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Capacitors and inductors are duals of each other.

A transformer is made of two inductors, and transfers power through mutual inductance, through the magnetic near-field (right?) Also, you can vary the ratio of voltages or currents by changing the ratio of turns on the core. You could think of this as coupling a single primary loop with many secondary loops, and then stacking the secondary loops so their output voltages are summed.

Is there an electric dual of the transformer? Something that uses capacitance and transfers power through the electric near-field over an isolation barrier? Some way to couple a single primary capacitor with multiple secondary capacitors and then stack them to do power conversion by summing their outputs?

I know that an isolated supply can be built using two capacitors, but I'm not sure if that's exactly a dual, or if there's an equivalent to adjusting the ratio of turns:

enter image description here source

Or maybe something related to this?

alt text source

For example, there are capacitive voltage dividers, but these only reduce voltage, they can't increase it like an autotransformer. There are charge pumps, but those require active elements like switches or diodes, which aren't present in a transformer.

More succinctly: Is there a way to transform power (1 V, 5 A on primary to 5 V, 1 A on secondary) using electric fields instead of magnetic fields, and passive components only? If not, why not? (Electric field screening?)

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    \$\begingroup\$ it would probably require a current made of magnetic monopoles to link the two, somehow. =P \$\endgroup\$ – JustJeff Dec 24 '10 at 21:46
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    \$\begingroup\$ This is a great question. I hope someone can answer this. I am leaning towards "no". At least nothing practical. Kortuk has provided a good response and he surely is correct that a magnetic analog to a transformer/inductor circuit can be constructed, but I am not sure this gets at the fundamental question that endolith is asking. (I have a few bones to pick with some of the points Kortuk makes and hopefully I can rebut them! @Kortuk, I mean no offense. But the magnetic flux does not need to be perpendicular to the wire loop in order for coupling to occur, for one thing.) \$\endgroup\$ – Adam P Apr 1 '11 at 0:55
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    \$\begingroup\$ It occurred to me that my previous comment might not have been totally fair... or clear... The perpendicular component of the B field is what induces emf in a loop of wire. So as long as the field is not exactly parallel to the plane of the loop, there will be coupling. In a transformer, this point is kind of moot since the field is assumed to be perpendicular to each loop in the winding. Also, just a thought: it might be worth asking this question on the physics forum. \$\endgroup\$ – Adam P Apr 1 '11 at 16:10
  • \$\begingroup\$ @Adam: meta.electronics.stackexchange.com/questions/603 :) \$\endgroup\$ – endolith Jun 1 '11 at 14:35
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    \$\begingroup\$ @avishva: That's not a dual. That's just a way to do power conversion using mechanical vibration and piezoelectricity. \$\endgroup\$ – endolith Apr 24 '13 at 13:51

11 Answers 11

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Actually, this is a common thing to wonder.

There is a dual to this. When you have devices sharing a common winding and magnetic flux(Magnetic "current") it makes a perfect dual for devices sharing a common electric conductor. Nice picture from wikipedia:

wikipedia image of flux in a transformer2

You can also take a look at "Magnetic Circuits". You can start learning some fun terms when you go into these concepts in detail, like "Magnetic Capacitance," it seems that my flux does have capacitance.

The way you can determine how much energy passes through a transformer can be broken down into a magnetic circuit that works just like an electric circuit with different units. Magnetic Circuits are an analog of Electric Circuits, which are much easier to work with for many many reasons.

Think of it like a voltage source or a current source. They are direct analogs, but when you build a voltage source it is a hell of a lot easier than a current source.

Side Note

The magnetic flux is shared in a core due to the fact that magnetic flux is perpendicular to the wire, the issue with electric flux is that it points between two surfaces, not looped around. If it looped around a dielectric would get the job done.

In relation to the capacitor inside the other

IF the smaller becomes bigger, it will end up acting like two coupling capacitors with a series resistor between them, as it gets smaller, the overall electric field will be minimal, but you could put a big big E-Field in there, not nearly as effective as a transformer.

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  • \$\begingroup\$ wow, arcane... :) \$\endgroup\$ – Count Zero Feb 26 '12 at 18:47
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I'll start by saying I don't know for sure. However, I'm inclined to say no. Transformers are not "elemental" electrical components. Capacitors and Inductors (and Resistors for that matter) are all fundamental (complex) impedence devices.

A transformer is a composition of two inductors. As you noted, it transforms energy through the principle of magnetic inductance. In particular, it operates on the basis of spatial side effect of current flowing through a coil (i.e. coupling of time varying magnetic field lines). All the "action" in a capacitor is sort of confined to what's going on between the plates, so to speak.

The closest thing I can think of to a dual-analogy to what's going on in a transformer is the idea of capacitative coupling that causes "cross-talk" between adjacent traces in high speed signalling buses...

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yes, you can step down at least, you can uses caps just like a resistive bridge - put two in series say in a 10:1 ratio (10nF and 1nf) across 110v AC and measure the AC voltage across the 10nF - you'll see roughly 11v AC - it's a rather inefficient way to make a lower voltage - but it's a cheap way if you only need a mA or so - but the more energy you need to tap (you need larger caps) the more inefficient it gets (just like a resistive divider)

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    \$\begingroup\$ It is not strictly speaking inefficient. It does not lose energy in the capacitors like a resistive divider would. A high current capacitive divider has large $\cos\varphi$ but that means reactive power and so no heating. \$\endgroup\$ – jpc Apr 1 '11 at 21:06
  • \$\begingroup\$ Is there a way to increase the output voltage, like an autotransformer, or just decrease it? Does the reactance go away when it's loaded, like a transformer with a shorted secondary? \$\endgroup\$ – endolith Jun 1 '11 at 14:19
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A transformer is electric and magnetic. It is not strictly magnetic, so it makes no sense to ask for an electric dual! We can instead ask, what is a transformer-like device in which magnetism and electricity change places. I give you:

enter image description here

A changing magnetic field coming in via the primary core induces current flow in the coil, which induces a changing magnetic field in the secondary core.

Now there is another duality, that between current and voltage. The transformer has no dual in that sense because it actually changes impedance. We could ask, what is a device which treats admittance like impedance (those two being duals). But that is really just the transformer itself, just with an inverted ratio of windings. That is to say, a device which steps up impedance by two and a device which steps up admittance by two are the same transformer, just used in the opposite direction.

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  • \$\begingroup\$ "A transformer is electric and magnetic." How so? You can screen the electric fields and it still works fine. \$\endgroup\$ – endolith Apr 24 '13 at 14:28
  • \$\begingroup\$ @endolith Electric fields are induced in the coils. There is no voltage without an electric field since it is a potential difference between two points in a field. I would be silly to suggest even to the layman that the transformer isn't en electric device. (What are the wires for, and why doesn't it do anything if it isn't plugged in?) If you chop the transformer in half, what do you have? An electro- magnet. \$\endgroup\$ – Kaz Apr 24 '13 at 14:47
  • \$\begingroup\$ No idea why this is downvoted -- sounds like an excellent answer to me. This is a device with a magnetic flux input and output, the ratios of the fluxes can be controlled by the turns ratio, the magnetic circuits are linked by an electric one. It is, in every sense, the dual of a transformer. \$\endgroup\$ – Phil Frost Jun 16 '14 at 13:43
  • \$\begingroup\$ @PhilFrost Finally, some appreciation for this! Thank you. \$\endgroup\$ – Kaz Jun 16 '14 at 13:59
  • \$\begingroup\$ I'd also point out that if one can make the magnetic flux travel in a helix, such as by making a coil out of a ferrite material, then the wire can be a torus and you have something that looks even more like a transformer. \$\endgroup\$ – Phil Frost Jun 16 '14 at 16:36
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Okay, I've been chasing this down in my head for months now. I've built a couple prototypes, as an exercise in understanding the fields involved. I finally have an answer I can believe.

Say you have the original concept, a capacitor inside a capacitor. Compare that to this:

schematic

I would argue that this circuit is identical to our four-plate arrangement. Each of the inner plates of our four-plate stack is still a conductor with a great deal of surface area, and large capacitance to the plates on either side. We've drawn them as two separate plates with no impedance between them, but that electrically changes nothing. Now the circuit looks more familiar. It's really just three capacitors. And the one across the secondary really doesn't add anything, it just creates a voltage divider. You'll get that when you attach a load anyway.

This has some very similar properties to a transformer. DC can't pass from primary to secondary, but AC can. This makes the system galvanically isolated. However, this does not necessarily make it isolated for practical purposes! If you put AC between the primary and secondary of an ideal transformer, nothing happens. If you put AC between the primary and secondary of this circuit, you get lots of current flow. So this would fail an AC hi-pot test, and common-mode noise on one side would transfer happily to the other.

If those aren't problems for an application, there may be some advantages to this over a magnetic transformer. For one, you can transfer more power at higher frequencies, somewhat the inverse of a transformer. (Depending on the transformer, of course.) There are no obscurities of core materials and geometries to deal with. I suspect it's more efficient than a transformer, though I have no data to demonstrate that. Instead of eddy currents, hysteresis losses, and winding losses, all we have is the ESR loss in the capacitors, which I'd expect to be much lower. And it's DC-safe! If you put DC on a transformer, the core saturates and you probably break something. Put DC on this, and absolutely nothing happens.

Now, why can't we step up, if it's truly the dual of a transformer? Because electric fields and magnetic fields have some fundamental asymmetries. An electric field starts on a positive charge and ends on a negative charge. You can't expose a conductor to another conductor's electric field; the electric field of a capacitor definitionally involves two conductors, and if you try to introduce a third, it just moves some of the termination points. (Cartoon version, I am not a physicist.) But a magnetic field always ends where it begins, so a single conductor can have a magnetic field that the secondary can be exposed to with varying geometry.

In other words, it's because electric fields are unipolar, with each end on a separate particle. Magnetic fields are dipolar, starting and ending on opposite poles of the same magnet, forming loops. So amusingly, @JustJeff 's comment was inverted! We really need an electric dipole, not a magnetic monopole!

If a transformer is two conductors sharing a magnetic field, its dual would be two conductors sharing an electric field. In other words, the dual of the transformer is a pair of capacitors.

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  • \$\begingroup\$ If you just want DC isolation between the two sides, you don't need the capacitor in the middle, and it's just ordinary capacitive coupling. \$\endgroup\$ – Phil Frost Jun 16 '14 at 13:37
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    \$\begingroup\$ If you want to get higher voltage, you just make the middle capacitor consisting of parallel connected and isolated multiple plates internally. And then connect those plates in series. N separate plates will multiply the voltage N times. \$\endgroup\$ – hkBattousai Jun 26 '14 at 20:41
  • \$\begingroup\$ @hkBattousai Are you sure? Can you draw a schematic? \$\endgroup\$ – endolith Apr 10 '17 at 18:42
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Yes, there is. It is "Slot coupled waveguide". Though it is not as pure as just 2 coupled capacitors, but it is nearly 100% capacitance based, and involves inherent inductance and magnetic air core and more conductors.

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    \$\begingroup\$ Can you elaborate? In what ways is it similar? In what ways is it different? Google search just finds academic articles. \$\endgroup\$ – endolith Jun 1 '11 at 14:24
  • \$\begingroup\$ Say horizontal slot in the wall is a capacitive (E-Field side of "E-Field transformer"). The top and bottom (ceiling and floor) plates of cavity or waveguide are the "high side" of "E-field transformer". So it is similar to transformer, shown on picture in original question. The question was "is this pattern known as existing device ?". Answer is yes. \$\endgroup\$ – user924 Jun 2 '11 at 14:23
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My idea of a dual is the dipole antenna, or, more broadly speaking, any antennae.

I see the main difficulty in finding a dual in the fact that magnetic field lines are always closed, whereas electric field lines are not. This means that while an inductor on its own is a self-contained system and needs not radiate energy, a capacitor armature will always be 'looking for its pair' and radiate to a greater or lesser extent. Put differently, If you have a wire and inject (high frequency) current, it is very likely that the current will actually be present, even if the circuit is not visibly closed. Just where the return path exactly is, depends on what large piece of conductor you have in its vicinity (e.g. file cabinet, plumbing, etc.). It is possible to define a mutual impedance, very much in the fashion the mutual inductance between coils in a transformer is defined.

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  • \$\begingroup\$ A capacitor's electric field lines are also always closed, no? But I think you're onto something here. The electric field lines end at the surface of the conductors, instead of passing through like magnetic field lines and coils. Transformers link multiple secondary coils with a single primary by passing the same magnetic field lines through all of them, but if you try to confine the electric field so it "passes through" multiple plates, it won't work. Electric fields don't pass through conductive objects. That might be the fundamental difference. \$\endgroup\$ – endolith Mar 19 '12 at 15:45
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    \$\begingroup\$ I was referring actually to something much more mundane: the fact that magnetic field lines need always to be closed is implied by that you can never ever separate a magnetic north pole from a magnetic south pole; a magnet will have them both (if you break one, in the fracture region you will have the opposite pole, again). Meanwhile, a positive and a negative electric charge (between which you have an electric field), can be separated in space. (In fact, if they're not, the total charge might get nullified, if the absolute values of the charges are equal.) \$\endgroup\$ – Count Zero Mar 21 '12 at 23:10
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    \$\begingroup\$ So the electric transformer would require magnetic monopoles? \$\endgroup\$ – endolith Jan 11 '17 at 4:17
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Does William Beaty answer your question in "RIGHT ANGLE CIRCUITRY - or - AC Electronics for Alien Minds" ?

A transformer is often drawn as a coil of copper wire on the left, a coil of copper wire on the right, and a ferrite ring in the middle going through the center of each.

That article suggests the possibility of ferrite "coil"on the left, a ferrite "coil"on the right, and a copper ring in the middle going through the center of each.

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    \$\begingroup\$ That article is pretty crazy... \$\endgroup\$ – Kevin Vermeer Mar 31 '11 at 21:04
  • \$\begingroup\$ That's kind of a dual, I guess, but I'm asking for something that couples electric circuits together using an electric field instead of a magnetic field. \$\endgroup\$ – endolith Aug 15 '11 at 15:13
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The main trouble is that inductors involve "multiply connected space." With purely electrostatic e-fields, if a charge moves from point A to point B, it always traverses the same potential drop, regardless of the crazy path it might take. But with changing b-fields added to the mix, if a charge traveling from A to B should make one, two, three circles which enclose changing magnetic flux, then the charge traverses 1x or 2x or 3x potential drop. When not passing through "simply connected space," the path it takes can affect the accumulated potential. So voltage-wise, if you go around a circle, you never get back to your starting point, and if you go around and around repeatedly, you end up farther and farther from the place you started. (And therefore, if you shove your hand through a coil with a fast-rising current, a FUNDAMENTALLY DIFFERENT HAND comes out the other side!)

If we had "magneticity" conductors full of mobile magnetic monopoles, then a coil wound from such a conductor would be a much better Dual of a conventional coil.

Here's a non-dual dual. Make a capacitor with a very long dielectric, like a PZT rod connecting the two plates. Now bend the rod and spiral it to form a coil. (Or perhaps cast it and then bake to harden.) Apply AC to the capacitor plates you've attached to the ends of the dielectric rod. Rats, it just generates a magnetic field, same as any coil, even though the "coil" is an insulator. Hmm. Not totally wasted though. We could probably connect similar ceramic rods to a neon sign transformer, then jump an arc between their ends. Might not work well at 60Hz, so use one of those 30KHz solid state neon drivers.

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A capacitor configured as a high pass filter, is transferring information (and energy) across the gap using the electric field.

It is worth noting in this context that the usual "capacitor" you put on a board has two poles, but there is nothing necessary about this arrangement. A loose conductor hangin out in space has a (small!) capacitance and is a capacitor.

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  • \$\begingroup\$ But is that actually a dual? You need two capacitors to get isolation, and there's no parallel to changing the voltage/current ratio like changing the turns of a transformer. \$\endgroup\$ – endolith Jun 1 '11 at 14:11
  • \$\begingroup\$ Yes, a loose conductor has a small capacitance -- to the rest of the universe. The other plate of the capacitor is everything else. Still two poles. \$\endgroup\$ – Phil Frost Jun 16 '14 at 13:41
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I don't know whether the three capacitor model above is indeed an analogue of the four plate concept. (Something that I have been puzzling over for the past 5 years or so without the opportunity do do any comprehensive experimental study.)

I would like to propose that the capacitive effect needs to surround the inner plates so as to ensure that the charge on the secondary (C1 in the original 4 plate diagram) is equal to the charge on the primary. This problem with the dual has been pointed out above with a mention of "multiply connected space" in reference to the magnetic coupling of the coils of the transformer. Here we need to have electrostatic coupling. (I'm throwing words around but I hope you get my meaning.)

When this is achieved (presuming the frequency of the supply is high to provide low reactances for the two capacitors) we could say that if Q = CV, and Q1 = Q2 then

C1V1 = C2V2 and you have something that is the dual of the Turns Ration for Transformers.

Inductive transformers, we know, are better at low frequency. The transformation - and transfer of energy through electrostatics - would be better at high frequency as the dual would imply.

As the transformation relies on the constant HF exchange of charge, you could call it a "Flux Capacitor" except that I think that name is taken! :)

My email address is jeffrey.stokes@tafensw.edu.au. I would invite any further discussion on this idea.

A late edit... If you want to step up the voltage, you just have to make the capacitance of the primary much higher than the secondary. Making the Primary the inner pair of plates would be the easiest way as the dielectric distance is naturally greater. If indeed C1V1 = C2V2 as my thought experiments have suggested to me then, in the primary, we would have a higher capacitance and lower voltage. In the Secondary we'd have lower capacitance and higher voltage.

schematic

simulate this circuit – Schematic created using CircuitLab

I have devised an experiment using thin aluminium sheet, plastic film and nylon screws to hold together a practical 4 plate device. Electrical connection will be made at the edge of each plate. I'll be using a 100kHz supply and a 1 kOhm load. I will publish my results here and include images of the waves as well as RMS current in and out. I will them halve the frequency and check the "coupling". Further, I will decrease the capacitance for the outer pair by inserting extra layers of film and determine if that has the effect of increasing the output voltage as I predict it should.

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    \$\begingroup\$ I've been thinking, for the past three minutes, about the three capacitor diagram above. It is definitely not equivalent. It is, in fact, simply three capacitors in series where the one in the middle drops part of the voltage. It is nothing more than a capacitive voltage divider; transforming, in a sense, and achieving isolation, which of course is a principal aim, but only dividing voltage; but never stepping voltage up!It comes down to what wbeaty said above. The dual, expressed as a practical device, needs to be based on equivalences between electromagnetics and electrostatics. \$\endgroup\$ – Jeff Stokes Jan 24 '15 at 4:36

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