EM Waves Generated by an LC Circuit?

[LarryS]

TL;DR Summary

Will E and B amplitudes be out of phase?

I am not an electrical or electronic engineer. I am trying to understand how a simple, series LC circuit running at it's resonant frequency can generate EM waves. I believe, based on what I have read, that the frequency of the generated EM waves will be the resonant frequency of the circuit and that the wavelength of those waves is based on the physical size of the circuit and the properties of its two components.

In the circuit, the electrical energy in the capacitor and the magnetic energy in the inductor will be 180 degrees out of phase. Does that mean that the E and B amplitudes of the generated EM wave will also be 180 degrees out of phase?

Thank you in advance.

 

[berkeman]

The EM wave is generated by the current in the antenna or wire, not the voltage that drives the current. Have you learned much about antennas yet?

https://en.wikipedia.org/wiki/Antenna_(radio)

 

[LarryS]

I have read very little about antennas so far. Thanks, I will continue reading.

 

[DaveE]

Summary:: Will E and B amplitudes be out of phase?

Does that mean that the E and B amplitudes of the generated EM wave will also be 180 degrees out of phase?

No. All EM waves propagate with the E-field in phase with the B-field. Although things may be confused near the source where you may actually have multiple sources.

 

[DaveE]

the frequency of the generated EM waves will be the resonant frequency of the circuit and that the wavelength of those waves is based on the physical size of the circuit and the properties of its two components.

This is kind of redundant. The wavelength and the frequency are determined by the speed of light in the material they are propagating through: ν⋅λ=c/n (n is the index of refraction).

 

[LarryS]

No. All EM waves propagate with the E-field in phase with the B-field. Although things may be confused near the source where you may actually have multiple sources.

Are you saying that the E wave and B wave may start out out-of-sync in the "Near Field" but eventually will become in-sync in the "Far Field"?

 

[Baluncore]

Summary:: Will E and B amplitudes be out of phase?

In the circuit, the electrical energy in the capacitor and the magnetic energy in the inductor will be 180 degrees out of phase. Does that mean that the E and B amplitudes of the generated EM wave will also be 180 degrees out of phase?

The E and I in a resonant tuned circuit are in quadrature, at 90 deg. When one is zero, the energy is stored in the other.

The magnetic component of the EM field that propagates through free space was generated by the current in the inductor.

The electric component of the propagating EM field is in phase with the magnetic component because the intrinsic impedance of free space is resistive = 377 ohms.

 

[DaveE]

Are you saying that the E wave and B wave may start out out-of-sync in the "Near Field" but eventually will become in-sync in the "Far Field"?

Not really. I would say that if it looks like they are out of sync, then you are probably seeing multiple waves from different sources.

It's not 100% accurate, but I would think in terms of the EM wave only being generated by moving electrons in the circuit, regardless of the component they are flowing through. So electrons that are flowing east-west will launch a different wave than electrons flowing north-south. Inductors are structured to accentuate the field because of their size and shape. Capacitors are very small with short current paths, so they don't radiate much. Good radiators (antennas, for example) will have a whole bunch of electrons moving the same way (like coils in the inductor) to launch waves with more energy.

However, there are as many ways of thinking about Maxwell's equations as their are readers. There will undoubtedly be others posting with their favorite rules of thumb. This is because none of us want to actually solve the equations precisely.

This subject, going from a circuit, usually with lumped elements, to an EM wave propagating away from the circuit is very complicated. Real world radio designers will usually design an antenna optimized to launch or receive EM waves and assume that the other circuitry doesn't. OTOH, people designing circuits that they don't want to radiate or receive, like everything that isn't a radio, will have to deal with radiation from the circuit with the emphasis on minimizing it. The latter subject is EMC or EMI, where nearly everything is an antenna.

 

[hutchphd]

If you want to spend some quality time on this subject I recommend Feynman's Lectures vol 2 lectures 20-25 about waves, waveguides, resonant cavities, lumped circuit elements and Maxwells eqns. Nontrivial of course.
https://www.feynmanlectures.caltech.edu/II_23.html

 

[sophiecentaur]

No. All EM waves propagate with the E-field in phase with the B-field

Are you saying that the E wave and B wave may start out out-of-sync in the "Near Field" but eventually will become in-sync in the "Far Field"?

To help you understand this, it's necessary to realize that there is a Resistive component in this, which accounts for the Energy lost into free space. The local E and H fields are in quadrature and are very high in value. They fall off rapidly and you are left with the far field which carries the energy and the fields in that wave are cophased

 

[hutchphd]

Are you saying that the E wave and B wave may start out out-of-sync in the "Near Field" but eventually will become in-sync in the "Far Field"?

In the "near field"there is a very complicated pattern of E and B fields and so defining what you mean by a wave and phase is problematic. The part that ends up being self-propagating has the phase relationship required for EM radiation. Far away it is all we can detect.

 

[Baluncore]

Why make it so difficult?
Because it is difficult.

You can think of a resonant parallel LC circuit as a λ/2 dipole. In turn, that dipole can be seen as a λ/2 long transmission line, open circuit at both ends.

When the voltage between the tips of the dipole is at a maximum, the current near the centre of the dipole will be zero. When the current at the centre is at a maximum, the voltage between the tips will be zero.
There is no voltage at the centre of the dipole except for the voltage gradient between the tips. There is no current at the tips of the dipole because the forward current cancels with the reflected current at the open circuit ends of the transmission line.

The quandary that defies a simple explanation is how the 90° quadrature-phase V&I : E&M in the dipole changes to become in-phase in the far field signal.
The one solid bridge is the concept that it is the current in the λ/2 dipole that radiates the magnetic field.

Now hold on tight, you will need a firm grip as the concept of resonant dipole radiation is as difficult to wrap your head around as is the way a transformer works.

Because of the quadrature phase, the resonant circulating energy in an LC circuit or a dipole is imaginary power. That is a parallel concept to the flux in a transformer, a flux that is created by the primary reactive magnetising current.

Now consider that the dipole is actually a resonant transformer. As real energy is radiated by the dipole current, it is replaced by in-phase energy from the transmission line. That is an identical concept to the way a transformer transmits real power through the in-phase components of the primary and secondary currents.

At this point you can view a centre fed dipole as a resonant auto-transformer. A Yagi has a driven element as a primary with several spaced resonant secondaries.

An antenna is simply a transformer that impedance matches the transmission line to the 377 ohms of free space. The circulating energy is just a distraction that makes it work more efficiently.

 

[sophiecentaur]

Now hold on tight, you will need a firm grip as the concept of resonant dipole radiation is as difficult to wrap your head around as is the way a transformer works.

There's always a risk of making people worry too much about resonance and radiation of waves. The radiation can be dealt with in terms of a current distribution over a conductor (by no means trivial) and you can get good answers by assuming a cosine distribution over the wire with a max at the middle. Dipoles are not actually operated at only their resonant frequency and the fractional bandwidth may (depending on your definition) be only a few percent of the design frequency. But you can use a thicker dipole and a fancy matching network and it will work over a whole range of Bands, with the same mechanism causing a wave to be radiated. Matching is more of problem when allowable transmitter power is a factor.

An antenna is simply a transformer that impedance matches the transmission line to the 377 ohms of free space. The circulating energy is just a distraction that makes it work more efficiently.

That's a good black box approach as it takes the magic out of it and makes the whole (magic) process approachable. But you said it's difficult and it is. We are lucky that the seat of pants approach that most people use works adequately in most cases.

 

[tech99]

We need only consider the current in the conductor. This alternates at the frequency of the generator or transmitter. It is helpful to use a capacitor to resonate the system, hence forming an LC circuit, as the transmitter can then see a pure resistance. For sure, there will be energy stored in the resonant system, which will create intense local fields, but this is not germaine to the radiation process. Radiation occurs because we accelerate electrons when we apply a voltage across the ends of the conductor and cause a current to flow. A stationary electron has static electric field lines pointing radially away from itself; when we accelerate the electron, these are swept back, creating a component parallel to the wire. This component is the electric radiation field, and spreads as a wave away from the wire and never comes back. As the electric radiation field passes an observer, they can notice an in-phase magnetic wave as well - this is a consequence of Relativity.
The electric field of the accelerated electron is swept back, so this opposes the motion of the electron, and our generator has to do work to move the electron. This work is noticed by the generator as resistance, called the radiation resistance.
To radiate efficiently, and antenna structure has to be commensurate with the wavelength, so for a very small LC circuit the radiation will be small. The generator will see significant loss resistance but very little radiation resistance.
As far as the phases of the local induction fields are concerned, the magnetic induction field is in phase with the radiated magnetic field, and the electric induction field is 90 degrees ahead of the radiated electric field. The two radiated fields are in phase with each other.

 

[LarryS]

I started this conversation several months ago and now I have a follow-up question.

Assume that our LC circuit under discussion consists of a parallel plate capacitor, a coil inductor and a battery all in series and that, obviously, it is running at its resonant frequency.

Now I’m assuming that the frequency of any EM radiation generated (free space), near or far, will be the resonant frequency of the circuit.

If that is the case, then the matching free space wavelength for that frequency can be calculated from the geometry of the two components alone and is independent of the length of the wires connecting them. This seems to be at odds with the above discussion in which the radiated EM waves are attributed more to the current-carrying wires than to the two components.

Comments?

 

[tech99]

There is no source of AC connected to the circuit, so I presume you are considering the radiation which occurs when the battery is first connected. The circuit wil have a resonant frequency determined by the L and C but influenced also by the connecting wires, depending on length, and we will see a damped wave train which soon dies out. For radiation to occur we require charges to be accelerated, and hence the charges must travel as much distance as possible during each half cycle. For this reason the radiation takes place from the wires, and very little from the L and C, which we presume to be compact.

 

[Baluncore]

If that is the case, then the matching free space wavelength for that frequency can be calculated from the geometry of the two components alone and is independent of the length of the wires connecting them. This seems to be at odds with the above discussion in which the radiated EM waves are attributed more to the current-carrying wires than to the two components.

The connecting wires form a transmission line that has a significant effect on the natural frequency of oscillation. A wire or transmission line can form a resonant circuit without additional lumps of L or C.
https://en.wikipedia.org/wiki/Lecher_line

 

[sophiecentaur]

Now I’m assuming that the frequency of any EM radiation generated (free space), near or far, will be the resonant frequency of the circuit.

If that is the case, then the matching free space wavelength for that frequency can be calculated from the geometry of the two components alone and is independent of the length of the wires connecting them.

This is your problem. The connecting wires are all part of the resonant circuit. Every real electrical circuit has 'parasitic' elements at work. Those wires will contribute to the Impedances and affect the resonant frequency.
When you discuss EM radiation from a circuit, you are letting a bit of reality into the simple circuit. To account for the lost RF power, you have to introduce a resistive element (as well as the resistance the wires) which is referred to as Radiation Resistance. There will be alternating (E and H) fields in close to the circuit which are largely in phase quadrature but, as the distance increases, there will just be in phase components of E and H fields

 

[DaveE]

Assume that our LC circuit under discussion consists of a parallel plate capacitor, a coil inductor and a battery all in series and that, obviously, it is running at its resonant frequency.

Now I’m assuming that the frequency of any EM radiation generated (free space), near or far, will be the resonant frequency of the circuit.

There is no such thing as a lossless LC resonance. So your statement that "obviously, it is running at its resonant frequency" isn't quite right. Transients, which you didn't specify, will die out. Granted, broadband transients, like a step function will excite a greater response at resonance, which will appear to die out slower. So, it's only running at its resonant frequency if it's driven at that frequency. An antenna absolutely can radiate when driven off resonance, in fact that's a common scenario. And yes, the frequency of the radiated waves will match the frequency of the electron motion in the circuit.

If that is the case, then the matching free space wavelength for that frequency can be calculated from the geometry of the two components alone and is independent of the length of the wires connecting them.

As @tech99 said, it's not the geometry of the components, it's the whole circuit, which is mostly wires. One (kind of sloppy) way to think of this is that the L & C components determine how the circuit works, and the wires between them act as the antenna. The resonant frequency of the circuit may not match the most effective antenna size for launching the radiation. I could make a very small circuit that resonates at, let's say, 30MHz but have really low radiation because my antenna, the wires connected to the LC tank, is way too small. This would kind of confirm your suspicion that the wavelength must be comparable to the size of the antenna. 1/4 wave dipoles can work OK, 1/128 wave dipoles don't.

At high frequencies there is the added complexity that the wires themselves have, or are, the L & C, you can have resonant circuits that don't really have isolated "components".

Also, I'm not at all sure about the relationship between the free space wavelength and the geometry of the antenna. My intuition says that the area near/at the antenna doesn't count as free space propagation. But, honestly, EM fields right near antennas are not my specialty, it's pretty complicated and depends on lots of specific details. I'm sure others can discuss this better. Anyway, I wouldn't assume that the wavelength doesn't change or that it's even well defined in that region. Free space propagation is a really useful simplistic benchmark case, it's great for understanding the fundamentals, and a really good model for talking to a spacecraft going to Jupiter, but it isn't always the case in real life.

 

[sophiecentaur]

Also, I'm not at all sure about the relationship between the free space wavelength and the geometry of the antenna. My intuition says that the area near/at the antenna doesn't count as free space propagation.

That's the thing. There will be standing waves in the proximity of any radiator and the wavelength in the near field is not the free space wavelength.
Because the OP has no transmitter in his model, the frequency is set by the natural resonance, which will (of course) determine the wavelength at a large distance but not in close.

 

[anorlunda]

If that is the case, then the matching free space wavelength for that frequency can be calculated from the geometry of the two components alone and is independent of the length of the wires connecting them.

Everyone else is telling you this is wrong. I'll say the same but in different words that may "resonate" with you.

You're using circuit analysis thinking. "The behavior of the circuit depends on the V, and L, and C of components, and does not depend on the length of path of the wires." But one of the fundamental assumptions of circuit analysis is that there are no fields outside of the circuit wires and components. That assumption is necessarily violated in the case of an antenna.

So you're confusing yourself by mixing different mental models.

 

[artis]

Because of the quadrature phase, the resonant circulating energy in an LC circuit or a dipole is imaginary power

Is this also the reason why there is max amplitude at resonance? As all the energy "sloshes" back and forth with minimum losses and opposition/damping?

I guess if you make the LC superconducting it's only loss is radiation.

 

[DaveE]

Is this also the reason why there is max amplitude at resonance? As all the energy "sloshes" back and forth with minimum losses and opposition/damping?

Oops, sorry I delete my reply (below) to edit it. That wipe out your reply too.

Not really, it's not that the losses are greater off resonance. It's just that the circuit doesn't respond as energetically. Like pushing a child on a swing, if you are pushing at the wrong times it's not that the energy is lost, it just isn't stored as well and doesn't build up as it would if you push at the right times.

 

[artis]

Yes I forgot to factor in that resistance (the main loss mechanism besides radiation) in the circuit doesn't change whether at resonance or off it.

 

[DaveE]

Is this also the reason why there is max amplitude at resonance? As all the energy "sloshes" back and forth with minimum losses and opposition/damping?

Anyway, yes your second take was better.

Consider just the capacitor's response first. If you drive it with a constant amplitude variable frequency voltage and measure the current, as you increase the frequency the current increases. But there aren't more losses, that's just how capacitors work. Then do the same with an inductor, and the current decreases at higher frequencies. When you combine them (in series), at some middle frequency you get a maximum. At resonance you also get a special effect where those impedances cancel and you get even more current. But that's just evident in the math of $|Z( \omega )| = |(\frac{1}{j \omega C} + j \omega L)| = \frac{1 - \omega^2 LC}{ \omega C}$. There are no losses (like R) in that equation but at resonance, you can make 0Ω.

It looks like this when you add some losses to it.

 

[sophiecentaur]

Is this also the reason why there is max amplitude at resonance? As all the energy "sloshes" back and forth with minimum losses and opposition/damping?

Hmm. There are two issues here. The maximum voltage swing of an LC circuit when fed from a signal source is because the Reactances of the L and C are equal and opposite. The amplitude is determined by the (real - in both ways) Resistive elements in the circuit.
The frequency of the oscillations when the input signal is removed will be that of the applied signal which may not be the natural frequency of the resonator. Only when the resonator is subject to a step change (on or off from a DC value) will the oscillations be at the natural frequency.

 

[DaveE]

The frequency of the oscillations when the input signal is removed will be that of the applied signal which may not be the natural frequency of the resonator. Only when the resonator is subject to a step change (on or off from a DC value) will the oscillations be at the natural frequency.

Yes, exactly. The key point is that if your excitation is a complex waveform, like a step function, it also contains the resonant frequency. The circuit responds, as it will, to all of the frequencies it's driven with. It's just that the frequencies at resonance make a bigger response, so we see them more easily or we care more about them.

OTOH, "when it's removed" could also be a step function depending on how you do it. The natural (or resonant) frequencies, precisely speaking, are really mostly related to the transient response. Otherwise it's more like a filter with it's own resonant frequency that you are also driving with whatever your frequencies are. Fourier/Laplace transforms are the key thing here.

 

[hutchphd]

The frequency of the oscillations when the input signal is removed will be that of the applied signal which may not be the natural frequency of the resonator. Only when the resonator is subject to a step change (on or off from a DC value) will the oscillations be at the natural frequency.

I would say this a little differently. The removal applied signal incorporates a host of different frequencies: if very abrupt it will look like a step function with basicly $\frac 1 {i\omega}$ Fourier transform. The magnitude of the subsequent oscillation at the free running frequency will depend upon the frequency content of this total applied external signal (in a calculable way) at the free running frequency
Also the free running frequency differs slightly from the resonant frequency if the resonance is not sharp (meaning some appreciable dissipation in the circuit because of radiation or resistance or core eddy currents or whatever takes energy to inaccessible degrees of freedom)

 

[sophiecentaur]

Haha you both sussed me out. A step change in the amplitude of the driving signal will introduce side bands ( it’s amplitude modulation) reduce the level slowly enough can reduce the side bands by an arbitrary amount.
Assume the driving signal is loosely coupled, the drive level can be reduced slowly without significant side bands. The content at natural frequency can be smaller than what a sudden ‘clang’ on or off would cause. I was too simple and PF got me!

 

[DaveE]

Haha you both sussed me out. A step change in the amplitude of the driving signal will introduce side bands ( it’s amplitude modulation) reduce the level slowly enough can reduce the side bands by an arbitrary amount.
Assume the driving signal is loosely coupled, the drive level can be reduced slowly without significant side bands. The content at natural frequency can be smaller than what a sudden ‘clang’ on or off would cause. I was too simple and PF got me!

Yea, not really. We knew you knew that.
I just wanted to stress to others that these linear circuits only respond to what you put in, frequency-wise.
Of course, the problem is, in the real world, things aren't always as linear as we would like. Then it's more complex but, even then, you only get harmonics of what goes in.

 

[LarryS]

Great discussion! Thank you all. My main takeaway is that electrical circuits, LC in particular, are much more complicated than indicated in Wikipedia entries.