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Transcript
```WIDEROE ACCELERATOR
CONCEPT (IN A BOTTLE)
Presented by:
Stephen, Leon, Daichi, Stefan
Outline of Presentation
Stephen

Aims and goals
 Wideroe
Accelerator
 Assumptions and Questions
Leon

Limits and Parameters
 Frequency
limit
 Size of bottle
Daichi

Stefan

Maths behind Accelerator
Results using MATLAB plots
Aims and goals



Investigate a simple linear accelerator
Design the accelerator into a standard wine bottle
Design to get the highest energy particles possible
Cathode
Vacuum
pump
High Potential Difference
Anode
Wine
bottle
Electrons are accelerated by a high potential difference (kV) from cathode to
anode. The blue diffusion given out by ‘Cathode rays’ are caused by fast moving
electrons exciting the gas molecules left in the bottle or striking the glass and
causing the excited molecules to decay emitting out photons of bluish wavelength.
http://www.instructables.com/id/DIY-Electron-Accelerator-A-Cathode-Ray-Tube-in-a-/ - Extracted 23/01/2014
Swapped
voltage
Ions come
out of
source
Wideroe Accelerator
•
Ion source sends our charged particles towards the first electrode
•
RF source flips the sign of the electrodes
•
Ions are at constant velocity in drift tubes as voltage flips
•
When Ions reach other side they are accelerated again and cycle repeats
Aim: Design a Linear Accelerator in
a Bottle

Assumptions:

Bottle's length is ~ 50 cm

Vacuum required ~ 1×10−3 torr
Questions to be discovered


How much energy can we get by changing these
parameters:

f = frequency of RF

U = Amplitude of alternating current

m = mass of particles

Length of drift tubes to be as small
as possible
Limits and Parameters



Voltage has to be in kV range
Length of drift tubes
Cannot exceed breakdown voltage
 3.0
MV/m – Air
 20-40 MV/m – High quality vacuum

Frequency of RF voltage < 10MHz
Why the frequency limit?
It can be seen that our system is like a parallel plate capacitor.
Where dQ = C(v)*dV
And C = (A/d) εr ε0
This presents another issue: The fact that our ‘d’ is increasing as we make our
tube longer, this decreases the capacitance.

The frequency limit can be shown by the simple
RC circuit. The transient response is shown on
the bottom left. What we see is τ = R*C,
where τ is the ‘time constant’ of the capacitor.
Due to this, it can be seen that at high
frequencies the material doesn’t have enough
time to fully charge to the given voltage.
Size, an issue?
•
•
As the length of each drift tube and the gap is
increasing for every iteration along electron path. It
can be seen that the structure will get very large,
expensive and difficult to build, extremely fast.
So what we’ve thought of is a systematic concept
which basically involves instead of increasing the
lengths incrementally but rather modulating the
initial RF carrier to keep in phase with the
accelerating electron.
Mathematics Behind the Design
➲
Constant velocity in the tube
V = V max sinωt
E i = iqU max sinψ 0
V = potential difference
E i = energy at ith tube
1 2
E i = mv i
2
➲
Equate the energy to solve for v.
Length of the Tube
√
2iqV max sinψ 0
vi=
m
The time that the particle travel: T = 1
2 2f
T = Period, f = frequency
➲
√
2iqV max sinψ 0 1
→ l i=
m
2f
Length of the Gap
➲
This time, velocity is not constant
E= E 0 sinωt
But also...
qE= F = ma
qE 0 sinωt
a (t)=
m
E= Electric field
−A
v (t)= a (t)dt =
cosωt+ v '
ω
qE 0
A=
m
∫
t= 0, v= v i
→v ' = v i + A/ ω
−A
∫
x (t)= v (t)dt = 2 sinωt+ v ' t
ω
qV max
Δx= x i = l i +
2
4mπf x i
1
Δt = τ / 2=
2f
V max
E 0=
xi
qV max
x − l i xi−
2= 0
4mπf
2
i
√
qV max
l i± l +
2
mπf
→ xi =
2
2 qV max
l i± l i +
2
n
n
mπf
L= ∑0
+ ∑1 l i
2
2
i
√
√
√
qV max n
qV max
2
∑
2 + 1 (3l i + l i +
2)
mπf
mπf
=
2
Varying Voltage of RF generator
(1)
5
9
Energy in tube vs Number of drift tubes
x 10
kV
8
1
10
25
40
70
100
7
Particle energy in tube (eV)
RF frequency = 5MHz
6
5
4
3
2
1
0
1
2
3
4
5
6
Number of tubes
7
8
9
10
Varying Voltage of RF generator
(2)
Length vs Number of drift tubes
2
10
RF frequency = 5MHz
1
Length (m)
10
0
10
kV
1
10
25
40
70
100
-1
10
-2
10
1
2
3
4
5
6
Number of tubes
7
8
9
10
Varying frequency of RF generator
(1)
5
2.2
Energy in tube vs Number of drift tubes
x 10
MHz
kV 0.003
2
Particle energy in tube (eV)
1.8
RF Amplitude = 25
0.01
0.1
1
10
100
1000
1.6
1.4
1.2
1
0.8
0.6
0.4
0.2
1
2
3
4
5
6
Number of tubes
7
8
9
10
Varying frequency of RF generator
(2)
Length vs Number of drift tubes
5
10
RF Amplitude = 25 kV
4
10
MHz
3
10
2
Length (m)
10
1
10
0
10
-1
10
-2
10
-3
10
1
2
3
4
5
6
Number of tubes
7
8
9
10
Final Idea!
5
Energy in tube vs Number of drift tubes
Length vs Number of drift tubes
5
2
4.5
1.8
4
1.6
3.5
1.4
3
Length (m)
Particle energy in tube (eV)
2.2
x 10
1.2
1
2.5
2
0.8
1.5
0.6
1
0.4
0.2
0.5
X: 2
Y: 4.207e+004
1
2
3
4
5
6
Number of tubes
7
8
9
10
0
X: 2
Y: 0.514
1
2
3
4
5
6
Number of tubes
7
8
9
10
Questions
```