Download PPT - KEK

Fast orbit bump magnet
• Use of magnetic field varying with time
Multi-turn septum injection
Orbit shift for phase-space painting of H- injection
• Use of pulse magnetic field at the peak value
Orbit shift close to the septum magnet for a fast extraction
• Use of pulse magnetic field at flat-top
Chicane bump for H- injection
Orbit shift close to the septum magnet for a slow extraction
1
Orbit shift multi-turn injection by septum magnet
Fig. 1 The principle of the multi turn injection
Use of decay field by critical damping
The principle of the power supply circuit and its waveform are shown in Fig.2
The critical damping of the circuit is given as, R  4L C
The excitation current is given by next equation,
i  V0
C
 Rt 
 exp   
L
 L
Fig.2 Principle of the circuit
Fast decay
1μs/div, 2V/div (50A/V)
Slow decay
Fig.3 Actual power supply circuit
1μs/div, 2V/div (50A/V)
Half sine wave by LC circuit for the use of peak value.
Short time orbit-shift within the td
Half sine
Voltage recover
Voltage recover
Combination of LC resonant circuit and LR damping circuit
Principle of the circuit
Actual power supply circuit
5μs/div, 5V/div (1kA/V)
2μs/div, 5V/div (1kA/V)
5μs/div, 5V/div (1kA/V)
5μs/div, 5V/div (1kA/V)
Fast orbit bump magnet for orbit shift multi-turn injection
• Fast decay time (3~6 μs)
• Ferrite is used for the core material.
• Swing of the magnetic field is not allowed (for injection)
Characteristics of Ferrite (Fe2O3)
Frequency characteristics
Temperature characteristics
Longitudinal field distribution (measured value)
Excitation characteristics
Caution !
For “window frame” and “H-type core”
• Shorted-magnetic circuit enclose
the beam.
• Magnetic resistance is very low.
• Strong magnetic field is induced
around bunched beams.
• Open-magnetic circuit
• Magnetic resistance is high.
• Magnetic field induced around
bunched beams is low.
C-type is better !
Orbit bump magnet
for Charge exchange injection
Stripping
Foil
Charge-exchange injection by chicane bump
magnets
Parameters of H- injection bump magnet for the KEK Booster
Cross section of the core with “end–slit”
Properties of core material
(0.1 mm Thick silicon steel, Nihon Kinzoku ST-100)
B-H characteristics
Iron loss
Excitation characteristics of the magnet
Longitudinal field distribution of
chicane bump magnets
Longitudinal field distribution
of single bump magnet
Waveform of injection bump magnets
(Use of magnetic field at the flat-top)
20μs/div, 2V/div, (1kA/V)
Pulse power supply by a pulse-forming-network (PFN)
Ladder-type
Line-type
Ladder-type
Rising phase of the wave form
V 
 z0  

i  1  exp   t  
z0 
 L 
Falling phase of the wave form
V
 z0

i  exp   t  td 
z0
 L

Pulse forming network for chicane bump magnets
(Using flattop field for injection)
PFN voltage, Magnet current and Magnet voltage
1ms/div, 5V/div (1kV/V)
50μs/div, 2V/div (1kA/V)
50μs/div, 0.5V/div
Fundamentals of Transmission Line Theory
“Exact transitional solution”
Let’s consider the part of transmission line as,
x
x + Δx
On the one side line, partial resistance and inductance per unit
length are (R/2) and (L/2) respectively. By the go and the return the
values become R and L.
The capacitance and conductance between two lines are defined as
C and G respectively.
Equations for v and i are given as,
finite difference equation.
v( x)  v( x  x)  i ( x) Rx  Lx
di ( x)
dt
i ( x)  i ( x  x)  v( x)Gx  Cx
dv( x)
dt
(1)
Divide both sides by Δx and in the limit of Δx→0,
we can get next differential equations.


v( x, t )
i ( x, t )
 Ri ( x, t )  L
x
dt
i ( x, t )
v( x, t )
 Gv( x, t )  C
x
t
(2)
These simultaneous partial differential equations are known as
“Telegraphy equation”
In the case of lossless transmission line, i.e. R = G = 0.
The telegraphy equation becomes
v
i
－ ＝L
x
∂t
i
v
－ ＝C
x
t
Here, v  v( x, t ) i  i ( x, t )
(3)
We can get wave equations. Here c  1
 2v
1  2v
＝ 2 2
2
x
c t
 2i
1 ∂2i
＝ 2 2
2
x
c ∂t
LC
(4)
The solution of Eq.(4) is given as,
v  v1 ( x－ct )  v2 ( x  ct )
(5)
v and i must satisfy the Eq.(3), we can get next solution for i,
i=
C
{v ( x－ct ) + v2 ( x + ct )}
L 1
(6)
Eq.(5) and (6) satisfies wave equation. Final solution can be obtained by
initial condition of “t” and boundary condition of “x”.
Here we define the initial value of “v” and “t“ as v(x,0) and i(x,0)
respectively. Then we perform Laplace transformation for Eq.(3) and (4).
dV
－
 sLI－Li( x,0)
dx
(7)
dI
－  sCV－Cv( x,0)
dx
2
∂v
2
2 d V
s V－c

sv
(
x
,
0
)

t 0
dx 2
∂t
d I
∂i

si
(
x
,
0
)

dx 2
∂t
2
s 2 I－c 2
(8)
t 0
For the case of initial values are zero. (or v(x,0)=0 and i(x,0)=0 )
V ( x, s) = V1 ( x, s) －( x c ) s + V2 ( x, s)  ( x c ) s
I ( x, s ) =
C
{V1 ( x, s) －( x c ) s－V2 ( x, s)  ( x c ) s
L
Eq.(9) is equivalent to Eq.(5) and Eq.(6).
(9)
In the Eq.(9), V1 and V2 are decided by boundary condition of the x.
When a voltage source e(t) is connected at x=0, The Laplace
transformation of e(t) is written as, L{e(t)}=E(s). For a current source i(t),
it is also as, L{i(t)}=I(s).
Those are,
at x=0, the voltage source e(t) is connected; V(0,s)=E(s)
at x=0, the current source i(t) is connected; I(0,s)=I(s)
The length of the transmission line is “ l ”
at x=l, the terminal is shorten; V(l,s)=0
at x=l, the terminal is open; I(l,s)=0
at x=l, Z(s) is connected; V(l,s) / I(l,s)=Z(s)
For example, a voltage source e(t) with internal impedance Z0(s) are
connected at x=0 as shown in Fig.
The conditional equation is,
V (0, s)  E (s)－Z0 (s) I (0, s)
Terminal is shorted-circuit as in Fig.
A electromotive force is
connected at x=0, and the terminal at x=l is shortened.
The boundary condition is,
at x=0 ; V(0,s)=E(s)
at x=l ; V(l,s)=0
From Eq.(9) first,
V (0, s)  V1  V2  E ( s)  L {e (t )}
V (l , s)  V1 
－( l / c ) s
 V2 
(l / c ) s
0
(10)
We can solve Eq.(10) for V1 and V2, and substitute them to Eq.(9),
the Laplace transform of the voltage v and current I is calculated as,
((l－x ) / c ) s－－((l－x ) / c ) s
V ( x, s ) =
E ( s)
(l / c ) s
－(l / c ) s

－
1 ((l－x ) / c ) s + －((l－x ) / c ) s
I ( x, s ) =
E ( s)
(
l
/
c
)
s
－
(
l
/
c
)
s
W

－
Here,
W=
L
C
(11)
(characteristic impedance)
(12)
After rearrangement of the Eq.(11), then expand it in a series,
V ( x, s )  E ( s )  －(l / c ) s ( (( l―x ) / c ) s－ (( l―x ) / c ) s )
(( 2 l―x ) / c ) s
 E ( s ) ( －( x / c ) s－ －
－
1
1－ －( 2l / c ) s
(( 2 l  x ) / c ) s
 －
－(( 4 l―x ) / c ) s

－(( 4 l  x ) / c ) s
－)
(13)
By the same procedure, we can get the I(x,s) as,
I ( x, s ) 
1
E ( s) (  －(l / c ) s   －(( 2l―x ) / c ) s   －(( 2l―x ) / c ) s )
W
(( 4 l―x ) / c ) s
 －
(( 4 l  x ) / c ) s
 －
 )
(14)
By inverse Laplace transformation
x
2l－x
2l  x
4l－x
4l  x
v( x, t )  e (t－ ) －e (t－
)  e(t－
) －e (t－
)  e (t－
)－
c
c
c
c
c
1
x
2l－x
2l  x
4l－x
4l  x
i ( x, t )  {e (t－ )  e (t－
)  e(t－
)  e(t－
)  e (t－
) 
W
c
c
c
c
c
(15)
Terminal is shorted-circuit
“For intuitive understanding”
Response for step voltage function
(Opposite phase reflection)
Terminal is shorted-circuit
Response for step current function
(Same phase reflection)
Terminal is open circuit as in Fig.
A electromotive force is
connected at x=0, and the terminal at x=l is opened.
The boundary condition is,
V (0, s)  V1  V2  E ( s)
(16)
1
I (l , s)  (V1 －(ｌ / c ) s－V2 (ｌ / c ) s）  0
W
We can solve Eq.(16) for V1 and V2, and substitute them to Eq.(9),
the Laplace transform of the voltage v and current I is calculated.
Then expand it in a series and next by inverse Laplace
transformation, we can get v(x,t) and i(x,t) as,
x
2l－x
2l  x
4l－x
4l  x
v( x, t )  e(t－ )  e(t－
)－e(t－
)－e(t－
)  e(t－
) 
c
c
c
c
c
1
x
2l－x
2l  x
4l－x
4l  x
i ( x, t )  {e(t－ )－e(t－
)－e(t－
)  e(t－
)  e(t－
)－}
W
c
c
c
c
c
(17)
Terminal is open-circuit
Response for step voltage function
(Same phase reflection)
Terminal is open-circuit
Response for step current function
(Opposite phase reflection)
Z(s) is connected to the terminal. The boundary condition is,
V (0, s ) = V1 + V2 = E ( s )
V1－(l / c ) s + V2 (l / c ) s
V (l , s )
=W
－(l / c ) s
(l / c ) s = Z ( s )
I (l , s )
V1
－V2
Here, we set Z(s)=R for the simplicity.
W is the characteristic impedance.
Z W
r
Z W
“reflection coefficient”
For Z = 0, the terminal is shorted circuit.
r = -1
For Z = ∞, the terminal is open circuit.
r=1
“Intuitive understanding”
Opposite phase
reflection
Same phase
reflection
Sum of the “go” and
“return” waves
Sum of the “go” and
“return” waves
Combined bump-septum magnet system for
negative-positive ion injection
Structure of combined bump-septum magnet
Magnetic field of the combined
bump-septum magnet
How to get a steep septum magnetic
field
Measured value of magnetic field
Change of the bump magnet field
by exciting the septum magnet
Change of the bump magnetic field
by exciting the septum magnet
Sep tum con ductor
1.02
1.015
Bum p-ON, Septum-ON
Bum p-ON, Septum-OFF
1.01
B/B
0
1.005
1
B=B
0.995
0
Normaliz ation point
(Center of bump magnet)
0.99
0.985
0.98
180
200
220
240
260
x (mm)
280
300
320
340
Figure of combined bump-septum magnet
Figure of combined bump-septum magnet
Mechanical structure of
the combined bump-septum magnet
Magnetic field distribution of
“Normal septum”and “Combined septum”
Magn etic fi eld distribution of
"Norma l se putu m" an d "Comb ined se ptum "
1.5
Combined septum
(B =0.775 T)
0
1
0.5
B/B
0
Normal septum
(B =0.729 T)
0
Le aka ge flux
0
-0.5
-1
-1.5
-5
0
X (cm)
5
Comparison of
“Normal septum”and “Combined septum”
Power supply system for
the H-injection bump magnets
Power supply system for
the combined bump-septum magnet system
Current waveform
of the combined septum conductor
(Superimpose rectangular waves)
20μs/div, 5V/div (1kA/V)
(a); Septum current
(b); Main bump current
DESIGN OF THE MAGNETIC FIELD
(For 400-MeV Injection)
• In the upstream of the stripping foil
The maximum magnetic field is estimated to be 0.55 T
The beam loss rate is less than 10-6
The injection beam power is 133 kW
Losses by Lorentz stripping is less than 1.3 W
• In the downstream of the stripping foil
The magnetic field of the bump magnet is set to be about 0.2 T.
Excited H0 with a principal quantum number of n ≥ 6 becomes the
uncontrolled beam
Yield of n ≥ 6 is 0.0136
The total H0 beam power is 0.4 kW
The maximum uncontrolled beam loss is about 6 W
The magnetic field at the foil is designed to be less than the value at which the
bending radius of the stripped electrons is larger than 100 mm.
Injection beam line (Horizontal)
• Injection line
– Lorentz stripping loss
• 0.14W/m (B<0.45T)
– H0,H- beam
• 0.4kW
(exchange efficiency
99.7%)
<0.45T
Main foil
(99.7%) 0.2T
0.4kW
– Excited H0 loss
• 5.5W (n6)
• H- beam and H0 beam are
exchanged to H+ beam by two
2nd foils ”A&B”
– Lead to beam dump
– 0.4kW
2nd foil “A”
2rd foil”B”
Schematic Layout of Beam Orbit
at Painting Injection Start
Fixed Closed-Orbit Bump Magnets ”SB-I~SB-IV”
•
•
•
Four dipole bump magnets named ”SB-I~SB-IV” are identical in construction
and are powered in series to give a symmetrical beam bump.
The dipoles are out of vacuum and ceramic vacuum chamber is included in
the magnet gap.
The structure of the magnet is composed of two-turn coils and window
frame core made by laminated silicon steel cores of which thickness is 0.1
mm.
Structure of the Split-type Bump Magnet
• The exitation current is supplied in the middle of the core trough the
split to form a symmetrical distribution of magnetic field along the
longitudinal direction.
• To insert the second foil
• Symmetrical power supply for a symmetrical field distribution along the
longitudinal axis
current
The Waveform of Magnetic Field
flat top level(k0)
 k k 0  0.5%
Beam injection
jitter  50ns
Unquestioned
reversal  k 0 5.0%
trigger
 552s
attack time
 500s
flat top time
600s
50s
release time
100 s
Fig.1 Current pattern of the power supply of the shift bump magnet in horizontal
Horizontal painting bump magnets
• Two sets of bump magnet pairs in the upstream of the F quadrupole
magnet and the downstream of the D quadrupole magnet.
• These four painting bump magnets will be excited individually.
• To form a local closed orbit include the F and D quadrupole magnets
current
Waveform of Horizontal Painting Bump Field
flat top level(k0
jitter  50ns
 k k 0  0.5%
Permissible error of the ideal waveform
±5%
±1%
Beam injection
reversal  k 0 5.0%
Unquestioned
trigger
 50s
attack time
 500s
flat top time
 50～100 s
decay time
300～550 s
•Ideal wave form
K0{ 1-sqrt( t/τ)}
•Design wave form
k0【1+[sqrt(ε/τ)-sqrt{( t+ε)/τ}]/[sqrt{(τ+ε)/τ}-sqrt(ε/τ)]】
•Differentiation same as the above
0.5k0/[sqrt{(τ+ε)/τ}-sqrt(ε/τ)]/sqrt{( t+ε)/τ}/τ
Fig.2 Current pattern of the power supply of the painting bump magnet in horizontal
Vertical Painting Magnets
360
10
400
160
• In the vertical plane, two steering magnets are installed on the
beam-transport line at a upstream point led by p from the foil.
• Painting injection in the vertical plane is performed by sweeping of
the injection angle.
• Both correlated and anti-correlated painting injections are available
by changing the excitation pattern of the vertical painting magnet
Waveform of Vertical Painting Bump Field
flat top level(k0)
 k k 0  1.0%
jitter  50ns
±1%
±5%
Unquestioned
attack time
 500s
flat top time
decay time
300～550 s
 30s
Unquestioned
flat top level(k0)
 k k 0  1.0%
±5%
±1%
50s
Unquestioned
Beam injection
decay time
jitter  50ns
attack time
flat top time
300～550 s
 30s
release time
 500s
Fig.2 Current pattern of the power supply of the painting bump magnet in vertical