RF cavity shunt impedance, PD and input power

[artis]

I still did not understand this from my previous threads, what is the way to determine the instantaneous voltage across the cavity plates at any given moment, in a klystron I suppose the voltage would be proportional to the arriving electron bunch density which would determine the overall bunch charge strength? I suppose there are additional complexities due to the constant DC component of the passing bunched electron beam?

Does the voltage that forms between the cavity plates which also determines the E field strength in the cavity is proportional to the amplitude/voltage of the cavity input sine amplitude?

I am reading this PDF, on page 7 it talks about cavity shunt impedance,
https://indico.cern.ch/event/109664...6/attachments/30663/44411/Progress_Report.pdf

So a given size cavity has given impedance and the only way to increase the E field strength in the cavity is to increase the voltage/amplitude of the cavity input waveform?

I am asking this because I am curious whether one can get a strong E field in the cavity by powering it from a low impedance, low voltage but very high current source at it's resonant frequency so that the generator and cavity resonant frequency match?
In theory at resonance the cavity impedance should be very low ? allowing for a low voltage high current waveform to create a strong E field within the cavity?

 

[artis]

One more thing, The beam voltage in a klystron is the potential difference between anode-cathode that accelerates the electrons in the beam increasing their kinetic energy , how is this kinetic energy then related to the output cavity sine amplitude/voltage? Is it directly related?

 

[tech99]

One more thing, The beam voltage in a klystron is the potential difference between anode-cathode that accelerates the electrons in the beam increasing their kinetic energy , how is this kinetic energy then related to the output cavity sine amplitude/voltage? Is it directly related?

The cavity will take energy from the kinetic energy of the electron bunches. Remember that a klystron works at high frequencies where electron mass is involved.
The cavity, if unloaded, will pesent infinite impedance across its plates. By altering the coupling of the load into the cavity, the impedance across the plates may be adjusted until max power is extracted. For a reflex klystron this condition will simultaneously allow the positive feedback to be optimised.

 

[artis]

So I take that as a yes, the cathode-anode voltage is directly related to the cavity output voltage , minus the losses ofcourse which for a 50% efficient klystron would mean that a PD of say 20Kv of cathode -anode would roughly translate into a 10kV amplified sine output amplitude?If the cavity when unloaded represents infinite impedance then how are accelerator cavities functioning because unlike a klystron where the electron beam drives the cavity , here the cavity is driven by a RF high power signal and the electrons/ protons arrive at the specified time intervals , so at the times when the cavity is empty the drive power sees a very high impedance but when the timed proton bunch arrives and passes through in phase the drive power supplied to the cavity now sees a much decreased impedance and can do useful work on the particles int he cavity ?

 

[sophiecentaur]

when the timed proton bunch arrives and passes through in phase the drive power supplied to the cavity now sees a much decreased impedance and can do useful work on the particles int he cavity

Could you try rewriting that in understandable terms? There is no "proton bunch" involved in the way a klystron works and what "particles" are in the cavity?
You are, once more, confirming my skepticism about what you have learned about the way a klystron actually functions. You have to start with the very basics and take it through to the end instead of picking superficial random questions to ask about it.

 

[artis]

Ok, ehh here we go again, once more.

Forget the word bunch, even though LHC papers themselves refer to proton bunches and even give the density and number of protons within a given bunch, and yes I know that unlike in a klystron in the accelerator tube and cavity there is no DC component or in other words no constant stream of particles like in a klystron but instead "a number of particles" (if you prefer such wording) this "bunch" arrives at timed intervals in order for it to match the moment when the E field is in (phase?) with the particles so that they can be accelerated instead of deaccelerated , so my questions is this then.
How does the cavity impedance as seen by the drive signal change when a) the cavity is empty and b) the cavity has particles passing through it or in other words the cavity is loaded?
My own guess is that when it's empty the impedance represented by the cavity is high and when loaded it drops.

Continuing on this line would it be fair to assume that an unloaded cavity has high impedance and wastes very little supplied RF power?

 

[artis]

I am also looking forward to answer to my question in the first post of this thread...
Quoting mysilg from the 1st. post

Does the voltage that forms between the cavity plates which also determines the E field strength in the cavity is proportional to the amplitude/voltage of the cavity input sine amplitude?

I am reading this PDF, on page 7 it talks about cavity shunt impedance,
https://indico.cern.ch/event/109664...6/attachments/30663/44411/Progress_Report.pdf

So a given size cavity has given impedance and the only way to increase the E field strength in the cavity is to increase the voltage/amplitude of the cavity input waveform?

I am asking this because I am curious whether one can get a strong E field in the cavity by powering it from a low impedance, low voltage but very high current source at it's resonant frequency so that the generator and cavity resonant frequency match?
In theory at resonance the cavity impedance should be very low ? allowing for a low voltage high current waveform to create a strong E field within the cavity?

PS. Sophie please stop constantly criticizing me for my approach, I know bit more than you actually would think or give me credit I'm just bad at expressing myself and if I had the option to simply go to a university instead of indulge in this painful online answer searching I would.

 

[sophiecentaur]

Sophie please stop constantly criticizing me for my approach

I am sorry but you start off with the same sort of misapprehensions and still hang on to them. Also you are hopping from klystron to accelerator in a random way and back again. Bunching in klystrons is there because of the drift spaces between cavities and it gives a chance for the KE of the DC beam current to transform into RF energy. An accelerator does precisely the opposite. It transforms RF Power into KE of the protons. There would be no point in having a long drift space if you want a long burst of fast protons. Bunch forming would not even be too desirable because of the mutual repulsion. Reading your post, I can see you are mixing the two up and that just involves more confusion.

curious whether one can get a strong E field in the cavity by powering it from a low impedance

You have asked this question several times before. Do you not understand that a cavity can behave as a Transformer? The Impedance of the cavity will be affected by the load which is presented by the power that's put into the proton beam so it will change, depending on the presence (and velocity) of the beam. Why supply the cavity with a low impedance feed? Because feeders of 50Ω are the best way to transfer REF power in most cases.
All this stuff is basic and it's discussed in many places and you should get this sorted before trying to go further and ask some questions that are not just easy alternatives to reading around.

 

[artis]

As for the first part of what you said, I already knew that, I apologize if my wording gave the impression that I didn't.

Sure an accelerator does the opposite, that is why I asked about the cavity impedance and how it changes under different circumstances , I was simply thinking of having a different RF power source than a klsytron which instead is low voltage but high current and was wondering could such a source be able to provide the same input power as a klystron in a given case, you partly answered my curiosity in the second part of your post for which I am thankful,
You say the cavity works like a transformer but does it ? at least in the ordinary sense of a transformer which can transform more current less voltage into more voltage less current and vice versa,
say I'm powering the cavity with high current low voltage supply, the E field between plates would be weak because voltage is low and it is essentially voltage that creates the field between cavity plates is it not?Well it may be basic stuff but when I tried googling the answers and reading the multiple PDF's I couldn't get a straight answer.

 

[sophiecentaur]

You say the cavity works like a transformer but does it ? at least in the ordinary sense of a transformer which can transform more current less voltage into more voltage less current and vice versa,

What do you mean by "the ordinary sense"?
If you had read the suggested information about transmission lines and cavities then you would see the direct correspondence between the magnetic coupling of a wound transformer and the electromagnetic coupling of a waveguide - style cavity. Perhaps you just don't like the idea but the theory is pretty well established and it's applied all over the World in many different forms so you just have to accept it, I'm afraid.

say I'm powering the cavity with high current low voltage supply,

. . . .and where are you going to find one of those and, in particular, where are you going to get hold of one that fits right up against the accelerating cavity? In fact, if you had on elf those, you wouldn't need any cavities; you could just use pairs of plates with holes in them and push your protons directly through your High RF Voltage. The reason for using accelerator cavities is that it does the job very well.

 

[artis]

I am reading multiple materials right as we speak about those matters, Ok let's take the coax cable attached to the accelerator cavity torus, the cable is most likely 50 Ohm? impedance and the cavity's impedance changes based on whether it's beam loaded or empty which is what I have understood so far, the coax is terminated at the cavity with it's shield (outer conductor) connected to the cavity torus itself while the inner conductor forms a loop within the torus and is left open ended? or connected to the cavity wall? hard to tell based on some pictures and diagrams.

so this loop couples the transmission line to the cavity torus through the B field that forms within the torus? The B field in the torus creates current that charges up the cavity plates and creates an E field between them and so this reverses back and forth (resonates) within the cavity.

Is it also true that due to transmission line characteristic impedance when the cavity impedance goes high (unloaded) the power source only "sees" the impedance of the line itself which would be 50 Ohms,
Here I do wonder what is the cavity impedance when it's loaded.
As for what you said in your last reply, I did not understand the second part, but let me ask a question and by the answer I could understand, so if we are speaking about the cavity as a transformer then if I could induce a stronger B field in the torus it would also translate into a higher voltage between the cavity plates and a stronger E field between them? And the thing that limits the B field strength is the amplitude/voltage of the drive signal for a given impedance, so either one has to increase voltage or decrease impedance to achieve a stronger B and E fields within the cavity, correct?PS. the thing that I am building if it works according to the theory that I have envisioned would essentially be a very high current low voltage source of RF, that is partially why I am asking all of this. I am learning and thinking at the same time.

 

[sophiecentaur]

I did not understand the second part,

Are you referring to this?
You are not likely to get hold of a high voltage, high power source that can fit directly onto the accelerator. Real Scientists need real solutions. They need to use feeder and that means they need a cavity to transform their 50Ω source.

if it works according to the theory

I'm not sure you mean "theory". I think you mean "as you would like it to work" and that's not a theory. At best it's a hypothesis.

Is it also true that due to transmission line characteristic impedance when the cavity impedance goes high (unloaded) the power source only "sees" the impedance of the line itself which would be 50 Ohms,
Here I do wonder what is the cavity impedance when it's loaded.

Have you really read about transmission line theory? If there is a mismatch on the line, the source can see a range of impedances, depending on the length of the line.
The cavity doesn't have 'an impedance'. It will have an input impedance (at its connector) and it will also have an identifiable output impedance which will transfer energy into the particle beam. The ratio of H and E Fields will be different in different parts of the cavity. As with all transformers, the impedance seen at the input connector will depend on the load on the output. The transformer function here is to Match source to Load as best as it can. The effective load on the gap will probably depend on the DC and AC components of the beam current and the I would imagine that the design of the cavity would achieve the best match under conditions where the maximum power it required.

 

[artis]

Ok I see, so the accelerator cavity is designed to have the closest impedance match between the torus input and it's output when loaded by particles (which makes sense) as that is the moment the energy is needed in the form of the E field between the plates in order to accelerate the particles. So an unloaded cavity has a high impedance and so the RF amplifier (klystron etc) sees the impedance of the waveguide/coax which is determined by it's length and termination type (open ended/short circuited)
So far so good ?What are you referring to with the word "feeder" , do you mean a coax cable/waveguide?

 

[sophiecentaur]

closest impedance match between the torus input and it's output when loaded by particles (which makes sense) as that is the moment the energy is needed

Yes

with the word "feeder" , do you mean a coax cable/waveguide?

Yes - a very common term.