm seriously asking.)

Atlant 23:53, 18 August 2005 (UTC)

Arent all capacitors really only transmission lines?

Consider this: Take a piece of 50 ohm coaxial cable with its far end o/c. Using your 50 ohm o/p impedance pulse generator, apply a pulse of voltage/current to the near end for just long enough so that the leading edge of the pulse reaches the far end of the cable. Disconnect the pulse gen leaving both ends of cable o/c. You now have a capacitor charged to half the o/c voltage of the generator. Right? So the capacitor has charged in the time it takes the pulse to travel to the end of the cable. NB PLEASE SEE MODIFIED CONDITIONS IN FOLLOWING POSTS (by me) BEFORE ATTEMPTING ANSWERS--Light current 13:42, 12 September 2005 (UTC) Questions:

1)If you now looked at either end of the cable with your oscilloscope, you would see only a dc voltage, so what has happened to the pulse of energy travelling at 2/3 c. Why does the pulse wave suddenly stop without reflecting off each open end of the wire continuously? (Does it suddenly stop - WHY?

2) what is the ESR of this capacitor?

3) what is the inductance of this capacitor?

AND the BIG ONE

4) If this tranmission line is a capacitor, are all capacitors transmission lines?

--Light current 00:54, 11 September 2005 (UTC)

No, they are not. How are they similar? User:Omegatron/sig 22:16, September 11, 2005 (UTC)

"You now have a capacitor charged to half the o/c voltage of the generator. Right? So the capacitor has charged in the time it takes the pulse to travel to the end of the cable."

I disagree with this. In the scenario you give above, the transmission line has not settled into steady state before the generator is disconnected. Thus, the voltage across the line is a function of time and position. In fact, for an ideal lossless T-line, the DC voltage at the ends is a square wave with a period of twice the propagation time of the line, a P-P amplitude equal to the open circuit voltage of the generator, and an average value of 1/2 of that voltage. Alfred Centauri 20:31, 11 September 2005 (UTC)

I hadn't considered that one. Do you realise that you've just invented a square wave oscillator!(you may have to keep kicking it - but so what?)--Light current 04:43, 12 September 2005 (UTC)

How is the steady state, as you put it, different from the condition when the pulse has just reached the end of the line. Remember, the pulse is only lon enough to fill the line, no longer.--Light current 20:45, 11 September 2005 (UTC)

In steady state, the voltage and current along the line are constant with time. For the scenario you describe above, steady state will be reached at 2T where T is the propagation time down the line. It takes one T for the initial pulse wave to reach the far end and another 1T for the reflected pulse to reach the source end. Because there is a match at the source end, there is no reflection so the transient is over - the line is in steady state. For t > 2T, the voltage along the line is equal to the open circuit generator voltage and the current in the line is zero. If you disconnect the generator after 2T seconds, nothing happens on the line. There is a charge stored on the line and there is a voltage across the line. The ratio of charge to voltage is then given by the product of the capacitance per unit length with the length of the line. Alfred Centauri 21:44, 11 September 2005 (UTC)

No, I said that the pulse generator was disconnected from the line after the line was filled with energy. So does the line reach a steady state?. What is that steady state? BTW lets make the pulse twice the electrical length of the line, then the 'capacitor' has charged completely in 2T? to the original o/c amplitude of the pulse generator.Yes?--Light current 21:58, 11 September 2005 (UTC)

Sorry AC, I misunderstood your reply initially. You are in fact Correct if t=2T. THere is charge stored and of course there is voltage across the line. But what has happened to the traveling EM Waves? I thought they could not be stopped!!.--Light current 00:08, 12 September 2005 (UTC)

So you are applying a voltage to one end of a transmission line, and then open-circuiting the source end when the pulse reaches the load? User:Omegatron/sig 22:21, September 11, 2005 (UTC)

Well, Originally I was but AC has shown that you need to allow the pulse to return to the sending end before diconnecting the pulse gen. Then you have a beautifully charged capacitor just sitting there. But what has happened to the travelling waves??? Mysterious eh?--Light current 22:38, 11 September 2005 (UTC)

You don't have to disconnect the generator at any given time, although maybe for your capacitor equivalence thing you would have to. Like maybe you could say "under certain conditions during which the generator is removed after x time period and the load is this and the source is that the transmission line will behave similarly to a capacitor".

I think the transmission line article needs some expanding.  :-)

OK but were just having a little fun right now - the serious stuff will come later, believe me!!--Light current 04:20, 12 September 2005 (UTC)

And if you open-source the sending component right as the traveling wave reaches it, the reflected wave will be double the incident wave, since the reflection coefficient is exactly 1, no? In a lossless line the chunk of pulse would just keep reflecting back and forth forever.

Yes! I believe that's what does happen, but you dont see it because the voltage is the same at all points on the line at all times. So the 2 oscillating pulses just add up to a steady dc voltage. Yes?--Light current 04:00, 12 September 2005 (UTC)

In a real, lossy line it tapers off.

Eventually--but since the EM waves are just guided by the conductors maybe you dont get much loss. But hey, its only dc on there isnt it? So what causes the loss mechanism? ;-)Light current 04:00, 12 September 2005 (UTC)

So you're saying that you send a pulse down a line with an open-circuited load and then open-circuit the source right when the pulse reaches it? then your incident pulse from the source would become zero, but the reflected pulse at the source would be 1 in the same direction, leaving the exact same voltage as originally (starts at 0, becomes 1, becomes 1+1, becomes 0+1+1, and will stay at 2 steady state). So the pulses are still there in a sense, just cancelling themselves out to the same value at any point along the line. If you didn't stop the source at exactly the right moment you would still be able to see the timing glitch bouncing back and forth. User:Omegatron/sig 03:30, September 12, 2005 (UTC)

Yep! Good isnt it?--Light current 04:00, 12 September 2005 (UTC) So, can you go on to answer questions 2,3 &4?--Light current 04:07, 12 September 2005 (UTC)

Questions what? User:Omegatron/sig 06:40, September 12, 2005 (UTC)

In my original post, I asked people four questions. You have answered the first question correctly. Can you, or anyone else, now answer the other three?--Light current 12:14, 12 September 2005 (UTC)

But we haven't agreed on your premise yet, LC, that the T-line is a capacitor. We have a length of line with oscillating electric and magnetic fields trapped inside, somewhat analogous to a resonating, lossless LC circuit. The voltage at either end of the line is a square wave (I think's it's a bit more complex in the middle). The electric field inside an open-circuited capacitor is d.c. The line is storing energy in a different way to the capacitor. --Heron 12:59, 12 September 2005 (UTC)

Apply the pulse(or battery via a switch if you like) for twice the delay time (T) of the cable. The front edge of the energy wave (amplitude V volts say) reaches the end of the cable (after time T). This incident wavefront is reflected back toward the source with preserved polarity and adds in voltage to the forward travelling wave. We therefore have a step of voltage (magnitude 2V) travelling backwards toward the source. (Dont forget that the transmission line equations allow for 2 independent oppositely travelling waves to exist on a line). After total time 2T, this step reaches the source end of the cable. All this time, though, the source has still been pumping energy into the line, so by the time the original wave front gets back to the source (2T) the line is completely filled with voltage at 2V. At this point, if the source is now instantaeously disconnected, voltages on the line are indistinguishable from steady dc on a capacitor. There are no oscillations! (PS please read the previous posts by AC and 'O' for more clarification )--Light current 13:38, 12 September 2005 (UTC)

There are still oscillations; they just sum together to equal a uniform voltage. The comparison with a capacitor is very sketchy. The only similarity I see is that you start by connecting a voltage source for a given period of time to a two-conductor object and end up with a uniform voltage across it when disconnected. You could say a battery is a capacitor by the same logic. The period of time in between these two events is not the same at all. The distribution of voltage across the objects is not the same at all. If you disconnect the source at any different time, the end results will be completely different for a capacitor and t-line. If you plot the voltage over time, the two curves are very different, and so on.

Although yes, all objects are transmission lines sort of, if you want to think of them that way.  :-) User:Omegatron/sig 14:24, September 12, 2005 (UTC)

So have you got oscillations or not? They cannot be detected by any instrument you care to think of. So this means that the incoming EM radiation has been perfectly captured by the transmission line. Yes? The comparison with a capacitor is very good actually. In fact it's perfect. There can be no better capacitor (apart from the amount of cable you would need for say 0.1uF). Have you ever seen bits of wire twisted together to form an adjustable capacitor (adjust by snipping bits off). In US its called a 'gimmick' or sucklike. Over here we use the 'posh' term a 'user adjustable cable capacitor'. They can be made with coax as well. The voltage distribution in the transmission line after time 2T is completely even. If you disconnect the source at a time later than 2T, what will happen?. Let's see.... The source has an o/c output voltage of 2V say, and is matched to the line so the amplitude of of the initial pulse is V. When after 2T the pulse has returned, 2V exists all over the line (as you have already seen) and a state of equilibrium exists. At this point you have a source of 2V connected via a resistor to a 'capacitor' (the t-line) charged to 2V. Does it now matter when you disconnect the source?? Also I dont agree with your statement that the results for a capacitor and a t-line will be different. Consider a battery charging a capacitor via a resistor. The voltage across the capacitor will rise logarithmically until it reaches the battery voltage. Do the same thing with a transmission line. The voltage will rise (in steps this time) approximating a logarithmic rise. THe macroscopic effects as shown on an osciloscope are the same. (unless you have avety long cable).--Light current 15:09, 12 September 2005 (UTC)

"There are still oscillations; they just sum together to equal a uniform voltage" I have to disagree 'O'. This business about disconnecting the generator 'just as the reflected wave arrives' is nonsense. Look at it this way, at the 'instant' the reflected wave arrives at the source end, there are only two possibilities - either the generator is there or it isn't. If the generator is there, the source end is matched and there is no reflection period - there are no waves on the line! Or, if you like, there are two waves on the line but the wave fronts have moved off the line so there is nothing to be reflected - the line is in steady state. If the generator isn't there, there is complete reflection at the source and far end so we have wave fronts reflecting off the ends forever if the line is lossless. The voltage along the line oscillates. Alfred Centauri 14:44, 12 September 2005 (UTC)

1. Why is it nonsense?

What does "disconnecting the generator at the instant the reflected wave arrives" mean? As I said there are only two possibilities - either the generator is there or it isn't at the instant the reflected wave arrives. Which one is it? Alfred Centauri 15:02, 12 September 2005 (UTC)

2. If the generator is there, the wave may be reflected varying amounts, depending on the source impedance. It's not necessarily matched.

"Take a piece of 50 ohm coaxial cable with its far end o/c. Using your 50 ohm o/p impedance pulse generator". The generator is matched in this setup you proposed. Did you change that along the way? Alfred Centauri 15:02, 12 September 2005 (UTC)

3. Yes, the generator won't be there at the instant the wave reaches it, which means complete reflection, as you said, and the wave reflects off the ends forever, as you said. But the total voltage when this is happening is uniform. The incident wave from the generator is 0, the reflected wave from the load at the source is V, and the reflected wave from the source at the source is V, leaving 2V everywhere. User:Omegatron/sig 14:52, September 12, 2005 (UTC)

LC, that's wrong and you should know that. If the generator isn't there when the reflected wave arrives, the generator was disconnected before the wave arrives - to say anything else is illogical. Even if it is just an instant before, there is nonetheless a second incident wave generated if the generator is disconnected before the reflected wave arrives. This wave travels down to the end and reflects so the voltage changes with time along the line. Alfred Centauri 15:02, 12 September 2005 (UTC)

AC theres been a bit of an edit conflict, so Im not sure what youre accusing me of being wrong about. Pleas see my last reply to 'O'--Light current 15:31, 12 September 2005 (UTC) ===Light current message for you at=== [2] also see discussion at [3] Scott 00:45:01, 2005-09-12 (UTC)