Klystron beam current, drive frequency, LHC CW example

artis
Messages
1,479
Reaction score
976
First of all I want to ask , do the LHC power klystrons work always in CW or are their frequency shifted to correct for timing of the bunches in the LHC tubes in case something isn't aligned or doesn't that ever happen?
Also how do they drive the klystrons at LHC or other CW operation from an RF amplifier that is fed from a oscillator or otherwise? I guess @mfb could know something about this, I searched the net but did not find much.Also I have noticed that in the RF field current is used less as a quantity instead parameters are expressed as power and less often voltage.
For klystrons I typically only see two quantities , namely the cathode filament current and the beam current , which is the bunched RF modulated beam of electrons from cathode to anode.
Now in my interest I have wanted to understand the amount of current running back and forth in the output cavity of a klystron. So can I say that the current in the cavity torus is proportional to the beam current? More precisely to the current/charge of the bunches as the DC portion of the beam doesn't induce RF current in the cavity.
I would tend to think so because I see the cavity beam gap as a capacitor and the electron bunch passing through the gap in the plates as a moving third plate. In any capacitor the amount of charge on one plate has to equal the amount of opposite charge in the other.

But here I have a few questions.
1) What quantity does the klystron beam current represent , just the bunch current or the bunch + the steady DC current together?
2) From my description above I make an analogy with a capacitor. Now since the bunches are moving through the cavity it would be similar to a parallel plate capacitor where two plates are stationary and in between them is a third plate that can move.
Normally when a standalone charged capacitor plates are separated the voltage between them will increase but charge will stay the same. In a klystron the two cavity plates are short circuited so as the bunch moves away from one plate the voltage should increase but can't because the plates are connected so instead what happens?
I am asking this to understand whether the current in the cavity is always proportional to the charge/current in the bunch.

Now I would tend to think like this. when the bunch first enters the cavity immediately an opposite charge is built up on the closest cavity plate. This is the moment when the RF sine wave starts from zero, then as the bunch is 1/4th of the way through the cavity or 1/4 away from the first plate but 3/4 away from the second the field reaches it's peak (positive half period), then as the bunch is now 1/2 way through the field drops back down to zero as the bunch is now an equal distance from each plate so there are equal number of charges now on each plate no current runs. Then as the bunch gets to 3/4 where it is 3/4 away from first plate and 1/4 from the second the sine wave reaches it's peak again (negative half period) and then finally as it almost exits the cavity the field returns back to zero again.
 
Physics news on Phys.org
artis said:
Also I have noticed that in the RF field current is used less as a quantity instead parameters are expressed as power and less often voltage.
The thing that matters is the "rate of flow of energy" = power. The ratio of voltage to current is determined by circuit impedance. If the impedance is not matched, that energy will be reflected back into the generator with unwanted consequences. Only the efficient transmission of energy is important.
 
The frequency is shifted a bit depending on the beam energy - at injection the particles need slightly longer per orbit. The difference is about 4 parts in a million. There is a feedback system from the accelerator cavity to the klystron control, see e.g. LHC Project Report 906.

At the frequency of klystrons concepts from low frequency electric circuits (like capacitors having to match charges) are not always applicable.
 
  • Like
  • Informative
Likes SBoh, berkeman, vanhees71 and 1 other person
Toponium is a hadron which is the bound state of a valance top quark and a valance antitop quark. Oversimplified presentations often state that top quarks don't form hadrons, because they decay to bottom quarks extremely rapidly after they are created, leaving no time to form a hadron. And, the vast majority of the time, this is true. But, the lifetime of a top quark is only an average lifetime. Sometimes it decays faster and sometimes it decays slower. In the highly improbable case that...
I'm following this paper by Kitaev on SL(2,R) representations and I'm having a problem in the normalization of the continuous eigenfunctions (eqs. (67)-(70)), which satisfy \langle f_s | f_{s'} \rangle = \int_{0}^{1} \frac{2}{(1-u)^2} f_s(u)^* f_{s'}(u) \, du. \tag{67} The singular contribution of the integral arises at the endpoint u=1 of the integral, and in the limit u \to 1, the function f_s(u) takes on the form f_s(u) \approx a_s (1-u)^{1/2 + i s} + a_s^* (1-u)^{1/2 - i s}. \tag{70}...

Similar threads

Back
Top