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Production of a Safe Drive Beam The Drive Beam pulse Which bunch goes where How to get rid of undesired bunches General Principle Operational Scenarios.

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Presentation on theme: "Production of a Safe Drive Beam The Drive Beam pulse Which bunch goes where How to get rid of undesired bunches General Principle Operational Scenarios."— Presentation transcript:

1 Production of a Safe Drive Beam The Drive Beam pulse Which bunch goes where How to get rid of undesired bunches General Principle Operational Scenarios

2 The Drive Beam pulse RF = 1GHz -> bucket == 1ns == 30cm A “header” (~121 bunches) needed to bring RF to a stable state. These bunches will be discarded after the Linac and before the “recombination complex” 24 x 24 “sub-pulses” of 121 bunches each (68696 bunches in total) Inside a sub-pulse, a bunch every second bucket (60 cm) In between sub-pulses, spacing of 30cm or 90 cm, to change bucket parity

3 Give a name to the sub-pulses... There are 24x24 sub-pulses of 121 bunches each The first 24 will be called 1.1, 1.2,...1.24 The second 24 will be called 2.1, 2.2,...2.24...The last 24 will be called 24.1, 24.2,... 24.24 The first index specifies the train number (i.e to which decelerator sector the sub-pulse goes) The second index specifies the path of the sub-pulse in the recombination complex

4 ... And to the bunches Inside every sub-pulse there are 121 bunches; we need to add a bunch index to the train and path ones The first bunch of the first sub-pulse will be 1.1.1 The last bunch of the last sub-pulse will be 24.24.121 Another way of identifying the bunches could be by their “sequential number” (train -1)  24  121+(path -1)  121 + bunch In this way, the first bunch of the complete pulse is #1, and the last is #68696, but we should not forget the header (bunches [-120..0])

5 Recombination paths All the bunches with the same path index follow the same path Earlier sub-pulses need to be delayed more If the path index is odd, the sub-pulse goes to the delay loop; if it is even, it does not go there The number of completed turns in the first combiner ring is given by 2 - ((path index -1) mod 6)/3 The number of completed turns in the second combiner ring is given by 3 - (path index -1)/6

6 Recombination paths summary table

7 How do the delay loop and the CRs work The delay loop entrance mechanism is an rf-deflector which changes state (IN/OUT) at every bucket; it lets enter (IN) all the bunches in the odd buckets, it repels (OUT) all the bunches in the even buckets. All the bunches in the same sub- pulse have the same parity; the parity is systematically changes in between two sub-pulses. The first combination ring opens its exit gate every 6 th sub- pulse, for a duration of one sub-pulse; it lets go the 6 sub- pulses it has accumulated. The second combination ring opens its exit gate every 24 th sub-pulse, for a duration of one sub-pulse; it lets go a decelerator train consisting of the 24 sub-pulses it has accumulated.

8 How to get rid of undesired bunches Options a)do not produce them b)kick them out before the Linac c)kick them out after the Linac d)kick them out after the recombination Granularity (and the most natural option to implement it) A.Can we get rid of any pattern of bunches ? (option a/b) B.Can we get rid of n consecutive bunches (several times inside a pulse) ? (option c) C.Can we only get rid of the end of a train ? (option d)

9 What can we do to produce a safer beam To produce a safer beam, we have to reduce the number of bunches of the initial pulse. Ideally this would be best done at the source, before sending the pulse to the Linac. But even if we had complete control at the source and therefore we could decide which bunches are part of the initial pulse, we are limited by the following....

10 General principle The Drive Beam Linac is able to accelerate correctly one or more consecutive bunches only if they are preceded by a “header” made of several bunches. The function of this header is to “prepare” the RF to accelerate the following bunches in the most possible homogeneous way. These header bunches will be discarded later – how? Automatically because of the energy mismatch, or in a controlled way. – where? Partly in between the two parts of the Linac, partly after the Delay Loop. The latest figure (Wegner) is a header of ~245 ns, i.e. equivalent to a complete 121 bunches sub-pulse

11 Consequences of the General Principle Consequence 1: immediately after the initial pulse leaves the Linac, we cannot have holes shorter that the header (for example, we cannot have just every second bunch). If we wanted to create such an hole, we would have to get rid of individual bunches after the Linac. Consequence 2: we can only make many holes if the system which gets rid of the header is able to tolerate many headers for pulse (rather than only one header) 2bis (if we need to do something to get rid of the headers): we can only make many holes if we are able to later get rid of the corresponding headers

12 Digression: how to get rid of the header(s) The bunches of the initial header (before sub-pulse 1.1) will have a “bucket parity” opposite to the bunches of 1.1. They will not enter the delay loop, while sub-pulse 1.1 will enter; therefore after the delay loop there will be a hole of 240 ns between the header and sub-pulse 1.1. This hole can be used to kick out the header there, without perturbing the sub-pulse. The same can in principle be applied to all “odd” sub-pulses. On the other hand, this cannot be applied to the “even” sub-pulses, which at the exit of the delay loop would be recombined with their headers (or with what is left of them) Therefore the production of a train consisting of an isolated “even” sub-pulse is problematic. (If required, a solution could be to install a kicker and a dump in the delay loop)

13 Scenarios Assumption: the header consists of 121 bunches Assumption: we have complete control on the “header-out” kicker in front of “odd” sub-pulses; but we cannot dispose of a header in front “even” sub-pulses. For each case we will show the “Reduction factors” for the entire drive beam and for each produced train. The nominal Drive Beam consists of 24x24x121 = 68696 bunches, and each train has 2904 bunches. Therefore: Drive Beam reduction factor = 68696/#bunches (total) Train reduction factor = 2904/#bunches per train

14 Case 1 Produce a short train (X bunches) probing only one recombination path (n, with n odd). Ex. X = 30 Path 1, Train 1 : send header + bunches 1.1.1 to 1.1.30 Path n, Train 1 : send bunches 1.(n-1).1* to 1.(n-1).121 (header), followed by 1.n.1 to 1.n.30 Reduction factors: Drive Beam ~ 2300, Train ~ 97 Note: this train would not be able to accelerate the main beam (its bunch density is too low). But this would be the safe beam to be sent to a decelerator sector when the machine is unknown *note: this number is 122-header length

15 Case 2 Produce 24 short trains (X bunches) probing only one recombination path (n, with n odd). Ex. X = 30 Path 1 : – Send header + bunches 1.1.1 to 1.1.30 – Send 1.24.1 to 1.24.121 (header) followed by 2.1.1 to 2.1.30 –..... – Send 23.24.1 to 23.24.121 (header) followed by 24.1.1 to 24.1.30 Path n: – send bunches 1.(n-1).1 to 1.(n-1).121 (header), followed by 1.n.1 to 1.n.30 – send bunches 2.(n-1).1 to 2.(n-1).121 (header), followed by 2.n.1 to 2.n.30 –..... – send bunches 24.(n-1).1 to 24.(n-1).121 (header), followed by 24.n.1 to 24.n.30 Reduction factors: Drive Beam ~ 97, Train ~ 97 Cannot accelerate the main beam Requires dumping 24 headers

16 Case 3 Produce 24 short trains (X bunches) probing all odd recombination paths (one path per train). Ex. X = 30 – Send header + bunches 1.1.1 to 1.1.30 (path 1) – Send 2.2.1 to 2.2.121 (header), followed by 2.3.1 to 2.3.30 (path 3) – Send 3.4.1 to 3.4.121 (header), followed by 3.5.1 to 3.5.30 (path 5) –..... – Send 12.22.1 to 12.22.121 (header) followed by 12.23.1 to 12.23.30 (path 23) – Send 12.24.1 to 12.24.121 (header) followed by 13.1.1 to 13.1.30 (path 1) –..... – Send 24.22.1 to 24.22.121 (header) followed by 24.23.1 to 24.23.30 (path 23) Reduction factors: Drive Beam ~ 97, Train ~ 97 Cannot accelerate the main beam Requires dumping 24 headers

17 Case 4 Produce 1 short train (X bunches) containing bunches from 4 recombination paths. Ex. X = 30, Paths 1,7,15,23 – Send header + bunches 1.1.1 to 1.1.30 (path 1) – Send 1.6.1 to 1.6.121 (header), followed by 1.7.1 to 1.7.30 (path 8) – Send 1.14.1 to 1.14.121 (header), followed by 1.15.1 to 1.15.30 (path 15) – Send 1.22.1 to 1.22.121 (header), followed by 1.23.1 to 1.23.30 (path 22) Reduction factors: Drive Beam ~ 575, Train ~ 24 Cannot accelerate the main beam Requires dumping 4 headers Maybe no advantages compared with using consecutive recombination path (which need only one header)

18 Case 5 To try an “even” sub-pulse, we have to send also the entire preceding odd sub-pulse Produce 1 train (121 bunches) containing bunches from 2 consecutive recombination paths. Ex. Paths 1,2 – Send header + bunches 1.1.1 to 1.1.121 (path 1) + 1.2.1 to 1.2.121 (path 2) Reduction factors: Drive Beam ~ 288, Train ~ 12 Cannot accelerate the main beam

19 Case 6 Produce a short train (X bunches) containing all recombination paths. Ex. Train = 1, X = 30 1.Send header + bunches 1.1.1 to 1.1.30 (path 1) 2.Send 1.1.1 to 1.1.121 (header), followed by 1.2.1 to 1.2.30 (path 2) 3.etc. note the overlap between 1) and 2): bunches 1.1.1 to 1.1.30, appearing as header in 2), should not be discarded. The scenario only works if we can control the timing of the “header-out” kicker. Alternatively (and more simply) we could use the header-out kicker just for the header before the initial pulse, and cut the tail of the train by some other mean (option d) Reduction factors: Drive Beam ~ 96, Train ~ 4 Unfortunately this cannot accelerate properly a shortened pulse of the main beam: the Main Linac RF filling time is ~ 100ns Requires dumping the tail of the train

20 Case 7 (minimal train usable for acceleration) Produce a short train (X bunches) suitable for acceleration (i.e. Containing all recombination paths). Ex. Train = 1, X = 55 1.Send header + bunches 1.1.1 to 1.1.55 (path 1) 2.Send 1.1.1 to 1.1.121 (header), followed by 1.2.1 to 1.2.55 (path 2) 3.etc. note the overlap between 1) and 2): bunches 1.1.1 to 1.1.55, appearing as header in 2), should not be discarded. The scenario only works if we can control the timing of the “header-out” kicker. Alternatively (and more simply) we could use the header-out kicker just for the header before the initial pulse, and cut the tail of the train by some other mean (option d) Reduction factors: Drive Beam ~ 50, Train ~ 2 Can accelerate a shortened pulse of the main beam (10ns) Requires dumping the tail of the train (lots of power) BUT: IS IT REALLY WORTH THE TROUBLE? JUST FOR A FACTOR 2? WE SHOULD BETTER SEND THE FULL TRAIN

21 From pilot to nominal: ramping up the intensity 1.Case 1, 30 bunches, try all odd sub-pulses one after the other 2.Case 1,121 bunches, try all odd sub-pulses one after the other 3.Case 5, 2 consecutive (odd/even) sub-pulses, try all of them 4.Case 3, 24 trains of 30 bunches each (1/96 of nominal) 5.Case 3, 24 trains of 121 bunches each (1/24 of nominal beam) 6.Case 5x24, 24 trains of 121x2 bunches each (1/12 of nominal beam) 7.Add other two sub-pulses to each train (1/6 of nominal beam) 8.Add other four sub-pulses to each train (1/3 of nominal beam) 9.Add other four sub-pulses to each train (1/2 of nominal beam) 10.Nominal beam (1-3 to the dump, 4-10 to the decelerator)

22 An additional problem with the header When the first bunches of the normal header (-120..0) arrive in the Linac, the RF is not yet providing full power. This is done on purpose by delaying the RF powering, to avoid that the first bunches get an energy too high and are prematurely lost before the end of the Linac (if this happened, the header will be shortened and the last cavities would not be in stable state when the real bunch arrives). If now we want to send other headers during the pulse (i.e. Every time we want to produce in parallel 24 trains of reduced intensity) the conditions are not the same, and things might not go as expected (i.e. Part of the header can be lost before the end of the Linac, and the last cavities might not be in a stable state). We need to be able to produce the 24 trains in parallel, otherwise some decelerator sectors will go back to the “Unknown” state. To mitigate this problem we have different options:

23 An additional problem with the header (2) Apply the same gymnastic to the RF: but this is not easy, and it might make things unstable Use longer headers (in the middle of the pulse); but we must be absolutely sure that the first part of the header will be lost in the Linac, otherwise we have to get rid of it later. If the first part (extension) of this longer header is made of standard bunches, it would have the odd parity and would go through the delay loop. A solution would be if we had complete control also on which buckets are populated. In this case, we would put the extension inside even buckets, and we would get rid of it like the standard header

24 Conclusion (requirements) We need to have control on the production of individual sub-pulses (or better, of individual bunches). We could also need to have fine control on which buckets are populated (to get rid of intermediate header extension). A header before a odd sub-pulse can be easily discarded after the delay loop; we need to be able to do it 24 times in a pulse. A header before an even sub-pulse cannot be easily discarded (we would need a kicker/dump in the delay loop). We probably do not need to be able to cut the tail of the trains after the Linac (lot of problems to gain just a factor 2). We need to check that different systems (measurements, Linac RF, Decelerator), optimized for a nominal beam, behaves meaningfully with our reduced beams.


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