tomb@hplsla.HP.COM ( Tom Bruhns) (02/10/88)
RE: Capacitor dielectrics. To do justice to a reply to your questions would take tooooo much net bandwidth -- but there is a fairly easy way to get the info. Look in any of a number of electronics text or reference books. One of my favorites is ITT/Howard Sams' Reference Data for Radio Engineers; its latest incarnation is just called Reference Data for Engineers. Chapter 5, pages 5-16 through 5-28 cover pretty much all the basics you asked about. Each dielectric (and its associated manufacturing technique) yeilds different characteristics in a number of areas, including, but not limited to: leakage, linearity w/applied voltage (read this as: possible distortion in an audio circuit), effective series resistance, self-healing, operating life, storage life, dielectric storage, cost, size, operating/storage temperature range, changes in each of these (but especially in capacitance) with temperature, . . . Each type of capacitor has a practical range of values over which it is competitive with other types. Given several possible choices for a particular value capacitance/voltage rating, arrange your priorities and make a choice! For example, if I want very low dissipation factor in a 1000pF cap, and don't care about operation at high temp, I might well choose polystyrene dielectric. If I want very small size, I might choose multilayer ceramic. . . The electrolyte in electrolytic caps is a conductor which allows effective very close "plate" spacing with attendant high capacitance per unit volume. The electrolyte forms/maintains an oxide coating on a metalic electrode; that coating is actually the dielectric. Again, see a ref. book. BTW, _Reference Data for Engineers: Radio, Electronics, Computer, and Communications_, seventh edition, Howard W. Sams & Co., Indianapolis, IN 46268, ISBN 0-672-21563-2
tycchow@phoenix.Princeton.EDU (Timothy Yi-chung Chow) (05/23/88)
The following problem appears in many books on elementary physics, elementary electromagnetism, and elementary circuit theory: Two capacitors, with capacitances C1 and C2 respectively, carry charges Q1 and Q2 respectively. Find the total energy stored. Now connect the positive plates and connect the negative plates. Show that the total energy stored decreases. Where has the energy gone? The energy calculations are trivial. What intrigues me is the question of the lost energy. The answer usually given is that a current flows when the capacitors are connected, and this causes energy to be dissipated as heat. However, suppose we idealize the problem and assume zero resistance. Surely energy is still lost somehow. How do we explain this? My own tentative explanations include: 1. It is meaningless to "idealize" the problem by assuming zero resistance, because then an infinite current would flow. 2. Charge would move from the plates of the capacitor that was initially at the higher potential to the other capacitor, but would not stop flowing the instant that the voltages equalized. Instead, it would "overshoot" and charge would keep flowing, although it would be slowed down by the opposing voltage. Eventually the current would stop, and then it would reverse. We would get an alternating current. Eventually the energy would get dissipated via electromagnetic radiation emitted by the accelerating charges. (The analogy of a pendulum slowed down by friction seems appropriate.) Any ideas, anyone? -- I hate long .signature files. -- T. Chow
max@trinity.uucp (Max Hauser) (05/23/88)
In article <2992@phoenix.Princeton.EDU> tycchow@phoenix.Princeton.EDU (Timothy Yi-chung Chow) writes: |The following problem appears in many books on elementary physics, |elementary electromagnetism, and elementary circuit theory: ... and, being a cliche, is also guaranteed to appear from time to time on the Usenet and similar media. I wish we could include it in the "answers to frequently asked questions" document and thence relegate it to undergraduate homework problems where it belongs. This being the Usenet -- where people argue to the echo about whether lightning strikes up or down; and argue about skin effect in conductors, breaking into two camps, both completely wrong; do you really think it is the proper forum to ask such a slippery question? Perhaps it was a meta-question, designed actually to generate volume. Stand by for exciting and informed postings. |Suppose we idealize the problem and assume zero resistance. |Surely energy is still lost somehow. How do we explain this? | |My own tentative explanations include: | |1. It is meaningless to "idealize" the problem by assuming zero | resistance, because then an infinite current would flow. No it wouldn't. But this being Usenet, why not start right, with technical gaffes at the very beginning, albeit tentative. |2. Charge would move from the plates of the capacitor that was | initially at the higher potential to the other capacitor, but | would not stop flowing the instant that the voltages equalized. | ...We would get an alternating current. Eventually the energy | would get dissipated via electromagnetic radiation emitted by | the accelerating charges. Ahh, much better. But I look forward to no end of creative alternative suggestions in the days to follow ...
greg@jif.berkeley.edu (Greg) (05/24/88)
In article <2992@phoenix.Princeton.EDU> tycchow@phoenix.Princeton.EDU (Timothy Yi-chung Chow) writes: >Two capacitors, with capacitances C1 and C2 respectively, carry >charges Q1 and Q2 respectively. Find the total energy stored. >Now connect the positive plates and connect the negative plates. >Show that the total energy stored decreases. Where has the >energy gone? It follows from either Maxwell's equations or consistency arguments in special relativity that any electrical system has both non-zero capacitance and non-zero inductance. If there are no resistors, the system is necessarily a collection of harmonic oscillators. If there is (linear) resistance, the system consists of damped harmonic oscillators. A (linear) DC system is one in which the resistances are large enough that the inductances are insignificant. In the above question, if there is non-zero resistance, the energy is lost to heat. If there is no resistance, the DC approximation fails and the system is a harmonic oscillator. -- Greg
leonard@bucket.UUCP (Leonard Erickson) (05/25/88)
In article <2992@phoenix.Princeton.EDU> tycchow@phoenix.Princeton.EDU (Timothy Yi-chung Chow) writes:
<The following problem appears in many books on elementary physics,
<elementary electromagnetism, and elementary circuit theory:
<
<Two capacitors, with capacitances C1 and C2 respectively, carry
<charges Q1 and Q2 respectively. Find the total energy stored.
<Now connect the positive plates and connect the negative plates.
<Show that the total energy stored decreases. Where has the
<energy gone?
<
<The energy calculations are trivial. What intrigues me is the
<question of the lost energy. The answer usually given is that
<a current flows when the capacitors are connected, and this
<causes energy to be dissipated as heat. However, suppose we
<idealize the problem and assume zero resistance. Surely energy
<is still lost somehow. How do we explain this?
The only time I came across this problem was in an old physics book
dug out of a storeroom when I was in high school. The text explicitly
stated that the energy was lost as electromagnetic radiation and
gave formulas for calculating the frequency(!). I lost my copies
of these and many other interesting formulas when my briefcase was
stolen late in my junior year :-(
--
Leonard Erickson ...!tektronix!reed!percival!bucket!leonard
CIS: [70465,203]
"I used to be a hacker. Now I'm a 'microcomputer specialist'.
You know... I'd rather be a hacker."keithe@tekgvs.TEK.COM (Keith Ericson) (05/26/88)
>Where has the energy gone?
Did you see that little flash, and hear the "snap?"
Any more (dumb) question?
Just hesitate to ask...
keithigp@camcon.uucp (Ian Phillipps) (05/27/88)
From article <271@snjsn1.SJ.ATE.SLB.COM>, by chuckc@trojan (Charles Crapuchettes): > In article <2992@phoenix.Princeton.EDU> tycchow@phoenix.Princeton.EDU (Timothy Yi-chung Chow) writes: >>Two capacitors, with capacitances C1 and C2 respectively, carry >>charges Q1 and Q2 respectively. Find the total energy stored. >>Now connect the positive plates and connect the negative plates. >>Show that the total energy stored decreases. Where has the >>energy gone? >>... However, suppose we idealize the problem and assume zero resistance. >> Surely energy is still lost somehow. How do we explain this? Energy IS lost - to electromagnetic radiation. Antenna designers use the idea of "radiation impedance" to allow for this - about 75 ohm for a half-wavelength and (I think) proportional for shorter bits of wire. So, even in the "ideal" case, energy is dissipated. (Not lost, of course!) -- UUCP: ...!ukc!camcon!igp | Cambridge Consultants Ltd | Ian Phillipps or: igp@camcon.uucp | Science Park, Milton Road |----------------- Phone: +44 223 358855 | Cambridge CB4 4DW, England |
strong@tc.fluke.COM (Norm Strong) (05/28/88)
In article <2992@phoenix.Princeton.EDU> tycchow@phoenix.Princeton.EDU (Timothy Yi-chung Chow) writes: }The following problem appears in many books on elementary physics, }elementary electromagnetism, and elementary circuit theory: } }Two capacitors, with capacitances C1 and C2 respectively, carry }charges Q1 and Q2 respectively. Find the total energy stored. }Now connect the positive plates and connect the negative plates. }Show that the total energy stored decreases. Where has the }energy gone? } }The energy calculations are trivial. What intrigues me is the }question of the lost energy. The answer usually given is that }a current flows when the capacitors are connected, and this }causes energy to be dissipated as heat. However, suppose we }idealize the problem and assume zero resistance. Surely energy }is still lost somehow. How do we explain this? } }My own tentative explanations include: } }1. It is meaningless to "idealize" the problem by assuming zero } resistance, because then an infinite current would flow. } }2. Charge would move from the plates of the capacitor that was } initially at the higher potential to the other capacitor, but } would not stop flowing the instant that the voltages equalized. } Instead, it would "overshoot" and charge would keep flowing, } although it would be slowed down by the opposing voltage. } Eventually the current would stop, and then it would reverse. } We would get an alternating current. Eventually the energy } would get dissipated via electromagnetic radiation emitted by } the accelerating charges. (The analogy of a pendulum slowed } down by friction seems appropriate.) } The energy is generally lost in the form of a brilliant flash and a loud bang that scares hell out of you (if the capacitors are fairly large). The heat generated is frequently sufficient to weld the two wires together. If you mange to short the leads quickly enough, current will oscillate back and forth between the capacitors until the damping from the residual resistance dissipates the required energy. (see 2, above) -- Norm (strong@tc.fluke.com)
ankleand@athena.mit.edu (Andy Karanicolas) (05/31/88)
In article <2992@phoenix.Princeton.EDU> tycchow@phoenix.Princeton.EDU (Timothy Yi-chung Chow) writes: >The following problem appears in many books on elementary physics, >elementary electromagnetism, and elementary circuit theory: > >Two capacitors, with capacitances C1 and C2 respectively, carry >charges Q1 and Q2 respectively. Find the total energy stored. >Now connect the positive plates and connect the negative plates. >Show that the total energy stored decreases. Where has the >energy gone? >... >I hate long .signature files. -- T. Chow This problem is a "classic" alright. Many answers are possible, each depending on the circuit model used. It is necessary to look one level higher to see the fundamental problem of asking "Where has the energy gone" Define: u(t) unit step function delta(t) Dirac delta function; unit impulse integral(a, b, whatever-kernel, dx) integral of kernel from a to b over dx Now, v1 = q1/c1 v2 = q2/c2 Since the total charge will be conserved, q1 + q2 = (c1)(v) + (c2)(v) where v is the final voltage. Thus, v1(t) = v1 + [v - v1] u(t) v2(t) = v2 + [v - v2] u(t) with the following convention for constituent relations +i(t) ---->-----[device]------ v(t) = f( i(t) ) + - v(t) i1(t) = c1 dv1/dt i2(t) = c2 dv2/dt Note we will be taking the derivative of a unit step function. Operationally, du(t)/dt = delta(t); as generalized functions, delta(t) and du(t)/dt behave equivalently: integral(-inf, inf, delta(t), dt) = integral(-inf, inf, du(t)/dt, dt) = 1 The function delta(t) has the "sampling" property: integral(-inf, inf, g(t)delta(t), dt) = g(0) as long as g(t) is a "smooth" function for t = 0. As we will see, it is this requirement that g(t) is smooth at t = 0 that is violated when the value of the energy change is asked for. It is very important to realize that delta(t) is not an ordinary function; an ordinary function can be assigned a value or it can be graphed. A common, and treacherous, definition of delta(t) is that it is infinity when t = 0 but it is 0 for any other t. This is convenient for visualizing such functions that are impulsive. However, delta(t) can be consistently defined only in terms of the above integrals and conditions; that is, delta(t) is defined in terms of what it does in these integrals. As will be seen, asking for the energy "lost" violates the definition of the dirac delta function. Taking the time derivatives above, i1(t) = c1 [v - v1] delta(t) i2(t) = c2 [v - v2] delta(t) The energy of the system is the time integral of the power: E = integral(-inf, inf, p1(t), dt) + integral(-inf, inf, p2(t), dt) p1(t) = c1 v1 (v - v1) delta(t) + c1 (v - v1)^2 delta(t) u(t) p2(t) = c2 v2 (v - v2) delta(t) + c2 (v - v2)^2 delta(t) u(t) Now, we see that E = c1 v1 (v - v1) + c2 v2 (v - v2) + [c1 (v - v1)^2 + c2 (v - v2)^2] integral(-inf, inf, delta(t)u(t), dt) The problem is now evident: we are attempting to assign a value to integral(-inf, inf, delta(t)u(t), dt). The unit step u(t) is not a smooth function at t = 0; u(t) is not defined for t = 0. That is, the DETAILS of how the capacitor plates are connected determine what the energy of the system will be. If the plates are suddenly connected in a step function manner ( u(t) ) in a lossless capacitive circuit then the energy E CANNOT be determined uniquely. The energy can be determined if, for instance, there is finite resistance in the circuit or if the switches connecting the plates take finite time to close. The particular model used for the switches will determine the outcome. If Kirchoff's voltage and current laws hold, then the system is electro-quasi-static; that is curl(E) = -dB/dt = 0. In this limiting case, energy is not "lost" to EM radiation. However, if EM radiation is to be worked into the system, then the structure of the capacitor, its dielectric constant and the circuit layout, among many other factors, will, in addition, determine what the energy E will be.
wern@utzoo.uucp (W.Thiel) (05/28/89)
Using capacitors to supply power to Cmos RAM
The Sharp pocket computer Model 1500 with about 8k of RAM
uses 2*220 microfarad to keep memory alive
during battery change.
It is certainly good for several minutes; based
on my experience; it may be as long as 20 minutes.
The measured current consumption on the Sharp Model 1200
is about 2 microampere when turned off. It has a 2.4k memory
This machine runs on one set of batteries for about
two years.
________________________________________________________________________
| Werner Thiel <> uunet!attcan!utzoo!wern wern@zoo.toronto.edu |
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