Understanding Klystron Energy Conversion and Impedance Transformation

In summary, the current in the output cavity torus is directly proportional to the current in the bunched electron beam.
  • #1
artis
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A while back I made a thread which helped me even though it went sideways at the end, now I have one (maybe few) questions.1) So in a klystron, the current running in the output cavity torus as the bunched electron beam passes through the output cavity center is directly proportional to the current in the bunched electron beam?
I assume this as equal in strength and number charges appear on the cavity center walls as the negative electron bunch comes close to it and passes by , the same then happens to the other cavity wall as the electron bunch passes it, so a current runs through the cavity torus from one side to the other as the electron bunch passes its center ?
 
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  • #2
artis said:
1) So in a klystron, the current running in the output cavity torus as the bunched electron beam passes through the output cavity center is directly proportional to the current in the bunched electron beam?
There are two issues here. Firstly there will be no current in the output cavity due to the DC component of the electron beam. The current waveform corresponding to the density of the traveling electrons will not be sinusoidal - particularly at high powers - there will be a number of harmonics.
Then the current running in the output torus is different all over it. It's a resonator and will have a voltage peak in one place and a current peak in another. Remember, the Power is taken from the cavity by a coupling loop, which will also affect the impedances in the cavity. The amount of Power that's coupled from the bunched beam into the cavity will depend on the tuning and hence the matching. I wouldn't even think the process is linear so that 'proportionality' idea doesn't seem to be appropriate.
I get the impression that you are trying to analyse, on your own, what goes on. Have you read many information sources about this topic? You could save a lot of time if you did.
 
  • #3
Yes I did read some info on the subject although not a complete book.
I understand there is a static (DC ) component to the electron current due to the fact that the input cavity cannot form all of the beam electrons into bunches entirely so some are left streaming along.

1)So the electron beam looks like a small constant electron beam which has amplitude peaks (electron bunches or maybe a better word areas along the beam where there is a much higher electron concentration),
but why did you say there is zero current in the output cavity , even though there is some nonzero constant electron beam passing through the output cavity there are peaks/bunches along this beam and shouldn't those create this charge buildup and movement on the cavity plates that I was asking about?

2) If the output cavity torus doesn't have a homogeneous current running through it either one way or the other in any given time instant then how does it produce a toroidal B field that loops around the electron beam inside the torus in closed field lines? The pictures I have seen show the output coupling as a coax attached to the output cavity toroidal outer surface and the coax having a sort of "hook" structure going some depth inside the torus , I assume this is so that it can interact with the cavity toroidal B field ?

It just doesn't make sense to me why would the current be different all over the cavity as you said because the electron bunch passing the cavity center is first passing the first opening and then the second , so while it is at the first opening that means the corresponding wall closest to it is positively charged while the fursthest should be negatively charged and then this reverses as the electron bunch passes the second opening or is this "capacitor" like action completely not what goes on in there?
And I'm asking this because I'm not gully sure how the DC part of the electron beam influences the capacitance and also inductance of the cavity, I would say it lowers it or minimizes it
 
  • #4
artis said:
zero current in the output cavity
Zero Direct Current because it's 'AC coupled' to the beam.
artis said:
a sort of "hook" structure
A loop which couples to the power in the cavity.
artis said:
why would the current be different all over the cavity
Why would it be the same? There is a great GAP that the electrons go past, which is an open circuit approximately then there is a conducting surface around the rest of the torus. This isn't an intuitive thing and you need more basics to make sense of it.
 
  • #5
at first you did not specify what kind of current, so I asked again, sure there is no DC current, I was thinking AC current from the very beginning.

so without the electron beam present the cavity torus is like a single turn that is not closed, then there is the always present DC portion of the beam, this DC portion partly loads the otherwise empty capacity of the cavity or does it have negligible effect on it? At least in ordinary circuit a DC loaded capacitor can transfer lower amplitude AC signals than it could otherwise as I think it then behaves like a capacitor with smaller capacitance.So when the electron bunch is passing through at that moment even though there is a gap as you say in the cavity and surely there is one, I was assuming that the charges with the E field close the cavity loop as the cavity walls become charged with opposite charges that then flow from one wall to the other with respect to the movement of the electron bunch and in this very moment when they do so I was thinking the current in the torus should be the same and in single direction is it not the case?

Otherwise it would seem weird for me because as an electron bunch approaches a conducting surface there should be a buildup of net positive charge on that surface which is proportional in strength to the electron bunch, but in order for this to happen there needs to be a net movement of current in a fixed direction , at least this applies in all ordinary cases where charges and E fields are involved , is klystron different?
Also how does then the cavity develop a toroidal B field inside the torus which is similar to that found inside a toroidal transformer core ? I can't imagine forming that field with currents that run oppositely within the torus so as to cancel any larger net homogeneous B field.
 
  • #6
artis said:
At least in ordinary circuit a DC loaded capacitor can transfer lower amplitude AC signals than it could otherwise as I think it then behaves like a capacitor with smaller capacitance.
Wherever did you get that from? A Capacitor is a linear device. Q = CV. whatever the Voltage - assuming it does not arc across.
Why not read something that has really good informationhttps://indico-lcagenda-2005-2011.s3.cern.ch/2010/C4480/c26/2153512233578930536/Lecture_B-3_Simrock_part_4.2a.pdf?response-content-disposition=inline%3B%20filename%3DLecture_B-3_Simrock_part_4.2a.pdf&response-content-type=application%2Fpdf&AWSAccessKeyId=4QR6RYTNSYFO7BN3I43N&Expires=1553464035&Signature=8tDFwUiCLEcjvPIINTQtphgVRlQ%3D? You will have a chance of getting to know the device from a valid standpoint.
 
  • #7
The link shows an error , maybe you can double check it?

If there is a constant electron beam present through the cavity doesn't that have some effect on the charge on the cavity walls? I was thinking in terms of that the walls are always DC biased due to the beam and the bunches cannot ever fully discharge he wall capacitance due to the DC component of the beam is it not so?
 
  • #8
The cavities are earthed to the frame of the unit and to the outer of the coax feed. What charge can buildup?
Also, the beam focus is adjusted to minimise spreading.
Why not follow the message of that link I gave you (and others), rather than trying to do your own analysis of the system? When you are an 'expert' you will have more useful questions.

Apologies - that link doesn't work for me either now (must be something I said). I will find another one - but then, so could you.
https://www.allaboutcircuits.com/technical-articles/introduction-to-the-two-cavity-klystron-amplifier/
 
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  • #9
Thanks, re-read the link, I think I read it some time ago, not sure, anyway.
The link talks a lot about the cavity walls as a capacitor and torus as inductor which makes sense to me but then you said that between the cavity wall peak potential as the capacitance reverses the current isn't in a single direction at a time through the torus and that kind of makes me wonder then how does the cavity is able to function if this is not the case?Also what seems interesting , something I haven't thought about before, say the the electrons approach the input cavity at a moment when the first cavity wall is positively charged and second wall is negatively charged, so the positive charge attracts the electrons and tends to accelerate them while the second wall repels those accelerated electrons and sort of keeps of accelerating them further as said in the link but as the electrons attracted by the first wall enter the cavity the same positive potential is now behind them and would in other situations tend to reverse this acceleration direction by the force it exerts and so pull them back or slow them down again like a slingshot, why this doesn't happen in the cavity scenario?
 
  • #10
I'll attempt to answer my own question with regards to the second part of my last post,
You said earlier that "the current waveform will not be sinusoidal" for the klystron so could it be that the only moment when the electron beam is really influenced is at and near the peaks of the input cavity sine when the E field in the cavity is at it's peak and can break or accelerate the incoming beam with enough force, and these faster and slower electrons then as they travel in the drift tube catch up with one another (faster ones catch the slower ones that were slowed down before the faster ones arrived at the cavity) and so this then creates a pulse like current at the output cavity where the pulse frequency matches the wavelength/frequency of the input signal but the waveform itself is not a "good classical sine" is this the case?

Thanks
 
  • #11
artis said:
The link talks a lot about the cavity walls as a capacitor and torus as inductor which makes sense to me
That is a 'lumped component' equivalent circuit of a resonator but the cavity has dimensions of the order of a wavelength so it's too simple, I think. Remember, the cavity is loaded with the output coupling loop too.
artis said:
say the the electrons approach the input cavity at a moment when the first cavity wall is positively charged and second wall is negatively charged, so the positive charge attracts the electrons and tends to accelerate
etc.
This is what the resonance of the cavity is all about. If the bunches arrive at the wrong frequency, there is no energy taken from the beam. Even when at resonance the current in the cavity will contain very few of the higher harmonics that exist in the variations of density of electrons with time.
You can read about all this stuff without trying to learn it in a Q and A session. Q and A is actually very inefficient in most cases - especially when there are so many sources of information at many different levels. If you think you have cracked a particular bit of this, why not read around and see if you have got it right? It's very hard work trying to direct you on the right path when you go about it this way. It's decades since I used Klystrons.
 
  • #12
To be honest I have searched google etc but since klystrons are very specific advanced scientific "niche" products , I only find academic articles that talk about specific band klystrons and their specific application for a very specific goal , in those papers there are only bits of the more basic and fundamental workings of the klystron itself since that is usually not the focus of the paper.Well I understood the bunching and some other functions now I think, to reverse this I can imagine the same cavity being pumped by a klystron being used for particle acceleration , in order to get any acceleration from the cavity they have to match the timing of the arrival of particles just right otherwise the E field slows them down instead of accelerates.Ok so the cavity is a lumped component and exhibits some "emergent properties" that individual L and C parts at lower frequencies don't exhibit, but is the current through the torus at any given time in multiple directions or is it running in one direction? Because I find it hard to imagine how a current having more than one direction at the same time could produce a toroidal field inside the torus of the cavity , even at RF frequencies I find this to not be possible to the best of my knowledge.
 
  • #13
artis said:
klystrons are very specific advanced scientific "niche"
I wouldn't say that. Klystrons are used in nearly all high power UHF TV transmitters because they are still (afaik) the only way to produce tens of kW of 'linear' UHF power amplification. They are also used as sources of high power microwaves for linac. That is hardly 'niche'.
What is happening here is that you have picked a technology that hangs on a lot of prior knowledge, like electron dynamics and transmission line theory etc. but you do not appear to want to go there. You will never find out what you want from simple Q and A on PF. You need to get down to the graft of formal EM work or be prepared to accept limits.
I don't know of a suitable modern textbook that you could use but you could perhaps post another thread, asking for PF opinions. Google may be demonstrating its limits as a source for you.
 
  • #14
Ok, one practical question maybe you remember, how deep on average is the output or input cavity coupling coax "hook" that extends and makes a turn inside the outer toroidal part of the cavity through an opening.
I am interesting say in cm as measured from the torus wall to the end of the coupling , in other words how far it extends inside?
 
  • #15
artis said:
I am interesting say in cm as measured from the torus wall
It would depend entirely on the wavelength involved and probably the dimensions of the feeder too. I have to ask why this would be so important? Or are you trying to fill in the blanks in some simulation program?
I have come across klystrons with cuboidal resonators too, if you think this approach is going to give you more understanding.
How's the general reading around exercise going?
 
  • #16
I am reading some papers , i still have one long PDF someone gave me here from the 40's about high frequency vacuum tubes and klystrons.

not exactly a simulation program more like a small high frequency generator , just an idea of mine but I'm having difficulty with figuring out how to produce a high strength magnetic field at RF frequency range, as I'm sure you know is a rather complicated aspect if not impossible. And by high strength I mean ideally somewhere up to 1 Tesla. For lower frequencies where ferrite or otherwise cores can be used this wouldn't be an issue.
And another thing I'm curious about, looking at any klystron I see that the power supply is always (not counting the heater filament psu) a high voltage DC supply which is logical in order to get a high strength E field for the electron acceleration, so would it be fair to say that a klystron is more of a RF voltage source than a current one?

I'm also wondering about this because at low frequencies it is easy to step up/down voltages/currents but at RF is it even possible as I don't recall there being anything similar to a transformer for RF and microwave?
Just imagine I have a RF very high current but low voltage source now since the power would be transmitted via a coax or other type of waveguide would I get high losses just as with low frequency high current/low voltage?
Also is there any apparatus for transforming RF high current/low voltage into higher voltage/lower current?

thanks.
 
  • #17
Did you consider the high level vacuum that’s needed for all electron tubes. A simple lab pump wouldn’t be adequate.
 
  • #18
This is not a vacuum tube, well the klystron is but not the device I'm trying to make so it needs no vacuum.
Its basically just an LC oscillator where I set the C and then supply the L with a high frequency B field which drives the whole circuit, anyways without a working model and a schematic it would be hard to explain. That is why I'm trying to get there and understanding as much as I can about other related topics like klystrons meanwhile. Much of the RF high frequency stuff applies the same way to what I'm trying to make with some exceptions.

PS. I would love to hear some answers to the questions in my previous post.
 
  • #19
artis said:
This is not a vacuum tube, well the klystron is but not the device I'm trying to make so it needs no vacuum.
Its basically just an LC oscillator where I set the C and then supply the L with a high frequency B field which drives the whole circuit, anyways without a working model and a schematic it would be hard to explain. That is why I'm trying to get there and understanding as much as I can about other related topics like klystrons meanwhile. Much of the RF high frequency stuff applies the same way to what I'm trying to make with some exceptions.

PS. I would love to hear some answers to the questions in my previous post.
OK, so the thread has moved totally off topic now you need to be finding out about Cavity Resonators, Waveguide Theory, Transmission Line theory etc. etc., neither of which is trivial
artis said:
at low frequencies it is easy to step up/down voltages/currents but at RF is it even possible
It is extremely common practice to use lengths of waveguide or RF feeder to produce Impedance Transformation. I really don't know where to suggest you should start but it's certainly way further back to basics than you seem to be operating at the moment. I have suggested using a good EM theory textbook but I suspect you are avoiding the rigour that would involve. You just cannot leap into a topic like in the way you are planning without a lot of Basics. Why do you think it's regarded as a specialist subject?
Needless to mention the cost of basic test equipment for this sort of study.
 
  • #20
I have the test equipment , have a few friends working in local university labs and a good friend who was a radio engineer back in his day.

Klystrons are more complicated due to electron properties etc, as far as my device goes I understand everything about it, the problem lies deeper I need to basically figure out a thing or two that have never been done at least to the best of my "google searching"
I need a homogeneous B field at RF frequency and berkeman or someone else suggested one of those pyramid shaped RF test cavities but sadly the way I need to implement the field can't be combined with such cavities at least the ones i looked at.

I could try to explain one of the ideas I'm having for producing such a field but I'm losing hope , first of all nobody seems interested here and second of all the numbers don't seem promising.
any air core coil needs large currents to get some decently strong fields so yeah.

One idea I had was to turn the capacitor into a sort of cavity like inductor, well for simplicity imagine two parallel plate as usual but one of the plates has many small few turn coils on the back of the plate, all connected in parallel , now as I would connect the capacitor together assuming a large enough RF current would flow through it the current would flow through the coils to reach the plate in each cycle producing a B field which at least on and near the surface of the plate would be homogeneous which is what I need. I need a homogeneous field in a rather thin area which can be very close or at the very surface of a capacitor plate or any plate or surface for that matter.
 
  • #21
artis said:
I need a homogeneous field in a rather thin area which can be very close or at the very surface of ... any plate or surface for that matter.
sophiecentaur said:
It is extremely common practice to use lengths of waveguide or RF feeder to produce Impedance Transformation.

A likely impossibe approach: make that plate the conductor and use the suggestion by @sophiecentaur to get the driving impedance low enough to supply the needed current across the plate.
 
  • #22
artis said:
but is the current through the torus at any given time in multiple directions or is it running in one direction?
artis said:
I don't recall there being anything similar to a transformer for RF and microwave
artis said:
so would it be fair to say that a klystron is more of a RF voltage source than a current one?
artis said:
Just imagine I have a RF very high current but low voltage source now since the power would be transmitted via a coax or other type of waveguide would I get high losses just as with low frequency high current/low voltage?
Then
artis said:
as far as my device goes I understand everything about it,

Are you sure about that?
What frequency would you be using? Fabricating your own resonator would require some specialised skill, if you want to be sure that the results you get are consistent with the theory. A 'suck it and see' approach is not usually successful in microwave Engineering.
It could help if you made it clear as to what you actually want to achieve here.
 
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  • #23
Fair point sophie but to be honest you actually answered none of those questions that you posted in your last post. The thing I can summarize from all of this is "go get a book" basically, which in itself is a good advice but not an answer to what I asked.I'm not building a cavity resonator, at least not in the sense that applies to klystrons or particle accelerators.
it would take me a lengthy thread to try to explain what I'm doing but right now I mostly want to realize whether I can get a homogeneous B field in the frequency range that starts roughly from 1Mhz up to microwave region , because if this cannot be done then everything else is not even worth to mention as it wouldn't work.

The problem is I can't have the field within some closed waveguide because I would need to introduce auxiliary gadgets like a rotor shaft and pickup wires etc which would render the waveguide useless.Essentially what I'm trying to do is I'm having a Faraday disc spinning at high rpm with a liquid metal contact (mercury) and I wan't to see whether I can produce an AC at its output, because the disc due to Lorentz force works as an electromechanical amplifier in the sense that the current waveform in the disc directly follows any changes in the exciting field, so in theory I could apply any high or low AC sine as its B field and collect a greatly magnified (mostly in current) output from the disc.
If the disc load circuit has a resistance and reactance (for AC ) that is close to or as low as that of the disc then in theory I could extract very large current at any desired frequency, for lower frequencies I could pass them through a transformer and get more useful higher voltage lower current output for higher frequencies well I don't know you tell me how can one take a large current low voltage RF output and make it in a high voltage low current one or vice versa
 
  • #24
artis said:
in the frequency range that starts roughly from 1Mhz up to microwave region
For a go - no - go decision on that, just look at what's available on the market. Something like that would be extremely useful in a calibration system. If you can find anything remotely like that then you could be in with a chance (when your knowledge and understanding is at the level of those Engineers).
What exact minimum wavelength would you be interested in?
artis said:
in theory I could extract very large current at any desired frequency
You want an 'amplifier' or just a transformer? "Large Current" or High Power"?
I notice that there are not many other respondents to this thread and it is getting deeper and deeper into word salad so I will dip out. Feel free to ask specific (but not 'wild') questions about part of this idea of yours in a separate thread.
 
  • #25
Well the idea is it could work as an amplifier, because the mechanical prime mover adds energy to the system while the properties of the system are such that input waveform is saved and amplified. The reason I asked for the possible options to transform is because the faraday disc being a short circuit essentially even at high frequencies develops very low voltage and tremendous current.
For lower frequencies this is easy simple use a large enough transformer and have any secondary voltage/current ratio needed but I can't imagine how one could achieve the same with RF.
The other problem with RF is that it becomes problematic to get a decent magnetic field which would also have strength and homogeneity which are needed.
This is why I was talking about high frequencies because doing this at lower ones where one can use a core material is far easier.
So from this I would say minimum wavelength would around 1Mhz because anything lower can be done through a core.
The other great part about the disc is that unlike a klystron ad RF cavities in general which for a certain size are rather frequency range limited the disc can work with anything from pure static B field aka DC up to Mhz maybe even higher..?
So I see this as an intriguing matter to check out what happens.
 
  • #26
Somehow I don't follow the logic behind stating that transformers for RF don't exist. Radio receivers have used them for signal coupling between amplification stages at least since the Superheterodyne radio was created in 1917. Radio, and TV, transmitters often use them to couple (impedance match) the output stage to the antenna, handling kilowatts of power.

The low power receiver transformers typically us a ferrite core, the high power ones use an air core to keep the losses down.

As fair as I know none of them come anywhere close to 1 Tesla magnetic field. MRI medical scanners operate in the neighborhood of 1 Tesla, but the current is so high that magnet windings are cyrogenically cooled.

Cheers,
Tom
 
  • #27
Tom.G said:
Somehow I don't follow the logic behind stating that transformers for RF don't exist.
There is no need to use a 'wound' transformer if you want to transform impedances. There is an overlap frequency around which it's still possible to use a wound component (and as long as the Power isn't too great) but, above the Hundreds of MHz range, it becomes more an more suitable to use transmission line / cavities to achieve the same effect.
I'm afraid the OP's ideas are a bit scattergun and probably based on light reading of not always too well informed publications. I have used the phrase 'word salad' and I think it's accurate.
 

1. What is Klystron energy conversion?

Klystron energy conversion is a process in which high-frequency electromagnetic energy is converted into direct current (DC) electrical energy. This conversion is achieved through the use of a specialized vacuum tube called a klystron.

2. How does a klystron work?

A klystron works by using a high-frequency electromagnetic signal to accelerate electrons through a series of resonant cavities. This causes the electrons to gain energy and produce a more powerful signal. The amplified signal is then extracted from the klystron and converted into DC energy.

3. What are the advantages of using klystron energy conversion?

Klystron energy conversion offers several advantages over other forms of energy conversion, such as high efficiency, high power output, and the ability to produce a stable and consistent energy supply. Klystrons are also capable of converting a wide range of frequencies, making them versatile for various applications.

4. What are some common applications of klystron energy conversion?

Klystron energy conversion is commonly used in microwave ovens, satellite communications, radar systems, and particle accelerators. It is also used in medical equipment, such as MRI machines and cancer treatment devices.

5. What are the potential drawbacks of using klystron energy conversion?

One potential drawback of klystron energy conversion is its high cost, as klystrons are complex and expensive devices. Additionally, klystrons require a high voltage power supply and need to be cooled, which can add to the overall cost and maintenance of the system. Another potential drawback is the potential for electromagnetic interference, which can affect the performance of nearby electronic devices.

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