A bunch charge measurement system based on Turbo-ICT for BEPCII

Ying Zhao; Yanhua Lu; Yaoyao Du; Jun He; Taoguang Xu; Di Yin; Huizhou Ma; Junhui Yue; Yanfeng Sui; Xiaohua Peng

doi:10.11884/HPLPB202133.210229

Journals >High Power Laser and Particle Beams >Volume 33 >Issue 11 >Page 114002 > Article

High Power Laser and Particle Beams
Vol. 33, Issue 11, 114002 (2021)

A bunch charge measurement system based on Turbo-ICT for BEPCII

Ying Zhao, Yanhua Lu, Yaoyao Du, Jun He, Taoguang Xu, Di Yin, Huizhou Ma, Junhui Yue, Yanfeng Sui, and Xiaohua Peng

Author Affiliations

Accelerator Division, Institute of High Energy Physics, Chinese Academy of Sciences, Beijing 100049, China

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DOI: 10.11884/HPLPB202133.210229 Cite this Article

Ying Zhao, Yanhua Lu, Yaoyao Du, Jun He, Taoguang Xu, Di Yin, Huizhou Ma, Junhui Yue, Yanfeng Sui, Xiaohua Peng. A bunch charge measurement system based on Turbo-ICT for BEPCII[J]. High Power Laser and Particle Beams, 2021, 33(11): 114002 Copy Citation Text

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Abstract

A new kind of sensor Turbo-ICT has been applied to accurate bunch charge measurement in BEPCII transportation line. The system improves the resistance ability against external interference and has high signal-to-noise ratio. This paper introduces the principle, the design ideas and the function of the system. Up to now, Turbo-ICT has been used to operate online for electron bunch charge quantity measurements successfully. This paper also presents the experiment results of the sensor.

Keywords

BEPCII bunch charge measurement Turbo-ICT

$ {{U}}_{{{\rm{BCM-RF}}}}={{U}}_{{{\rm{cal}}}}{}{\lg}\frac{{{Q}}_{{{\rm{in}}}}{ \times }{K}{ \times }{D}}{{{Q}}_{{{\rm{cal}}}}} , \;\;\;\;{{Q}}_{{{\rm{in}}}}={{Q}}_{{{\rm{cal}}}}{ \times }{}\frac{1}{K}{ \times }\frac{{1}}{{D}}{ \times }{{10}}^{\tfrac{{{U}}_{{{\rm{BCM-RF}}}}}{{{U}}_{{{\rm{cal}}}}}} $

(1)

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$ \text{20lg}\frac{{{U}}_{\text{x}}}{{{U}}_{\text{Turbo-ICT}}}\text{= cable attenuation} $

(2)

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$ {{S}}_{\text{1}}\text={{Q}}_{\text{cal}}\text{×}\frac{1}{{K}}\text{×}\frac{{{U}}_{\text{Turbo-ICT}}}{{{U}}_{\text{1}}}\text{= 0.062 28} $

(3)

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$ {{S}}_{\text{2}}\text={{Q}}_{\text{cal}}\text{×}\frac{\text{1}}{{K}}\text{}\text{×}\frac{{{U}}_{\text{Turbo-ICT}}}{{{U}}_{\text{2}}}={{Q}}_{\text{cal}}\text{×}\frac{\text{1}}{{K}}\text{×}\frac{{{U}}_{\text{Turbo-ict}}}{{{U}}_{\text{1}}/\text{1.412}}\text{= 0.087 97} $

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