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Can you please explain why almost nowhere do I see such a parameter as the output voltage from a klystron, it's easy to find the DC accelerating voltage of the beam and the electron bunch densities etc but I can't find what would be the voltage developed between the cavity torus inner plates that form the capacitor as the electron bunch travel through it.

Now whatever that voltage might be am I right in thinking that the cavity torus is a very low inductance inductor and also a shunt of the cavity plates so whatever voltage is developed across the plates is quickly turned into high current low voltage that travel through the torus back and forth creating a toroidal B field which is then coupled to the RF transmission line by a "hook" like wire ending extending from the transmission line into the RF cavity?

So would it be fair to assume that the klystron's electron gun creates and accelerates electrons to high velocities by a high voltage DC potential and then the beam gets velocity modulated aka" bunched" and these bunches arrive at the output cavity creating a strong E field in the cavity between the cavity plates and voltage potential (high?) between them which is then turned into current through the torus walls which then creates a toroidal B field from which the output power is coupled, so essentially the high voltage input of the klystron is transformed to a RF frequency B field which is coupled to a transmission line,

but why I can't find any relevant info on the voltage or current such an output would yield,?

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Microwave engineering tends to work with power, not voltage or current. It’s the power of the created microwaves that propagates through the system, eventually accumulating in the accelerating cavity. Along the way, as the wave’s mode is transformed, the current and voltage distributions can change, but the power (except for losses) remains the same.

Once that power reaches the accelerating cavity, one does start to talk about the voltage that’s generated by the power. That accelerating voltage is interesting there. For an example, see this paper about PEP at SLAC. The accelerating cavities are resonant cavities: the energy in them grows until the losses are balanced by their input power, so the fields can be many times as large as those seen in the input waveguides or microwave power generators.

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  • $\begingroup$ I suppose the reason they normally don't talk about voltage or current is because of the short wavelengths in the microwave spectrum one can have voltage swings across a short length of waveguide so only the total power is meaningful while in a cavity for acceleration the E field is crucial so it is then expressed as voltage gradient in volts between the cavity plates $\endgroup$
    – Girts
    Commented Dec 15, 2019 at 6:49
  • $\begingroup$ one more question if the field in the accelerator cavity located within the tube of the accelerator is say 1MV, would that then mean that also the original E field developed within the powering klystron was that high in the output cavity or is there some transformer action going on here where the klystron produces a higher B field within it's cavity which is coupled and through the waveguide and receiving accelerator cavity it somehow turns into a weaker B field stronger E field situation? $\endgroup$
    – Girts
    Commented Dec 15, 2019 at 6:51
  • $\begingroup$ Added a sentence about resonant cavities. There’s a little bit of transformer action due to mode shifts, but the majority of the increase is due to resonant energy storage. $\endgroup$ Commented Dec 15, 2019 at 7:46
  • $\begingroup$ so basically there is constructive interference going on in the cavity where a smaller ac signal (microwave input) is in phase with the resonant signal (in the cavity case I suppose the torus B field) and so the two signals add in amplitude? Now I normally know cases where two in phase AC signals of same magnitude can produce a single signal of twice the amplitude but how many times the amplitude increases in the resonant cavity? Because I suppose the klystron output cavity doesn't produce a 1MV E field between it's plates from each of the electron bunches passing through $\endgroup$
    – Girts
    Commented Dec 15, 2019 at 10:54

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