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2 nd harmonic RF perpendicular biased cavity update C.Y. Tan, W. Pellico, G. Romanov, R. Madrak, and D. Wildman 04 Apr 2014
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2 nd harmonic RF perpendicular biased cavity update

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2 nd harmonic RF perpendicular biased cavity update. C.Y. Tan, W. Pellico , G. Romanov, R. Madrak , and D. Wildman 04 Apr 2014. Proposed cavity. solenoid not shown here. Ferrite. 70. 200. 220. 390. 4 0. BeO. 490. - PowerPoint PPT Presentation
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Page 1: 2 nd  harmonic RF perpendicular biased cavity update

2nd harmonic RF perpendicular biased cavity update

C.Y. Tan, W. Pellico, G. Romanov, R. Madrak, and D. Wildman

04 Apr 2014

Page 2: 2 nd  harmonic RF perpendicular biased cavity update

04 Apr 2014; C.Y. Tan2

Proposed cavity.

Ferrite disk: 380 mm outer diam., 230 mm inner diam., 25 mm thicknessBeO disk: 380 mm outer diam., 230 mm inner diam., 5 mm thickness

490

7040

200

220

390

Ferrite

BeO

solenoid not shown here

Page 3: 2 nd  harmonic RF perpendicular biased cavity update

Some possible parameters

• Tuning range 76.7 − 107 MHz.• Gap voltage. 100 kV per cavity.• Ramp profile determines losses in the garnet.

04 Apr 2014; C.Y. Tan3

Page 4: 2 nd  harmonic RF perpendicular biased cavity update

CST Model (done by G. Romanov)

04 Apr 2014; C.Y. Tan4

Complete cavity model with magnetic field generated by solenoid

Solenoid coil

Page 5: 2 nd  harmonic RF perpendicular biased cavity update

04 Apr 2014; C.Y. Tan5

R11

0R

205

190 mm

Yoke, steel 1008

Coil, 12 turns

Water cooling channels, 10x5 mm

Ferrite G810, R=190 mm, r=115 mm, l=25 mm

Ceramic AlN, l=5mm

230 mm90

mm

20 m

m

This is old picture, not properly scaled. But the marked dimensions are current.

Ferrite tuner details

Page 6: 2 nd  harmonic RF perpendicular biased cavity update

Static field distribution in ferrite

04 Apr 2014; C.Y. Tan6

Separate solenoid model

Complete cavity model

Field non-uniformity is about 25-30%

Page 7: 2 nd  harmonic RF perpendicular biased cavity update

Suppression of fields with steel insert.

04 Apr 2014; C.Y. Tan7

Yoke

Coil

Ferrite

Copper drift tube

Steel 1008 insert

Radial distribution of H field at tuner median plane. H field on the

axis can be as high as 85 kA/m. With steel drift tube it is

suppressed.

Page 8: 2 nd  harmonic RF perpendicular biased cavity update

RF magnetic field distribution in ferrite and losses

04 Apr 2014; C.Y. Tan8

f=75.6 MHz

These power losses spikes are not real. They are due to the singularity of low frequency mesh that is used for thermal simulations

Page 9: 2 nd  harmonic RF perpendicular biased cavity update

Tuning curves

04 Apr 2014; C.Y. Tan9

0 5000 10000 15000 20000 25000 30000 3500070

75

80

85

90

95

100

105

110

115

Solenoid current, Ampere·turns

Freq

uenc

y, M

Hz

Conversion of the solenoid current to the equivalent uniform field. We can continue to use uniform magnetization – the results are very close.

Page 10: 2 nd  harmonic RF perpendicular biased cavity update

Thermal analysis

04 Apr 2014; C.Y. Tan10

AlN cooling disks. Thermal losses in the ferrite are 14 kW for V=100 kV. Max T ≈ 75°C with cooling water temperature of 25°C.

020

040

060

080

010

0012

0014

0016

0050

100

150

200

250

300

Saturation magnetization, Gauss

Cur

ie te

mpe

ratu

re, °

C

AL-400-30

Curie temperature

Page 11: 2 nd  harmonic RF perpendicular biased cavity update

Mulitpacting with maximum solenoidal field

04 Apr 2014; C.Y. Tan11

Range: up to 2.5 % of nominal field level

Location of multipacting Exponential growth of particle number

Page 12: 2 nd  harmonic RF perpendicular biased cavity update

Mulitpacting without solenoidal field

04 Apr 2014; C.Y. Tan12

Range: up to 1 % of nominal field level

Exponential growth of particle number

Location of multipacting

Page 13: 2 nd  harmonic RF perpendicular biased cavity update

Magnetic permeability (Gyrotropic model)

04 Apr 2014; C.Y. Tan13

Page 14: 2 nd  harmonic RF perpendicular biased cavity update

Measuring AL400 (R. Madrak and D. Wildman)

04 Apr 2014; C.Y. Tan14

Page 15: 2 nd  harmonic RF perpendicular biased cavity update

Measured losses

04 Apr 2014; C.Y. Tan15

0 20 40 60 80 100 120-1

0

1

2

3

4

5

6

7

8 Measured Losses Vs Solenoid Bias

76 MHz

106 MHz

53 MHz

solenoid bias (A)

Loss

es (-

dB)

method looks at s11 and from there calculate the loss in the garnet.

This number will scale with the length of the garnet.

Page 16: 2 nd  harmonic RF perpendicular biased cavity update

Model in ADS used to calculate μ’ from s11 phase data

04 Apr 2014; C.Y. Tan16

Page 17: 2 nd  harmonic RF perpendicular biased cavity update

Fits to the s11 phase data

04 Apr 2014; C.Y. Tan17

Page 18: 2 nd  harmonic RF perpendicular biased cavity update

Measured μ

04 Apr 2014; C.Y. Tan18

15 35 55 75 95 1151.001.502.002.503.003.504.004.505.005.506.00

-0.1

0.1

0.3

0.5

0.7

0.9Mu and Losses, 76 MHz

Mu Losses

solenoid bias (A)

Mu

Loss

es (-

dB)

15 35 55 75 95 1151.001.502.002.503.003.504.004.505.005.506.00

-0.1

0.1

0.3

0.5

0.7

0.9Mu and Losses, 106 MHz

Mu Losses

solenoid bias (A)

Mu

Loss

es (-

dB)

recall μe = μ’ – iμ’’.

Back of the envelope requiresμmax/μmin = (fmax/fmin)2 = (106/76) 2≈ 2.

Sims say ratio is 2.5, then if μmin=1.5, then μmax=1.5×2.5 = 3.75

μ’

prop to μ’’

3.75

24

-0.4 dB loss @ μ’=3.75

s11 = (Pin - Pref )/Pin = 1-Pref/Pin.

In dB = 10 Log10(1- Pref/Pin)≈10(-0.43 x – 0.21 x2)

Tan[δ] = μ’’/μ’=1/Q

Page 19: 2 nd  harmonic RF perpendicular biased cavity update

Todo (partial list) to be discussed

• Check that solenoid can be built. PS for solenoid -- kA

• Coupler?• PA?• Dimensions OK?• Impedance of cavity? Shunt impedance > 1

MΩ?• How to incorporate μ measurements into

simulations?

04 Apr 2014; C.Y. Tan19