Talk:Equivalent series resistance
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[edit] ESR in capacitors
How does leakage thro' the dielectric of a capacitor affect its ESR, I wonder??--Light current 04:26, 12 September 2005 (UTC)
- Yes, it does. [1] (see Figure 8) But, the leakage resistance is in megaohms and the ohmic resistance is in milliohms. I am starting to see where the differences over ESR are coming from. When a circuit designer (at least the ones I know) discusses ESR (in a capacitor or inductor), he/she almost exclusively refers to the metallic ohmic resistances, which are small resistances (typ. below 1 Ω). ESR is properly the real component of Z=R+jX of whatever device is being analyzed, and that can be a whole bunch of R L and C's. The distinction between the two uses of ESR probably lies in the application how the cap and coil are used. The metallic resistance are significant in some applications. F'rinstance, the metallic resistance of a coil is important to power supply design, the metallic resistance of a tantalum capacitor is important in certain amplifier designs, because it is actually used as a resistor in the frequency compensation to create a zero. And at the frequencies of these applications, this metallic resistance is the only significant component of ESR. Once the circuit reaches frequencies where the other resistances significantly affect Z, the metallic resistance has insignificant effect on the ESR. It is a rare circuit where all resistive components of ESR are important, so the unimportant elements are very often neglected in the discussion of ESR in a circuit. Snafflekid 17:38, 14 October 2005 (UTC)
Oh yes. Nearly forgot. What are the other factors affecting the ESR?--Light current 04:29, 12 September 2005 (UTC)
[edit] Isolating transformer- is this statement really true??
An isolating transformer uses its ESR to limit current. I have never heard this one before. Does anyone have any references please before I alter this statement??--Light current 04:33, 20 September 2005 (UTC)
- There won't be a reference for that. A standard isolating transformer will have a low ESR to keep output voltage steady with load changes. On short circuit it will burn out unless protected by a fuse. Meggar 05:05, 20 September 2005 (UTC)
Yeah. thats what I thought. but I thought I'd check-- just in case!--Light current 11:48, 20 September 2005 (UTC)
- I suppose that the author may have been thinking about why an isolation transformer is used - to isolate the Earth reference from the chassis so that a fault current to Earth will be very small. So, in this broad sense, you could say that the isolation transformer limits the fault current. This, however has nothing to do with ESR. Alfred Centauri 13:14, 20 September 2005 (UTC)
[edit] Is hysteresis reflected in increases esr for inductors?
I dont know about this one. Does hysteresis of a core affect coils esr?--Light current 21:12, 20 September 2005 (UTC)
Interesting idea but I don't think so. Basically this hysteresis uses a bit of energy each time the direction of the current changes in the coil. Power = energy x freq. so the power demanded by this hysteresis increases with frequency of current direction change. Maybe this power could be somehow represented as a frequency dependent resistance, but it is not to my knowledge. The resistances that are part of the Q factor are all small-signal resistances, and this would be a large-signal resistance, which is definitely not part of DCR. Snafflekid 07:23, 14 October 2005 (UTC)
- after putting my thinking cap on I remembered that the hysteresis is represented by magnetizing inductance, which is in parallel with the coil. The magnetizing inductance requires magnetizing current. So no, hysteresis is not reflected in ESR. Snafflekid 16:53, 14 October 2005 (UTC)
- All practical inductors exhibit losses due to the resistance of the wire or absorption by materials within the magnetic field surrounding it. It is possible to model these losses as a resistance, R, in series with a perfect or loss free inductance L. - From[2] Meggar 01:57, 21 September 2005 (UTC)
THis ref [3] only quotes resistive losses in inductors. It doesnt mention the term ESR--Light current 21:23, 25 September 2005 (UTC)
As far as I know, ESR is not quoted for inductors. What is quoted is the Q. I believe ESR is limited to capacitors- unless someone else knows different!--Light current 06:39, 25 September 2005 (UTC)
[edit] Clean up
Page needs some serious work I feel. Coiuld do with a diagram or two and a bit more detail. Any offers?--Light current 15:07, 23 September 2005 (UTC)
- Adding diagrams and more detail is not "cleanup". Gene Nygaard 15:58, 23 September 2005 (UTC)
Clean up is covered under my mention of 'serious work' & 'more details'--Light current 16:36, 23 September 2005 (UTC)]
- No, it isn't (as I already pointed out in the case of "more details"). Tell me what is currently here and in need of cleanup. If nothing, or nothing that cannot be easily fixed, go find the proper tag to add here and remove the cleanup tag, or I will. Gene Nygaard 17:01, 23 September 2005 (UTC)
If you actually read the cleanup tag: {{cleanup}} you will see that it refers to the article as being of insufficient quality. Its the name of the tag thats wrong, not my usage of it.--Light current 14:23, 25 September 2005 (UTC)
- I don't see any real problem with the tag; cleanup is just one factor affecting quality. Gene Nygaard 19:50, 25 September 2005 (UTC)
[edit] Merge or redirect?
There isnt really enough meat in this subject at the moment to sustain a separate article I feel. Any thoughts?--Light current 06:36, 25 September 2005 (UTC)
Does Meggar have any more details on the application of ESR to circuits in general?. I'm not familiar with this particular use of the term. Could Meggar also say how this measurement is done? As I said I think the term Q covers it for coils-- unless youre talking about dc resistance which is not the same as ESR --Light current 21:21, 25 September 2005 (UTC)
- Yes, the ESR is not the DC resistance. I did put up the entry about hysteresis and eddy current losses adding to the DC resistance. Do you see that small equivilent circuit in dashed lines under the quote in the above link? The one showing an equivilent resistor representing the sum of the various losses in series with an ideal inductor? Both coils and capacitors have ESR and Q ratings, the two are inseperable. A drawing could be added to expand this, but it is futile since there is also a merge tag. If you are intent on merging it there is no point in making it too big to fit back into the capacitor article. How about picking one of these incompatable tags and removing the other? Meggar 04:42, 27 September 2005 (UTC)
I ve looked at the link quite carefully, but I cant see the diagram to which you refer. Could you say exactly where it is please?--Light current 04:59, 27 September 2005 (UTC)
Yes I see it. ESR is not mentioned but Q is. Thats what Ive been saying about coils!--Light current 06:28, 10 October 2005 (UTC)
I recommend that the page be turned into a disambiguation page, because ESR refers to both capacitors and inductors. I know that inductors are not spec'ed in terms of ESR, but it is still correct to speak of the ESR in an inductor, in circuit analysis. A redirect just to capacitor is too simplistic IMO. I can do this, in the free time I have between my 3 chip projects and 2 grad classes, ugh. Snafflekid 22:17, 12 October 2005 (UTC)
No I disagree!. ESR is NEVER referrred to WRT inductors. Give me a quote!.--Light current 02:19, 13 October 2005 (UTC)
- There are some here [5]. ESR isn't a term, only an abbreviation. The full phrase is used for many things. ESR is also a common spec for quartz crystals. Meggar 03:00, 13 October 2005 (UTC)
It seems that ESR is used in the SMPS world to describe the non idealites of inductors. OK. In other circuits however the term Q is exclusively used. If we mention ESRs narrow applicability to electronic power supplies, I suppose I could drop my objection to its use for certain types (not all) of inductor application.--Light current 00:42, 14 October 2005 (UTC)
You are correct (BTW that's what I do, switch-mode power supply ICs). Some background--"Q" of an inductor is important in resonant circuits, because the "R" that is part of the "Q" could be two or three R's, has frequency dependence and does not lend itself to the concept of lumped elements and is not necessarily a series resistance either. "Q" factor is also a function of frequency. Inductors have DC resistance spec'ed because when they saturate, they turn into resistors, or worse fuses, and also to power rate them. DCR is an important spec for SMPS because significant current is in the inductor, Q factor isn't. This DCR value is the value of the ESR element when shown in circuit diagrams. I think this naming convention is historical. Calling it DCR must be intuitive because it's measured with an ohmmeter. Have I beaten this dead horse enough? Snafflekid 06:48, 14 October 2005 (UTC)
Yes I submit. Enough already!--Light current 16:55, 14 October 2005 (UTC)
[edit] Persistent error? about hysteresis and eddy current losses
I posed a question here quite some time ago about how eddy current losses and hysteresis losses manifest themselves in incresed ESr for cored inductors. No reply has been forthcoming. The article still has this statement included even tho' User:Snafflekid has agreed that it is nonsense. Can some one give a quote of how these phenomena are represented in ESR please? Otherwise I will remove offending statement.--Light current 04:51, 18 October 2005 (UTC)
- Snafflekid did not say that. Once he conciders the matter a bit he will recall that both hystereses and eddy current losses (and skin effect) all add to the DC resistance to make for higher equivalent series resistance, as you would yourself if you read the article in that link. Perhaps there is something in the way of background knowledge that I can help with here. Q factor pertains to resonant circuits. As it happens, in the average resonant circuit, the Q of the capacitor will be many times higher than the Q of the inductor. The inductor Q therefore is the determining factor in the design. Both L and C have Qfactors as well as ESRs, Q and R being inversely related at a fixed F. In non-resonant pulse circuits there is no F to use in a calculation, so there is no Q either. In pulse circuits such as power supplies the ESR matters both in C and L. Core losses in the L increase the ESR at high frequencies. The key term here is LOSS, alternating current goes through, gets turned into heat by the losses, kind of equivalent to a resistance!
- I don't know about Snafflekid but I am finding this tedious. Meggar 06:42, 18 October 2005 (UTC)
Why not give an equation relating eddy current and hysteresis loss to ESR then. Then we can get it over with!--Light current 15:17, 18 October 2005 (UTC)
-
- I do believe that eddy current is represented by a resistive lossy element, but I don't think hysteresis is. Though I think that a discussion of either on the ESR page is not very useful. Better on transformer or inductor, IMOSnafflekid 17:17, 18 October 2005 (UTC)
I tend to agree with you so can we alter the text to reflect this?--Light current 18:34, 18 October 2005 (UTC)
- Snafflekid. Would you now kindly arrive at an answer to the question? Does hysteresis in inductor cores represent a loss equivalent to a resistance or not? Here is an equation showing that a loss in Watts applies ( Ph = Kh×f×Bn watts m-3 ) from the reference above. A loss of power of course cannot occur in a reactance. At the same time you will be telling me whether my edits are nonsense as suggested. Light-current will of course feign incomprehension, apparently to string us along as entertainment. I am sure you have better things to do, but there can be nothing less worthwhile than keeping up this conversation indefinitely. I think that after this we will both be more cautious. Meggar 03:47, 21 October 2005 (UTC)
Do I detect a sign of slight impatience here Meggar? (Its a virtue you know!)--Light current 03:54, 21 October 2005 (UTC)
- Actually power can be expended in a reactive element–when it is switching. In a cap, i=C(dv/dt). say you are switching a voltage difference across a cap repeatedly, f=1/dt, you can rewrite this, i = C*Δv*f and since the cap is seeing this voltage, Δv, across it, P = Δv*i = C*Δv^2*f. (It's the main source of power use in CMOS logic, charging and discharging the gate capacitances.) The similar principle applies to inductors, but with switching currents. P= v*Δi = L*Δi^2*f. I wish I knew more about the Kh term, but anyway the equation shows power density as a function of frequency. Excellent link BTW. Quoting the site:
"You can see that the energy required to 'pump up' the core by moving from P1 to P2 is more than that which it returns when going from P2 to P3. This is evocatively termed inelastic behavior. You could look at this another way by thinking about the 'back emf' which opposed the initial increase in coil current. The emf generated is always proportional to the change in flux; but the flux changes less on the 'way down' than it does going up."
- So I think why hysteresis is represented by an inductor in parallel with the ideal inductor is that it represents this opposition to the change in coil current when the coil current changes direction. Any current built up in this inductive element needs to be removed before current will enter the actual inductor, effectively setting a minimum current needed between switching cycles in order to get real work done by the coil. Similarly a magnetizing inductance needs to be satisfied before magnetic flux can become usefully stored in the core, effectively demanding current each cycle. This current per cycle and the voltage across the inductor amounts to power used by the magnetizing inductance.
"In any resistive circuit the power is proportional to the square of the applied voltage. The induced voltage is itself proportional to f×B and so the eddy losses are proportional to f^2B^2."
- So eddy current is a function of frequency as well, so I don't think it is going to be represented as a resistive element. However if the circuit is linearized around an operating point and the frequency of operation is fixed, then the eddy current could be represented as a resistive element. I think this whole subject requires painful excruciating preciseness so that we are always talking about the same thing. Too easy to mean one thing and say another. Snafflekid 01:01, 22 October 2005 (UTC)
[edit] Book quotation
OK heres what Ive got on this from Power Electronics 2nd ed, B.W. Williams, Macmillan, 1992
Because of core losses, a coil can be represented by
- a) a series Ls - Rs cct for inductors
- b) a parallel Rp,Lp cct for a transformer.
Core losses are modelled by including a resistance, and associated losses are accounted for by considering the core mu as a complex variable. mu bar.
Series equ cct of inductor:
- Z = Rs + jwLs = jw(mubar)cN^2 ohms
where mu(bar) = mu' - j mu =Ls/cN^2 - j Rs/wcN^2
For a transformer, the parallel eq cct is
- 1/Z = 1/Rp + 1/(jwLp)
So 1/mu = 1/mu' - 1/(jmu)
- = 1/(Lp/cN^2) - 1/(jRp/wcN^2)
- series and parallel ccts are eqivalent so:
musub s/u'sub s = mu'sub p/musub p = tan delta
where tan delta is the core loss factor
tan delta = Rs/wLs = WLp/Rp = 1/Q
So pick the bones out of that lot!
Notice that ESR is not mentioned but Q and tan delta are!!!
--Light current 08:51, 19 October 2005 (UTC)
[edit] did you know...
1 + 2 is 3 :P