Basic question about RLC circuits

[Malamala]

I am really new to this so please let me know if I am doing anything wrong. I have an RLC circuit which contains a capacitor with $C = 0.1$ nF. I need to drive it resonantly at $\omega_0 = 2\pi\times 100$ MHz and I want the voltage across the capacitor to be $V_C = 500$ V. In this case the required voltage that I need to drive the circuit with is $V_0 = V_CZ\omega_0C$, where $Z$ is the impedance. I initially assumed $Z = 50\Omega$, as this is the value used in many applications, but with this I get $V_0 \approx 3V_C = 1500 V$. Is this right? I naively assumed that in a resonant RLC circuit the voltage drop on various elements would be bigger that the applied voltage. The issue is that I can't apply such a high voltage in my setup, so I was wondering if there is anything I can do. I guess the obvious question is: can I change the impedance to a smaller value? And if so, what limits that value (beside the obvious non-zero resistance in the capacitor, inductance and wires)? Thank you!

 

[tech99]

First of all the capacitance value is ridiculous for such a high frequency! However, I assume you have a series RLC combination and you want to obtain 500V across the capacitor. So can you first tell us the resistance value, upon which the obtainable voltage depends. Thank you.

 

[Malamala]

First of all the capacitance value is ridiculous for such a high frequency! However, I assume you have a series RLC combination and you want to obtain 500V across the capacitor. So can you first tell us the resistance value, upon which the obtainable voltage depends. Thank you.

Thank you for the reply. I am not sure I understand the question. I do plan to have a series RLC circuit (unless a parallel one would be better for some reason?), but I don't have any constraints on my circuit other that the capacitor itself. I did the math and the resistance of each of the plates is $0.1 \Omega$. Is there another resistance your are asking about?

 

[Baluncore]

I am really new to this so please let me know if I am doing anything wrong.

You have not given a circuit diagram.
We need to see the RLC circuit and the way you drive the resonance, or couple the input energy.

 

[DaveE]

I highly recommend Khanacademy.org for learning about reactive circuit elements. It sounds to me like you need to study a bit more before you can ask a clear question and understand our answers.

For example, what value of inductance will you need to make 100pF resonate at 100MHz? What is Q, and what is the relationship between Q and the steady state voltage/current in the resonant circuit?

Have you studied calculus yet? That will have a big effect on how we answer you follow up questions.

 

[Malamala]

I highly recommend Khanacademy.org for learning about reactive circuit elements. It sounds to me like you need to study a bit more before you can ask a clear question and understand our answers.

For example, what value of inductance will you need to make 100pF resonate at 100MHz? What is Q, and what is the relationship between Q and the steady state voltage/current in the resonant circuit?

Have you studied calculus yet? That will have a big effect on how we answer you follow up questions.

Sorry I am a bit confused. I did read about RLC circuits (at least some basic stuff), but now these questions make me doubt myself. For reference the capacitance in my circuit are 2 plates with an area of $100$ cm$^2$ and located $500$ microns apart. This is done by coating a thin layer of metal on microfabricated silica glass (I don't think this is relevant to my question but I can go into more details if needed). In order to have a resonant circuit I need an inductance of $\sim 10^{-8}$ H. Based on this it seems like a coil with with 10 turn, 5 mm diameter and 10 cm length would do that (that seems like a reasonable size, but again I don't have practical experience with it, so please let me know if this doesn't make sense).

Now, if we are on resonance, the impedance is only given by the resistance. My question is, is there any reason to not make that as small as possible, such that the require applied voltage is reduced compared to a 50 $\Omega$ impedance.

But I am not sure if I am missing some deeper issues, based on your questions.

For reference the source will be similar to Fig. 6.6 in this PhD thesis (I won't need a $\lambda/2$ resonator, just the $\lambda/4$). And yes I do have experience with calculus (I am doing a PhD in the US, I just never worked in practice with RLC circuits and the project is also out of the expertise of my Professor).

EDIT: Just for reference, in my experiment I have some atoms trapped in an optical dipole trap (OPT) in a volume with a diameter of $\sim 50$ microns. What I need is to have them experience a very uniform electric field, with an amplitude around 10 kV/cm and oscillating at $100$ MHz. Using a capacitor as described above was the first idea that came to my mind, but if there is an easier way, I'd also be happy to look into other approaches too (my group is an AMO group but we don't really have experience with RLC circuits especially in such high frequency regime).

 

[Baluncore]

For reference the source will be similar to Fig. 6.6 in this PhD thesis (I won't need a λ/2 resonator, just the λ/4 ).

Re: Figure 6.6: "The rf setup consists of a coaxial λ/4 electrical resonator that produces a large voltage for the rf trap electrodes. A portion of this resonator is inside the vacuum system.

The resonant LC circuit that provides the intense electric field for the trap is not made from discrete inductor and capacitor components, but is implemented as a coaxial resonator. That resonator looks electrically like a coaxial cable transmission line, shorted at one end, open at the other. RF energy is injected through a coupling loop at the shorted-end. The open-end of the central conductor forms one small capacitor plate of the trap, while the other capacitor plate is a wall across the open end of the resonator. Coaxial resonators, are often tuned by screw adjustment of the end capacitance between the inner and outer coaxial electrodes. That is where your ion trap should be placed.

An adjustable disc capacitor is shown in this example, along with input and output coupling links. Your coaxial resonator will only need the input link. Your ion trap will be located inside what is shown here as the disc capacitor.
https://www.changpuak.ch/electronics/Coaxial_Tank_VHF_Filter_Designer.php

 

[tech99]

Thank you for the reply. I am not sure I understand the question. I do plan to have a series RLC circuit (unless a parallel one would be better for some reason?), but I don't have any constraints on my circuit other that the capacitor itself. I did the math and the resistance of each of the plates is $0.1 \Omega$. Is there another resistance your are asking about?

If you have a resonant circuit consisting of an L, R and C in series, I will tell you how to find the voltage across the capacitor. You say the capacitor is 0.1nF (100pF) so first calculate the reactance of this at 100 MHz. X=1/2 pi f C, which gives 16 Ohms. To obtain 500 V across this reactance, we apply Ohms Law and get V=16 x I, so the current must be I=500/16 = 31.25 Amps. When we connect L and C in series, they have a total reactance of zero, so the only limitation of current in the LCR circuit is the resistance. If we assume 1 Ohm, for instance, you will see that to obtain 31.25 Amps we will require 31.25 Volts.
When analysing a series LCR circuit, a good way is to assume a current of 1 Amp and then work out the voltage across each component. You can also work out the total voltage, and perhaps the easiest way is to draw a phasor diagram.

 

[Malamala]

Re: Figure 6.6: "The rf setup consists of a coaxial λ/4 electrical resonator that produces a large voltage for the rf trap electrodes. A portion of this resonator is inside the vacuum system.

The resonant LC circuit that provides the intense electric field for the trap is not made from discrete inductor and capacitor components, but is implemented as a coaxial resonator. That resonator looks electrically like a coaxial cable transmission line, shorted at one end, open at the other. RF energy is injected through a coupling loop at the shorted-end. The open-end of the central conductor forms one small capacitor plate of the trap, while the other capacitor plate is a wall across the open end of the resonator. Coaxial resonators, are often tuned by screw adjustment of the end capacitance between the inner and outer coaxial electrodes. That is where your ion trap should be placed.

An adjustable disc capacitor is shown in this example, along with input and output coupling links. Your coaxial resonator will only need the input link. Your ion trap will be located inside what is shown here as the disc capacitor.
https://www.changpuak.ch/electronics/Coaxial_Tank_VHF_Filter_Designer.php

Thank you so much for this! This is extremely helpful. I have a few more questions (please let me know if there is any paper/book that would help answer this kind of questions; I haven't found much useful stuff online).

1. I am not sure I understand the connection from outside to inside. Is the ground of the SMA isolated from the outer wall of the cylinder (it kind looks soldered to it)? And is the central SMA wire of both SMA connectors soldered to the bottom of the cylinder? Also are the 2 capacitor plates soldered?

2. How can I think of the flow of current in this setup? I assume the two plates would act still as a normal capacitor (i.e. shown in textbook examples), but I am not sure how to imagine the flow of current. In my case I need to account for any possible magnetic fields perpendicular to the plane of the 2 electrodes. In the ideal case such a magnetic field would be zero, but I would like to estimate the effect for possible tilts, and thus knowing the currents around is needed.

3. Is there an easy way to estimate the impedance (even assuming ideal conditions)? Again, I would need this for current estimates.

4. In my case (and I saw the same in the ion trapping community) the electrodes are painted on a silica glass, such that the 2 capacitor electrodes are very close to each other and as parallel as possible. I am not sure how this is done here, as I can't attach the glass to the inner rod. Can I just have two wires coming from the rod and outer wall connected to my actual electrode plates (I can perhaps make a holder for the glass structure out of non-conductive materials), instead of connecting the plates directly?

 

[Baluncore]

2. How can I think of the flow of current in this setup?

You are starting a fascinating journey. To begin, you will need to understand short λ/4 transmission lines, and resonant λ/2 dipole antennas.

Two equal and opposite waves of current flow. One is on the outside of the inner-conductor, the other is on the inside of the outer-conductor. That is a closed chamber. The currents flowing on the outside of the outer-conductor are external and unrelated to the internal currents.

The short-circuit end is a plane of mirror symmetry, so what is at the open end of the resonant line, (your ion trap), is seen as a negative image of the trap, λ/2 away from the real trap. You can then see the inner conductor as being a resonant λ/2 dipole antenna. You will couple energy into that transmission line, to excite a standing wave on the virtual dipole.

The coupling loop is very close to the virtual centre of the dipole, where the dipole has a voltage null, but a current node. The SMA is a virtual wormhole between the outer world, and the inner resonator. The SMA connector screen is connected to the outer conductor, inside and out, at that point. The signal from the SMA is connected inside, through a loop, to the mirror. That loop is the primary of a transformer, with the secondary being the inner conductor of the resonator. Notice how the small loop is also imaged in the mirror, so it feeds the resonator current, at the voltage null of the dipole. The SMA is 50 ohm impedance, so the loop will match the inner conductor at a 50 ohm point, close to the centre of the virtual dipole. The impedance at the open end of the dipole will be virtually infinite, so it will be all electric, no magnetic.

The ion trap will be connected between the end of the dipole, and the conductive end wall of the resonator, the end plate. The ion trap is therefore at the current null of the resonator, where the voltage node is at a maximum. Since the standing wave current is zero at the tip, there is almost no magnetic field near or in the ion trap.

You can save several weeks in the laboratory now, by studying transmission lines and standing waves, for one day in the library. Where you start will depend on your background in studying electromagnetic waves.

 

[Malamala]

You are starting a fascinating journey. To begin, you will need to understand short λ/4 transmission lines, and resonant λ/2 dipole antennas.

Two equal and opposite waves of current flow. One is on the outside of the inner-conductor, the other is on the inside of the outer-conductor. That is a closed chamber. The currents flowing on the outside of the outer-conductor are external and unrelated to the internal currents.

The short-circuit end is a plane of mirror symmetry, so what is at the open end of the resonant line, (your ion trap), is seen as a negative image of the trap, λ/2 away from the real trap. You can then see the inner conductor as being a resonant λ/2 dipole antenna. You will couple energy into that transmission line, to excite a standing wave on the virtual dipole.

The coupling loop is very close to the virtual centre of the dipole, where the dipole has a voltage null, but a current node. The SMA is a virtual wormhole between the outer world, and the inner resonator. The SMA connector screen is connected to the outer conductor, inside and out, at that point. The signal from the SMA is connected inside, through a loop, to the mirror. That loop is the primary of a transformer, with the secondary being the inner conductor of the resonator. Notice how the small loop is also imaged in the mirror, so it feeds the resonator current, at the voltage null of the dipole. The SMA is 50 ohm impedance, so the loop will match the inner conductor at a 50 ohm point, close to the centre of the virtual dipole. The impedance at the open end of the dipole will be virtually infinite, so it will be all electric, no magnetic.

The ion trap will be connected between the end of the dipole, and the conductive end wall of the resonator, the end plate. The ion trap is therefore at the current null of the resonator, where the voltage node is at a maximum. Since the standing wave current is zero at the tip, there is almost no magnetic field near or in the ion trap.

You can save several weeks in the laboratory now, by studying transmission lines and standing waves, for one day in the library. Where you start will depend on your background in studying electromagnetic waves.

Thank you so much for this! It's a lot more complicated than initially thought, but at least I have a place to start! However, I am afraid that given the setup, this won't work to me. For my experiment, beside the large electric field, I also need:

1. The ability to apply a static magnetic field at the location of the atoms (i.e. in between the 2 electrodes). I usually do this by using two coils of current around the setup, but I am not sure how the magnetic field will interact with the external shield (it might shield the field lines?). Do you think static magnetic fields applied on the outside would be able to reach in (I only need few Gauss)?

2. I need laser access, including a hole to bring the atoms in and out. These holes will need to be $0.5-1$ cm in diameter (probably at least 6 of them, as I need at least 3 laser beams). Would this significantly affect the setup? I expect that the resonant frequency will change, but I can adjust for that. Is there anything else that might be significantly affected?

 

[Baluncore]

Do you think static magnetic fields applied on the outside would be able to reach in (I only need few Gauss)?

A static magnetic field could be generated by a Helmholtz coil arrangement, with one coil on each side of the ion trap.

These holes will need to be 0.5 - 1 cm in diameter (probably at least 6 of them, as I need at least 3 laser beams). Would this significantly affect the setup?

You will need to design the equipment. I have only considered generation of the RF electric field. That field is available behind the plane of the ion trap. Access to the ion trap will be from outside the resonator, inside the vacuum chamber.

I expect that the resonant frequency will change, but I can adjust for that.

The resonator can be physically adjusted in length, to a frequency just above the operating frequency. The centre frequency of the resonator, can then be fine-tuned, by screw adjusting a small capacitor plate, located between the inner and outer conductors. That increased capacitance will lower the frequency of the resonator.

You need to get more detailed information about how other experimenters have assembled similar ion traps.

 

[Malamala]

A static magnetic field could be generated by a Helmholtz coil arrangement, with one coil on each side of the ion trap.


You will need to design the equipment. I have only considered generation of the RF electric field. That field is available behind the plane of the ion trap. Access to the ion trap will be from outside the resonator, inside the vacuum chamber.


The resonator can be physically adjusted in length, to a frequency just above the operating frequency. The centre frequency of the resonator, can then be fine-tuned, by screw adjusting a small capacitor plate, located between the inner and outer conductors. That increased capacitance will lower the frequency of the resonator.

You need to get more detailed information about how other experimenters have assembled similar ion traps.

Thank you for the reply. I am actually having still issues understanding the need for 2 SMA connections... In the PhD thesis I mentioned earlier (https://www.nist.gov/system/files/documents/2017/05/09/blakestad2010thesis.pdf) in figure 6.6 it appears like they have only one cable connected to the outer wall of the resonator (which I assumed it would be one SMA connector). Am I missunderstanding their diagram? Where are the 2 SMA's (or their equivalent) in that figure?

Basically, based on my still very very basic knowledge of RF, I thought that Fig. 6.6 means that you have a coaxial cable (SMA for example) going from the source (HP 8640) to the RF amplifier, then another SMA from the RF amplifier to the quarter wave resonator (I am ignoring the other elements in the figure for now). I am not even sure where a second SMA would by physically connected to.

 

[Baluncore]

I am not even sure where a second SMA would by physically connected to.

There will be no second SMA or coupling loop on your resonator.

I am actually having still issues understanding the need for 2 SMA connections...

Coaxial resonators are usually employed by radio engineers to select narrow-band signals from the RF spectrum. That is why they have two 50 ohm SMA couplers, one for input and one for output. The example I linked comes from a HAM radio experimenters site, inspired by an article in the amateur magazine QST, from October 1964.

The RF drive to your resonator, is coupled in through the one SMA connector. No low-impedance output is required, as the high-impedance, high-voltage, RF electric field, is available at the open end of the resonator, where your ion trap will be located.

 

[Malamala]

There will be no second SMA or coupling loop on your resonator.

Coaxial resonators are usually employed by radio engineers to select narrow-band signals from the RF spectrum. That is why they have two 50 ohm SMA couplers, one for input and one for output. The example I linked comes from a HAM radio experimenters site, inspired by an article in the amateur magazine QST, from October 1964.

The RF drive to your resonator, is coupled in through the one SMA connector. No low-impedance output is required, as the high-impedance, high-voltage, RF electric field, is available at the open end of the resonator, where your ion trap will be located.

Thanks a lot! That makes more sense. I am now trying to do some basic simulations in Ansys to understand everything a bit better visually (I found our university provides his software, but please let me know if there is anything better). Does the location of the SMA port matters (I assume it will affect the resonant frequency, which can be latter adjusted). Also, for the simulation, I found this on youtube:

which looks similar to what I have (well they have 2 SMA connectors), but in that setup the central wire of the SMA doesn't physically touches the inner part of the outer cylinder (still trying how I'd do that in the simulation, as I'd need a curved shaped cylinder). It comes in as an antenna only. Is that an option? Doesn't the wire need to be in contact with the wall? Sorry for asking so many questions...

 

[Baluncore]

Does the location of the SMA port matters (I assume it will affect the resonant frequency, which can be latter adjusted).

Not so much the resonant frequency, but the input impedance and the degree of coupling. For high voltages you want high resonator Q, so you need only a small coupling. The resonator is then not circulating energy back to the RF PA.
Since the resonator is only lightly coupled, the resonant frequency is independent of the cable and connector, which are driven by the RF PA.

... , but in that setup the central wire of the SMA doesn't physically touches the inner part of the outer cylinder (still trying how I'd do that in the simulation, as I'd need a curved shaped cylinder). It comes in as an antenna only. Is that an option?

Yes. It is probably a virtual capacitive coupling, between the wire antenna and the inner conductor of the resonator. It may be seen as a quarter wave whip antenna, above the ground-plane of the outer wall or shorted end.

The PA will be impedance matched to that small coupler antenna. Most losses will be in the coaxial cable. By matching the coupler loop or capacitive patch to the 50 ohm cable, there is minimum reflected energy in the cable, so minimum cable loss. The PA will then be happy, since the PA load will look like a 50 ohm cable. You will probably need to tickle the coupler style and size, to optimise the output voltage from your resonator to the ion trap.

 

[Baluncore]

Sorry for asking so many questions...

You have been metaphorically thrown in at the deep end, where the learning curve is an overhang.

Ask more good questions, that is what we are here for. With experience, you will get better at asking questions, until your questions are so good, that you will be able to answer them yourself.

“The only interesting answers are those which destroy the questions.” ― Susan Sontag.

 

[Malamala]

You have been metaphorically thrown in at the deep end, where the learning curve is an overhang.

Ask more good questions, that is what we are here for. With experience, you will get better at asking questions, until your questions are so good, that you will be able to answer them yourself.

“The only interesting answers are those which destroy the questions.” ― Susan Sontag.

Hello! I think I have a better understanding of what is going on now. I was able to simulate the setup I am interested in (I am using Ansys) and it behaves as expected. One thing I wanted to ask which is not totally clear to me (but based on what I found online, the behavior is expected) is the following. I am simulating a quarter wave resonator (I attached some figures of the simulated setup) where the signal is injected using a loop of current (at the bottom right of the attached figure). Normally, the length of the inner resonator should be $\lambda/4$. However, in my case, the end of the inner resonator is very close (in my case 0.5 mm) from another metal plate (the atoms will be between these 2 plates, experiencing a large oscillating electric field). If the distance between these 2 plates is large, the length needed is pretty close to $\lambda/4$. However, in my case, (at a 0.5 mm spacing), the required length to achieve a given resonant frequency is much lower (in my case around 3 times smaller). Why is this happening? I assume it has something to do with the fact that I create some huge capacitance between these 2 plates? But I am not really able to write down some sort of effective circuit that would explain this significant length reduction. Could you please help me understand this effect better?

 

[Baluncore]

However, in my case, (at a 0.5 mm spacing), the required length to achieve a given resonant frequency is much lower (in my case around 3 times smaller).

It is almost certainly that close spacing, at the open-end, that is lowering the frequency, by increasing the end-capacitance.

If you have a dielectric material inside the resonator, then it will have a lower frequency, as the transmission line has a lower velocity factor; vf ∝ 1/√Er. Another way to lower the frequency, is to place a lump of capacitance at the open end. Radio engineers will have two signal ports at the shorted end, then use the open-end capacitance to tune the resonator. But you must reduce the end capacitance, to that of the ion trap capacitance, by keeping the central conductor smaller in diameter than the typical radio filter example. You will need to shorten the resonator, but certainly not by a factor of three. If the central conductor is made too thin, it will have higher surface resistance, so the Q of the resonator will be lowered.

I assume you are designing for air or vacuum inside the resonator. Where the resonator penetrates the wall of the vacuum chamber, there must be a glass, or maybe polystyrene wall, somewhere. It is important that the presence of that dielectric does not mismatch the impedance of the resonator line, or greatly magnify the end capacitance.

 

[Malamala]

It is almost certainly that close spacing, at the open-end, that is lowering the frequency, by increasing the end-capacitance.

If you have a dielectric material inside the resonator, then it will have a lower frequency, as the transmission line has a lower velocity factor; vf ∝ 1/√Er. Another way to lower the frequency, is to place a lump of capacitance at the open end. Radio engineers will have two signal ports at the shorted end, then use the open-end capacitance to tune the resonator. But you must reduce the end capacitance, to that of the ion trap capacitance, by keeping the central conductor smaller in diameter than the typical radio filter example. You will need to shorten the resonator, but certainly not by a factor of three. If the central conductor is made too thin, it will have higher surface resistance, so the Q of the resonator will be lowered.

I assume you are designing for air or vacuum inside the resonator. Where the resonator penetrates the wall of the vacuum chamber, there must be a glass, or maybe polystyrene wall, somewhere. It is important that the presence of that dielectric does not mismatch the impedance of the resonator line, or greatly magnify the end capacitance.

Thank you for this! Are you saying that the factor of 3 I am getting is a mistake? Or that I can reduce the frequency without having to reduce the size by a factor of 3, if I want to (actually a smaller size is better for me, so the factor of 3 is great, actually, for my setup)?

I am not sure I understand the last part. I was planning to put everything inside the vacuum chamber and I thought that the inner rod would be soldered (or something similar) to the bottom of the outer wall. What is this polystyrene wall needed for? Also, when you say "resonator", do you mean the inner rod only, or the whole setup (inner rod + outer wall and the vacuum in between)?

 

[Baluncore]

Are you saying that the factor of 3 I am getting is a mistake?

It is a big change, so I am concerned that the Q will be low, or the centre frequency will be highly dependent on the small separation, so very temperature and pressure sensitive.

I would expect the resonator to penetrate the wall of the vacuum chamber, so the RF connection, coupling, and possibly the tuning could be done externally, from behind.

What is this polystyrene wall needed for?

It is the lowest-loss common material that can be used to make capacitors or mountings for high-voltage, high-Q circuits.

In your case, the open end of the resonator needs to be held in position, so the ion trap can be mounted.

You will also need to fine tune the resonator by sliding a dielectric sleeve inside, or screw adjusting a capacitive patch, near the open-end of the resonator. Maybe you can do that through an access window.

 

[Malamala]

It is a big change, so I am concerned that the Q will be low, or the centre frequency will be highly dependent on the small separation, so very temperature and pressure sensitive.

I would expect the resonator to penetrate the wall of the vacuum chamber, so the RF connection, coupling, and possibly the tuning could be done externally, from behind.


It is the lowest-loss common material that can be used to make capacitors or mountings for high-voltage, high-Q circuits.

In your case, the open end of the resonator needs to be held in position, so the ion trap can be mounted.

You will also need to fine tune the resonator by sliding a dielectric sleeve inside, or screw adjusting a capacitive patch, near the open-end of the resonator. Maybe you can do that through an access window.

Thank you for the clarifications. Sorry for not providing more details about the simulated setup. The resonance I am aiming for is ~35 MHz, corresponding to $\lambda/4 \approx 2.1$ m. The size of my setup in the simulation, reaching a resonance at 35 MHz is ~ 0.65 m, so 3.2 times smaller (it seems I can make it even smaller by adjusting the diameter of the rod and/or the diameter of the outer wall). I did not check the Q-factor, but in the second picture above, based on the simulation, the field between the plates is ~10 kV/cm (for 1 W of input power), which is what I need for my experiment. I don't care too much about the Q value itself, I just need to reach 10 kV/cm electric field and I assumed that 1 W of input power is reasonable (I saw papers, including the PhD thesis I shared before, that go up to 15 W). Am I missing something else here? I will look into the variation of the resonant frequency and Q with the gap between the 2 plates, but again, from literature, people claim they can have that distance controlled at the ~ 1-2 micron level.

When you say "I would expect the resonator to penetrate the wall of the vacuum chamber", do you mean the outer wall, or the inner rod? I imagined that the inner rod is completely encased in the outer wall, then everything is placed inside a vacuum chamber (I will need a way to support the outer resonator wall on the chamber walls, but that won't affect what is happening inside i.e. that won't affect the resonant frequency or the Q factor) and use a vacuum feedthrough to send the RF signal from air to vacuum (i.e. I thought everything in the figure above would be fully placed inside vacuum).

 

[Baluncore]

When you say "I would expect the resonator to penetrate the wall of the vacuum chamber", do you mean the outer wall, or the inner rod?

I mean the outer tube.

Now that you inform me the frequency is close to 35 MHz, not hundreds of MHz, I would expect the resonator to be made from a helical coil with a capacitor.
There are two ways to implement such a resonator:

1. As a lumped LC circuit, a tank circuit, enclosed, external to the vacuum chamber, with a short, (in wavelengths), high-voltage coaxial cable to the ion trap. The RF input coupler would be a tap onto the first turn of the tank coil.
https://en.wikipedia.org/wiki/LC_circuit

2. As a transmission line, a "helical resonator", which is a shortened version of the cavity resonator we have been discussing.
https://en.wikipedia.org/wiki/Helical_resonator

 

[Malamala]

I mean the outer tube.

Now that you inform me the frequency is close to 35 MHz, not hundreds of MHz, I would expect the resonator to be made from a helical coil with a capacitor.
There are two ways to implement such a resonator:

1. As a lumped LC circuit, a tank circuit, enclosed, external to the vacuum chamber, with a short, (in wavelengths), high-voltage coaxial cable to the ion trap. The RF input coupler would be a tap onto the first turn of the tank coil.
https://en.wikipedia.org/wiki/LC_circuit

2. As a transmission line, a "helical resonator", which is a shortened version of the cavity resonator we have been discussing.
https://en.wikipedia.org/wiki/Helical_resonator

Thanks a lot for this! I will have to look more into these 2 options. Why is the quarter wavelength resonator we discussed so far suitable at 35 MHz (at least based on the simulations I made, the parameters required for what I need seem quite reasonable - in terms of sizes and input power)? Also, initially my question was at 3 x 35 = 105 MHz. Why does a factor of 3 in the wavelength makes such a big difference in the required setup?

 

[Baluncore]

Why does a factor of 3 in the wavelength makes such a big difference in the required setup?

Because, for lower frequencies, the size of the resonant circuit can be independent of wavelength, while maintaining a high Q.

With higher frequencies, a (long in wavelengths) stub connecting the end of the resonator, to the ion trap, would have a significant effect on the resonator Q. At low frequencies, that (short in wavelengths) stub looks more like one lump of capacitance.

I expect you will be able to wind a high-Q inductor from copper tube, or flat copper strap. Calculate or measure the inductance, then compute the total capacitance needed for resonance. Part of that total capacitance will be made up from the stub, the ion trap, and a frequency adjuster.

 

[Malamala]

Because, for lower frequencies, the size of the resonant circuit can be independent of wavelength, while maintaining a high Q.

With higher frequencies, a (long in wavelengths) stub connecting the end of the resonator, to the ion trap, would have a significant effect on the resonator Q. At low frequencies, that (short in wavelengths) stub looks more like one lump of capacitance.

I expect you will be able to wind a high-Q inductor from copper tube, or flat copper strap. Calculate or measure the inductance, then compute the total capacitance needed for resonance. Part of that total capacitance will be made up from the stub, the ion trap, and a frequency adjuster.

Actually, the simple LC case takes me back to my very initial question. In such a circuit, the Q factor is $Q = \frac{1}{\omega_0 R C}$. For $C = 0.1$ nF (which is my case) and $\omega_0 = 2\pi\times 35 MHz$ I get $Q = 45/R$. Won't this be too small? For example for an impedance of $R = 50 \Omega$ my Q factor would actually be below 1. Am I miss understanding these calculations?

 

[Baluncore]

Won't this be too small? For example for an impedance of 50 Ω my Q factor would actually be below 1. Am I miss understanding these calculations?

You will not connect the 50 ohm cable directly to the circuit.

As I understand it, you want to generate a 500 volt field, at 35 MHz.
You have a fixed capacitance of 0.1 nF = 100 pF, that you will neutralise with an inductor.
The reactance of the capacitance, at 35 MHz is; XL = 45.473 ohms.
The inductance needed to neutralise the capacitance will be = 206.78 nH.
The current in the circuit is = 11.00 amp.
The reactive power circulating is = 5.50 kVAR .
The higher the Q, the less real power will be needed to maintain oscillation.
Connect the inductor in parallel with the capacitive plates.
If I remember, you have about 0.25 ohms of loss resistance in the circuit.
The Q will then be (45.473 / 0.25) = 182.
Now you must loosely couple energy into the inductor from your 35 MHz generator.
You can do that by tapping onto the inductor, or by having a small loop nearby.
The next step is to design a 206.78 nH inductor that will fit in parallel with the plates.

 

[Malamala]

You will not connect the 50 ohm cable directly to the circuit.

As I understand it, you want to generate a 500 volt field, at 35 MHz.
You have a fixed capacitance of 0.1 nF = 100 pF, that you will neutralise with an inductor.
The reactance of the capacitance, at 35 MHz is; XL = 45.473 ohms.
The inductance needed to neutralise the capacitance will be = 206.78 nH.
The current in the circuit is = 11.00 amp.
The reactive power circulating is = 5.50 kVAR .
The higher the Q, the less real power will be needed to maintain oscillation.
Connect the inductor in parallel with the capacitive plates.
If I remember, you have about 0.25 ohms of loss resistance in the circuit.
The Q will then be (45.473 / 0.25) = 182.
Now you must loosely couple energy into the inductor from your 35 MHz generator.
You can do that by tapping onto the inductor, or by having a small loop nearby.
The next step is to design a 206.78 nH inductor that will fit in parallel with the plates.

Thank you for this! These are the numbers I am getting, too, I was not sure if I actually needed $50 \Omega$ or not (that was part of my initial post question). I have one more practical question: in practice (e.g. account for the temperature variations, vibrations etc.) how much should I expect the resonant frequency (and hence the power amplification at the fixed applied frequency) to change in time? For example, by just changing the spacing between the 2 capacitor plates from 500 to 510 microns (and keeping the inductance and everything else constant), I am getting a resonant frequency of 35.384 MHz, which is quite far away from the needed one (for Q=200 the linewidth of the resonance peak is only 175 kHz). Is this right, as it seems like in practice implementing this is almost impossible (i.e. controlling the lengths in the circuit to below 10 microns). Am I missing something?

 

[Baluncore]

I have one more practical question: in practice (e.g. account for the temperature variations, vibrations etc.) how much should I expect the resonant frequency (and hence the power amplification at the fixed applied frequency) to change in time?

You will need to build and test it.
Frequency stability, can be gained by adding a Phase Locked Loop, PLL, to the LCR resonator.

Drive the resonator with the required frequency from a signal generator and power amplifier. By measuring the phase shift between the resonating signal and the generator, (multiply and low-pass), you get a DC phase error signal. That can be used to change the resonator frequency by adjusting a sensitive parameter. The possibilities I would consider are:
1. Heat or cool the capacitor or inductor.
2. Use pairs of variable capacitance diodes, (difficult at 500 volts).
3. Mechanically move a brass or ferrite slug, in or out of the inductor, with a motor, or the moving coil from a speaker or a meter movement.
4. Mechanically move a dielectric wedge in the capacitor's plate space.

 

[Malamala]

You will need to build and test it.
Frequency stability, can be gained by adding a Phase Locked Loop, PLL, to the LCR resonator.

Drive the resonator with the required frequency from a signal generator and power amplifier. By measuring the phase shift between the resonating signal and the generator, (multiply and low-pass), you get a DC phase error signal. That can be used to change the resonator frequency by adjusting a sensitive parameter. The possibilities I would consider are:
1. Heat or cool the capacitor or inductor.
2. Use pairs of variable capacitance diodes, (difficult at 500 volts).
3. Mechanically move a brass or ferrite slug, in or out of the inductor, with a motor, or the moving coil from a speaker or a meter movement.
4. Mechanically move a dielectric wedge in the capacitor's plate space.

Thanks a lot for this! Do you have an idea (or can you point me towards some readings) about how well a properly implemented feedback loop can keep the frequency of the setup stable (i.e. what changes in the resonant frequency should I expect in a given amount of time)? I have experience with PDH locking of laser to optical cavities and I assume the principle is similar here, but I don't know how stable would the frequency be in this case.

 

[Baluncore]

Do you have an idea (or can you point me towards some readings) about how well a properly implemented feedback loop can keep the frequency of the setup stable (i.e. what changes in the resonant frequency should I expect in a given amount of time)?

The operating frequency of the resonator, would be as stable as the signal generator crystal, say 1:10⁵, (which could be GPS locked if needed to 1:10¹²). The frequency would always be correct, only the phase could change, because the resonator is driven by the signal generator through the PA.

If the peak of the resonator moved away from the signal generator frequency, the resonator continues at the signal generator frequency. The thing that changes with the tuning of the resonator, is the amplitude of the resonance. For a low-Q resonator that is not a problem, but for a high-Q resonator it could significantly reduce the amplitude of the voltage. By measuring the resonator phase shift deviation, the PLL would pull the resonator back onto the reference frequency, restoring the amplitude, with zero phase error.

When the PA is loosely coupled to the resonator, there will be a phase shift across the resistive coupling network. That occurs because the reactance of the resonator is no longer zero at the operating frequency. The task of the PLL is to recognise a non-zero phase shift, and to pull the resonator back to zero reactance.

 

[Malamala]

The operating frequency of the resonator, would be as stable as the signal generator crystal, say 1:10⁵, (which could be GPS locked if needed to 1:10¹²). The frequency would always be correct, only the phase could change, because the resonator is driven by the signal generator through the PA.

If the peak of the resonator moved away from the signal generator frequency, the resonator continues at the signal generator frequency. The thing that changes with the tuning of the resonator, is the amplitude of the resonance. For a low-Q resonator that is not a problem, but for a high-Q resonator it could significantly reduce the amplitude of the voltage. By measuring the resonator phase shift deviation, the PLL would pull the resonator back onto the reference frequency, restoring the amplitude, with zero phase error.

When the PA is loosely coupled to the resonator, there will be a phase shift across the resistive coupling network. That occurs because the reactance of the resonator is no longer zero at the operating frequency. The task of the PLL is to recognise a non-zero phase shift, and to pull the resonator back to zero reactance.

I am sorry for the confusion. What I meant to ask is how much does the resonant frequency of the resonator circuit is expected to change (for a well done feedback loop)? Of course the actual frequency will be the driving one, but I would like to know how much should I expect the Q-factor at that driving frequency to change. For reference, I need the amplitude of the electric field in between the parallel plates to be as stable as possible, thus I need the change in the difference between the resonant frequency of the circuit and the driving frequency to be minimized (and this difference should ideally be zero).

For example, for a PDH lock of a laser to a cavity, you can lock the laser frequency to 1/1000 of the cavity linewidth. I am looking for a similar estimate here, when locking the circuit itself to the fixed driving frequency. Basically, for a driving frequency of 35 MHz, should I expect a variation of the circuit resonant frequency (under the feedback loop) on the order of tens of kHz? Can I go lower than that?

 

[Baluncore]

What I meant to ask is how much does the resonant frequency of the resonator circuit is expected to change (for a well done feedback loop)?

The physical stability of the resonator's self-resonant-frequency, is decided by materials, construction, and environment.

The aim is to operate the resonator on the flat top of the Q curve. A PLL is able to bring the self-resonant-frequency, to the driven operating frequency, and to lock it there. By monitoring the PLL output voltage, you can confirm that the resonator has been phase locked to the drive signal.

Of course the actual frequency will be the driving one, but I would like to know how much should I expect the Q-factor at that driving frequency to change.

To minimise the variation in Q, over the range of PLL regulated operation, identify if it is the inductor or the capacitor that is most significant in causing the deviation. Then use the PLL output to correct that component.