Doubling Main Injector Beam Intensity using RF Barriers
Doubling Main Injector Beam Intensity using RF Barriers
K.Y. Ng
K.Y. Ng
Fermi National Accelerator Laboratory, P.O. Box 500, Batavia, IL 60540
Fermi National Accelerator Laboratory, P.O. Box 500, Batavia, IL 60540
Abstract. Using rf barriers, 12 booster batches can be injected into the Fermilab Main Injector continuously,
Abstract.
Using
barriers,
booster After
batches
canadiabatic
be injected
into of
thethe
Fermilab
Main
continuously,
thus
doubling
the rfusual
beam12
intensity.
that,
capture
beam into
53Injector
MHz buckets
can be
thus doubling in
theabout
usual10beam
intensity.
that,
adiabatic
capture
of theare
beam
into
MHz
can be
accomplished
ms. The
beam After
loading
voltages
in the
rf cavities
small
and53they
canbuckets
be eliminated
accomplished
in about
10 ms. The beam
loading voltages
by
a combination
of counterphasing
and mechanical
shorts.in the rf cavities are small and they can be eliminated
by a combination of counterphasing and mechanical shorts.
THE INJECTION METHOD
THE INJECTION METHOD
Instead of slip-staking [1], a method to double beam inInstead of slip-staking [1], a method to double beam intensity
in the Main Injector by rf barriers without adiatensity in the Main Injector by rf barriers without adiabatic bunch compression was proposed by Griffin [2]. A
batic bunch compression was proposed by Griffin [2]. A
booster batch of protons of length Tb and full fractional
booster batch of protons of length T and full fractional
momentum spread ∆, denoted by 1 inbFig. 1(a) is injected
momentum spread A, denoted by 1 in Fig. l(a) is injected
into the Main Injector at a negative momentum offset, so
into the Main Injector at a negative momentum offset, so
that
that the
the highest
highest fractional
fractional momentum
momentum offset
offset is
is δ8i1tl and
and the
the
lowest
fractional
momentum
offset
is
δ
=
δ
−
i2
i1
lowest fractional momentum offset is Si2 = 8tl — ∆.
A. The
The
longitudinal
longitudinal position
position of
of the
the injection
injection is
is so
so chosen
chosen that
that
the
just touches
of aa
the right
right side
side of
of the
the batch
batch just
touches the
the left
left side
side of
square
square barrier
barrier B
B of
of width
width T7\1 and
and magnitude
magnitude V
V,, which
which is
is
moving
to
the
left
at
the
speed
of
Ṫ
=
T
/(2T
moving to the left at the speed of T22 = Tbb/(2Tcc)) where
where
TTcc is
is the
the booster-cycle
booster-cycle time.
time. The
The momentum
momentum offset
offset of
of the
the
batch
injection
is
so
chosen
that
protons
of
highest
batch injection is so chosen that protons of highest enenergy
right
ergy on
on the
the left
left1 side
side of
of the
the batch
batch will
will drift
drift to
to the
the right
at
Tb per
at the
the speed
speed of
of 2\T
per booster
booster cycle
cycle or
or −
—ṪT22.. Thus,
Thus, after
after
one
one booster
booster cycle,
cycle, another
another batch
batch marked
marked 22 in
in Fig.
Fig. 1(b)
l(b)
can be injected again with its right edge touching the left
side
side of the barrier. In other words, at every booster cycle,
aa new
new booster
booster batch can be injected with the injection
point moved half a booster-batch length to the right. The
linear beam density can therefore be doubled.
The only parameter here is the strength of the barrier,
Β
(a)
(a)
which is so chosen that after the booster batch emerges
whichthe
is so
chosen
after and
the booster
batch emerges
from
barrier,
thethat
highest
lowest energies
of the
from the barrier, the highest and lowest energies of the
batch become symmetric about the nominal energy of
batch become symmetric about the nominal energy of
the accelerator ring. This is necessary because we need
the accelerator ring. This is necessary because we need
to place another stationary barrier on the right side of the
to place another stationary barrier on the right side of the
moving barrier in order to limit the longitudinal motion
moving barrier in order to limit the longitudinal motion
of the protons after passing through the moving barrier
of the protons after passing through the moving barrier
so as to guarantee empty spaces along the Main Injector
so as to guarantee empty spaces along the Main Injector
for
future batch
batch transfer
transfer from
from the
the Booster.
Booster. We
We use
use aa
for future
rectangular
barrier
here
just
for
simplicity,
although
the
rectangular barrier here just for simplicity, although the
results
depend
only
on
the
integrated
strength
of
the
results depend only on the integrated strength of the
barrier.
barrier.
To derive
of the
the barrier,
barrier, let
let us
us go
go to
to aa more
more
To
derive the
the strength
strength of
general
scenario
of
having
the
barrier
moving
to
the
left
general scenario of having the barrier
moving to the left
1
at
rate of
xTbb/T
/Tcc (x
at the
the rate
of Ṫ
T22 =
= xT
(x =
=\
above).Suppose
Supposethe
the
2 ininabove).
first
proton with
with momentum
momentum spread
spread δ5i1a exits
exits the
the barrier
barrier
first proton
in
the A^
N1 turns.
turns. Its
Its phase
phase drift
drift in
in time
time towards
towards the
the right
right is
is
in the
Z N
1
T0 (nδ+nA8)dn,
[\Tη
i1 + n∆δ )dn ,
0(S
JO0
(1)
(1)
where η
factor, TTQ0 is the
the revolution
revolution period,
period,
7] is the slip factor,
22 E), e is the proton charge, E is the on∆
δ
=
eV
/(
β
A5 =
/(f$ E\
E
β c its longitudinal
longitudinal velocvelocmomentum proton energy with /3c
ity and c the velocity of light. At this moment, the barrier
barrier
left by the
the phase A^
N1TṪ22TTQ0.. Because
has moved towards the left
T1 , we must have
the width of the barrier is 7\
Z N
1
r]T
T1 − N1 Ṫ2 T0 = (Nl η
T00((8δni1 + n∆δ )dn ,
(2)
J0
8,7
δ
i1
1
δ
8*2
i2
t
!
Β
(b)
(b)
δ f1
1
δ f2
2
FIGURE 1.
1. (a)
(a) Booster
Booster batch
batch 11 is
is injected
injected at
at negative
negative momoFIGURE
mentum offset
offset with
with the
the right
right edge
edge just
touching aa left-moving
left-moving
mentum
just touching
barrier B.
B. The
The vertical
vertical arrows
arrows are
are spaced
spaced one
one booster-batch
booster-batch
barrier
length TTb apart.
apart, (b)
(b) After
After aa booster
booster cycle,
cycle, the
the barrier
barrier moves
moves by
length
by
b
half
a
booster-batch
length
while
all
particles
in
batch
drift
half a booster-batch length while all particles in batch 11 drift
past the left edge of the barrier. Exiting the barrier, the batch
past the left edge of the barrier. Exiting the barrier, the batch
has maximum and minimum momenta, 5^ and 5/- , equal and
has maximum and minimum momenta, δ f 1 and δ f 22, equal and
opposite. Another booster batch 2 is now injected.
opposite. Another booster batch 2 is now injected.
from which A^
from
N1 can be solved. The exit momentum spread
spread
of this particle is therefore
s
2
Ṫ2
Ṫ2
2T ∆δ
(3)
−
.
δf1 =
δi1 −
− 1
|η |
|η |
|η |T0
The final
final fractional
of the protons
The
fractional momentum
momentum offset
offset δ5™
f 2 of the protons
with
the
lowest
initial
momentum
spread
can be
with the lowest initial momentum spread 8
δi2
i2 can be
obtained
similarly.
Equating
the
two,
we
arrive
obtained similarly. Equating the two, we arrive at
at the
the
strength
of
the
barrier
strength of the barrier
" #
#"
T0 Tb2 β 2 E
x∆|η | 2 2 , (4)
∆|η | 2
V T1 =
1− 2t
1+
−x , (4)
2|η |Tc2
2Ṫ2
2Ṫ2
where use
where
use has
has been
been made
made of
of the
the fact
fact that
that the
the protons
protons with
with
initial
momentum
spread
3
must
drift
to
the
tl
right at
at aa
initial momentum spread δi1 must drift to the right
rate
of
(1
—x)T
b/Tc so that a next batch can be injected
rate of (1 − x)T /T so that a next batch can be injected
b
c
= −(1
-(l-x)t
2/(x\ri\).
− x)
Ṫ2/(x|η |).
or δSi1n =
th
CP642, High Intensity and High Brightness Hadron Beams: 20 ICFA Advanced Beam Dynamics Workshop on
High Intensity and High Brightness Hadron Beams, edited by W. Chou, Y. Mori, D. Neuffer, and J.-F. Ostiguy
© 2002 American Institute of Physics 0-7354-0097-0/02/$ 19.00
226
Obviously, a larger initial fractional momentum spread
a larger
momentum
∆ of Obviously,
the beam will
leadinitial
to a fractional
larger final
fractionalspread
moA
of
the
beam
will
lead
to
a
larger
final
fractionalmomomentum spread. However, when the initial fractional
mentum
spread.
However,
when
the
initial
fractional
momentum spread is large enough, protons with the highest
mentum spread
is largeso
enough,
with the
momentum
will acquire
much protons
energy from
thehighest
movsospeeds
much energy
from
the movingmomentum
barrier thatwill
theiracquire
drifting
to the left
exceed
the
ing barrier
theirbarrier.
driftingThese
speedsparticles
to the leftwill
exceed
the
speed
of the that
moving
not be
speed
of the from
moving
not be
able
to emerge
thebarrier.
movingThese
barrierparticles
and thiswill
injection
able tofails.
emerge
from
the moving
injection
method
This
happens
when barrier
Ṫ2 < |ηand
|δ fthis
1 , resulting
fails.strength
This happens
when Tbarrier
2 < \T]\Sfl9 resulting
in method
the critical
of the moving
in the critical strength of the moving
barrier
T0 Tb2 β 2 E
,
(5)
(V T1 )c =
2e|η |Tc2
(5)
which is independent of x, and the maximum allowable
which
independent
of ;c, andspread
the maximum
allowable
initial
fullisfractional
momentum
of the beam
is
initial full fractional
spread of the beam is
rmomentum !
1 1 2Tb
21; .
x2 + −
∆c =
(6)
4 2 |η |Tc
Ac =
(6)
For a given full fractional momentum spread ∆ of the
For a given full fractional momentum spread A of the
beam, the speed of the moving barrier must be faster than
beam, the speed of the moving barrier must be faster than
xTb /Tc where
xTb/Tc where s
|η |Tc ∆ |η |Tc ∆
l\r]\TcAf\rj\TcA + 1 .
(7)
x=
x = 2Tb
+1
(7)
2Tb
2T
2T
EMITTANCE
EMITTANCEINCREASE
INCREASE
The
Theposition
positionofofinjection
injectionalong
alongthe
theaccelerator
acceleratorring
ring
slips
to
the
left
by
xT
for
each
injection
b b for each injection or
slips to the left by xT
or for
for each
each
booster
cycle.
bb
booster
cycle,implying
implyingthat
thata awhole
wholebatch
batchofoflength
lengthTT
of of
beam
can
be
injected
in
the
length
xT
.
Therefore
b b. Thereforethe
beam can be injected in the length xT
the
ratio
ofof
final
initial
ratio
finaltoto
initialemittances
emittancesisis
δf1 × 1
|η |Tc ∆
re r= =∆ 1 = 1 +
,
(8)
1
(8)
e
2T27;
×
b
2
x
which
is is
independent
which
independentofofx.x.Here,
Here,we
weassume
assumecircumfercircumference
of
the
accelerator
ring
is
infinitely
long
ence of the accelerator ring is infinitely longsosothat
thatinjecinjection
can
be
continued
forever.
On
the
other
hand,
tion can be continued forever. On the other hand,the
thefinal
final
linear
density
linear
densitydoes
doesincrease
increasebybythe
thefactor
factorrdrd==1/x.
1/x.For
For
this
reason,
wewe
should
this
reason,
shouldchoose
choosethe
thelowest
lowestbarrier
barriermoving
moving
speed
speedallowable
allowablebybythe
theinitial
initialmomentum
momentumspread
spreadofofthe
the
batch.
batch.InInother
otherwords,
words,given
given∆,A,we
weshould
shouldstick
sticktotothe
the
critical
x given
bybyEq.
T1 )c)cfor
critical
x given
Eq.(7)(7)and
andthe
thecritical
criticalsize
size(V(VTj
for
thethe
barrier
given
bybyEq.
barrier
given
Eq.(5).
(5).
with the lowest energy are injected with offset ∆Ei2 =
1
with
the lowest
energy are injected with offset AEi2 =
∆E
i1 − 2∆E 2 i = −21.6276 MeV. After exiting the movAE
2AE
i,=
-21.6276
After
exiting
the moving barrier,
their final
energyMeV.
offsets
are ∆E
a
f 1 = −∆Ei1 =
ing
barrier,
their
final
energy
offsets
are
AE=
— AE
11.8285 MeV and ∆E f 2 = −∆E f 1 = −11.8285
MeV.
a =
11.8285
MeV
and
A£,
=
-AE
=
-11.8285
MeV.
Thus, the energy spread has
been increased
by the factor
2
fl
Thus,
the1energy spread has been increased by the factor
∆E
f 1 /∆E 2 i = 11.8285/4.8995 = 2.419 and this is also
AE^/AEi,= 11.8285/4.8995
2.419 andtothis
also
the
ratio of the
final longitudinal=emittance
theisinitial
the ratio of the
final longitudinal
emittance
to the initial
longitudinal
emittance.
Figure 2 shows
the situation
one
longitudinal
Figure 2 shows
one
booster
cycleemittance.
after the injection
of the the
12thsituation
batch. The
booster cycle
after the
of the
batch.colors
The
consecutive
bunches
areinjection
represented
by 12th
different
consecutive
bunches are
by different
colors
for
easy identification.
Therepresented
vertical dashed
lines separate
for ring
easy into
identification.
vertical
lines
separate
the
7 boosterThe
lengths
anddashed
the first
batch
was
the
ring
into
7
booster
lengths
and
the
first
batch
injected at the location between 5.5 and 6.5Tb . Thewas
reinjectedbarrier
at the islocation
between
5.5solid
and square
6.57^. The
reflecting
represented
by the
at locaflecting
is represented
by theissolid
square atby
location
7Tb , barrier
while the
injection barrier
represented
the
tion
7T
,
while
the
injection
barrier
is
represented
by
b
hollow square with the arrow pointing to its directionthe
of
hollow Theoretical
square with reasoning
the arrow pointing
to its direction
of
motion.
and simulations
show that
motion.
reasoning
simulations
show
about
10 Theoretical
ms will be required
to and
capture
the beam
intothat
the
about
10
ms
will
be
required
to
capture
the
beam
into
the
53-MHz buckets.
53-MHz
buckets.
Barriers are usually used to confine a beam within a
Barriers are usually used to confine a beam within a
designated region along the accelerator ring. The intedesignated region along the accelerator ring. The integrated sizes of the barriers need not be accurate as long
grated sizes of the barriers need not be accurate as long
as they are large enough to confine the particles of the
as they are large enough to confine the particles of the
highest and lowest energies. However, this is not true
highest and lowest energies. However, this is not true
here. Although the injection process does not depend on
here. Although the injection process does not depend on
the
has to
to
theshape
shapeof
ofthe
themoving
moving barrier,
barrier, its
its integrated
integrated size
size has
be
accurate
in
order
to
have
the
final
positive
and
negbe accurate in order to have the final positive and negative
ativeenergy
energy spreads
spreads be
be equal
equal and
and opposite.
opposite. Otherwise,
Otherwise,
after
reflection
by
the
stationary
barrier,
enafter reflection by the stationary barrier, the
the positive
positive energy
spread
will
be
modified,
resulting
possibly
in
an
ergy spread will be modified, resulting possibly in an
increase
in
the
longitudinal
emittance.
Here,
we
would
increase in the longitudinal emittance. Here, we would
like
plot of
of
liketotostudy
study the
the effect
effect of
of such
such an
an error.
error. The
The left
left plot
Fig.
3
shows
the
situation
when
V
T
is
10%
larger
than
Fig. 3 shows the situation when VT^1 is 10% larger than
its
of
its critical
critical value,
value, 22 booster
booster cycle
cycle after
after the
the injection
injection of
the
12th
bunch.
Actually
the
final
momentum
spread
bethe 12th bunch. Actually the final momentum spread becomes
emerge
comessmaller.
smaller. However,
However, many
many particles
particles cannot
cannot emerge
from
The
fromthe
thestrong
strongbarrier
barrier resulting
resulting in
in aa loss
loss of
of 10.6%.
10.6%. The
situation
T1 by
situationof
of aareduction
reduction of
of VV7\
by 10%
10% from
from its critical
value
will
valueisis shown
shown in
in the
the right
right plot.
plot. Now all particles will
emerge
emergefrom
fromthe
themoving
movingbarrier
barrier earlier
earlier and their energies
APPLICATION
APPLICATIONTO
TOMAIN
MAININJECTOR
INJECTOR
Simulations
were
performedfor
fordoubling
doublingthe
theMain
MainInInSimulations
were
performed
jector
intensity
injectionenergy
energyofofEE==8.938
8.938GeV
GeV
jector
intensity
at at
thethe
injection
critical
conditionforforbarrier
barriermovement
movementofof12 \T
per
Tbbper
at at
thethe
critical
condition
1
booster
cycle
withthe
theinitial
initialhalf
halfcritical
criticalhalf
halfenenbooster
cycle
(x(x
==2 )^)with
1 i i,ergy
spreadofof∆EAE
=4.900
4.900MeV.
MeV.Here,
Here,TbTb==1.59
1.59µjus,
=
ergy
spread
s,
2
0.667
ms,
T}
=
-0.00892
and
the
criti0 =
bT,c T=
c =
T0 T=
7T7T
,
0.667
ms,
η
=
−0.00892
and
the
critib
barrier
strengthis is(V(VTjc
3.1421kVkV-jUs.
Particles
calcal
barrier
strength
T1 )c ==3.1421
µ s. Particles
withthethehighest
highestenergy
energyare
areinjected
injectedwith
withthe
theoffset
offset ofof
with
2 2£/|T]| = -11.8285 MeV while particles
a = -f 2 /3
∆EAE
i1 = −Ṫ2 β E/|η | = −11.8285 MeV while particles
2
4
6
8
1O
Distance Along Ring in /u.s
FIGURE 2.2. (color)
(color) Longitudinal
Longitudinal phase
phase space
space one
one booster
booster
FIGURE
cycleafter
afterthe
theinjection
injection of
of the
the 12th
12th batch
batch between
between
cycle
227
I
"3
o.ooo - — —
O
2
4
6
8
—O.OO2
2
4
6
8
Distance Along Ring in JLLS
Distance Along Ring in
FIGURE3.3. (color)
(color)Plots
Plotsshowing
showing22booster
booster cycle
cycle after
after injection
injection of
of 12th
12th bunch,
bunch, with
with the
the strength
strength of the moving barrier (left)
(left)
FIGURE
increasedbyby10%
10%and
and(right)
(right)decreased
decreasedby
by10%.
10%.
increased
aretherefore
thereforereduced,
reduced,which
whichresults
resultsininan
anincrease
increasein
inmomoare
mentumspread
spreadby
by9.8%
9.8%after
after reflection
reflection by
by the
the stationstationmentum
arybarrier.
barrier.AAsmall
smallamount
amountparticles
particleshave
haveenergies
energieshigh
high
ary
enoughthat
thattheir
theirleft-drift
left-drift velocities
velocitiesafter
after reflection
reflection are
are
enough
largerthan
thanthat
thatofofthe
themoving
movingbarrier.
barrier. Eventually,
Eventually, they
they
larger
maycatch
catchup
upwith
withthe
themoving
movingbarrier
barrierand
andget
getlost.
lost.
may
BEAMLOADING
LOADING
BEAM
Because ofof the
the high
high beam
beam current,
current, the
the problem
problem of
of
Because
beamloading
loadingmust
mustbe
beexamined.
examined.At
Atthe
the passage
passage across
across
beam
theNNc c==1818rfrfcavities,
cavities,aavery
very short
short bunch
bunch containing
containing
the
1010protons
6.0×x1010
protonswill
willinduce
induceaatotal
totalinstantaneous
instantaneousbeam
beam
6.0
loading voltage
voltage VVbQ == qN
qNccωcor RrRLL/Q
/QLL =
= 5.5
5.5 kV,
kV, where
where
loading
b0
500kΩ,
kQ,QQL L==5000,
5000,and
andωco
52.8MHz
MHzare,
are,
r/(2n)
π ) ==52.8
RRL L==500
r /(2
respectively,the
theloaded
loadedshunt
shunt impedance,
impedance, loaded
loaded qualqualrespectively,
ityfactor,
factor,and
andresonant
resonantfrequency
frequency of
of each
each cavity.
cavity. For
For aa
ity
batch
of
84
such
bunches
injected
into
the
Main
Injecbatch of 84 such bunches injected into the Main Injector,the
thebeam
beamloading
loading voltage
voltage after
after the
the passage
passage of
of the
the
tor,
last
bunch
increases
to
444
kV.
Thus
the
difference
in
last bunch increases to 444 kV. Thus the difference in
beam
loading
voltages
experienced
by
the
last
and
first
beam loading voltages experienced by the last and first
bunchesisis 439
439 kV.
kV.Taking
Taking into
into account
account of
of the
the finite
finite
bunches
lengths
of
the
bunches,
this
difference
becomes
388
kV
lengths of the bunches, this difference becomes 388 kV
when
steady
state
is
reached.
During
normal
operation,
when steady state is reached. During normal operation,
thetotal
totalrfrfvoltage
voltageatatinjection
injectionisis1.2
1.2MV.
MV.IfIfthe
thedesigned
designed
the
synchronousphase
phaseφ05== 00isissynchronized
synchronized to
to the
the midmidsynchronous
s
dle
bunch
of
the
batch,
the
rf
phase
errors
introduced
bedle bunch of the batch, the rf phase errors introduced become
A
0
=
±9.18°
for
the
first
and
last
bunches.
Even◦
come ∆φs 5= ±9.18 for the first and last bunches. Eventually,the
thelongitudinal
longitudinalemittances
emittancesof
ofthe
thebunches
buncheswill
will be
be
tually,
increased
up
to
18%,
which
is
considered
still
tolerable.
increased up to 18%, which is considered still tolerable.
Duringslip-stacking,
slip-stacking, although
although the
the bunch
bunch density
density isis
During
doubled,
the
beam
loading
voltage
may
not
increase
doubled, the beam loading voltage may not increase
because
the
bunch
spread
out
to
almost
fill
the
buckets
because the bunch spread out to almost fill the buckets
at
the
very
low
rf
voltage
of
~
64
kV,
almost
times
at the very low rf voltage of ∼ 64 kV, almost 77 times
less
than
the
difference
in
beam
loading
voltages
seen
by
less than the difference in beam loading voltages seen by
the first and last bunches. Such low rf voltage is required
the first and last bunches. Such low rf voltage is required
in order to allow two series of buckets to stay within
in order to allow two series of buckets to stay within
the momentum aperture of the ring. The rf phase errors
the momentum aperture of the ring. The rf phase errors
now become so large that the most beam particles will
now become so large that the most beam particles will
be driven out of the buckets. Feedforward and feedback
be driven out of the buckets. Feedforward and feedback
have been planned to compensate for the beam loading.
have been planned to compensate for the beam loading.
In the
the present
present continuous
continuous multiple injection
injection scheme
In
using barriers,
barriers, the
the booster bunches before injection
using
injection will
be made
made very
very long
long in the Booster so that the momentum
be
momentum
spread can
can be
be reduced. These bunches start to debunch
spread
debunch
immediately after
after injected
injected into the Main Injector.
immediately
Injector. For
For
completely uniformly
uniformly distributed
distributed beam, there is no
aa completely
53 MHz
MHz rf
rf component
component and therefore no beam loading
53
loading at
all.
Here,
although
the
distribution
is
not
uniform
all. Here, although
uniform (see,
(see,
for
example,
Fig.
2),
the
5
3-MHz
rf
component
of the
for example,
53-MHz
beam current
current at
at any
any moment of the injection
injection is found
beam
found
to be
be less
less than
than 0.25%
0.25% of twice the dc beam current,
to
current,
implying that
that the
the beam loading voltage will be reduced
implying
reduced
at least
least 400
400 times.
times.
at
Actually no
no rf
rf voltage
voltage is
is required
required in
Actually
in this
this injection
injection
scheme. The
The Main
Main Injector
Injector is
is presently
presently equipped
equipped with
scheme.
with
fast mechanical
mechanical shorts,
shorts, which
which can
can be
be inserted
100 ms
fast
inserted in
in 100
ms
and removed
removed in
in 50
50 ms
ms [4].
[4]. To
To cope
cope with
with beam
and
beam loading
loading
and arrive
arrive at
at zero
zero rf
rf voltage,
voltage, first,
first, the
the rf
16 caviand
rf drive
drive of
of 16
cavities
is
turned
off
and
shorts
are
inserted.
Second,
counterties is turned off and shorts are inserted. Second, counterphasing isis used
used for
for the
the two
two remaining
remaining cavities
phasing
cavities to
to arrive
arrive at
at
zero
rf
voltage
with
the
tiny
beam
loading
voltage
zero rf voltage with the tiny beam loading voltage taken
taken
into consideration.
consideration. Further
Further fine
fine adjustment
adjustment can
into
can be
be impleimplemented
using
a
fast
low-level
feedback.
Notice
that
mented using a fast low-level feedback. Notice that councounterphasing isis not
not possible
possible for
for slip-stacking
slip-stacking because
terphasing
because the
the
two rf
rf controls
controls have
have been
been employed
employed already
already to
two
to produce
produce
the two
two series
series of
of buckets
buckets at
at slightly
slightly different
the
different frequencies.
frequencies.
REFERENCES
REFERENCES
C. Ankenbrandt,
Ankenbrandt, “SLIP-STACKING”:
"SLIP-STACKING": A
1.1. C.
A New
New Method
Method
of Momentum-Stacking,
Momentum-Stacking, Fermilab
Fermilab Report
Report FN-352,
1981;
of
FN-352, 1981;
F.E. Mills, Stability of Phase Oscillations under two
F.E. Mills, Stability of Phase Oscillations under two
Applied Frequencies, BNL Report AADD 176, 1971.
Applied Frequencies, BNL Report AADD 176, 1971.
2. J. Griffi n, Momentum Stacking in the Main Injector using
2. J. Griffin, Momentum Stacking in the Main Injector using
Longitudinal Barriers, Fermilab Report, 1996.
Longitudinal Barriers, Fermilab Report, 1996.
3. K.Y. Ng, Continuous Multiple Injections at the Main
3. K.Y. Ng, Continuous Multiple Injections at the Main
Injector, Fermilab Report FN-715, 2002.
Injector, Fermilab Report FN-715, 2002.
4. D. Wildman, Transient Beam Loading Reduction during
4. D. Wildman, Transient Beam Loading Reduction during
Multi-Batch Coalescing in the Fermilab Main Ring,
Multi-Batch
the Fermilab
Main Ring,
Proceedings Coalescing
of the 1991inIEEE
Particle Accelerator
Proceedings
1991 IEEE
Accelerator
Conference, of
Santhe
Francisco,
MayParticle
6-9, 1991,
p. 410.
Conference, San Francisco, May 6-9, 1991, p. 410.
228