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The CLIC IP Kicker & QD0

The CLIC IP Kicker & QD0. A quick look to see if magnetic forces on IP feedback kicker would be a significant design consideration. For more details see: IP Kicker Supplementary Notes.pdf. C.Perry - Oxford - 21 April 2010. 2: System Geometry.

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The CLIC IP Kicker & QD0

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  1. The CLIC IP Kicker & QD0 A quick look to see if magnetic forces on IP feedback kicker would be a significant design consideration. For more details see: IP Kicker Supplementary Notes.pdf C.Perry - Oxford - 21 April 2010

  2. 2: System Geometry • Magnetic field assumed axisymmetric about detector axis • - from the detector magnet & bucking coil cancelling field at QD0 • neglects fringing field from QD0: off-axis but should be negligible • file: Magneticfield_aroundKickerBPM_CLIC.txt • originally from H.Gerwig, 15mar2010 • Kicker Position: from 325 to 350cm from IP - given • Angle from magnet axis to beam: 10mrad - half crossing angle • Horizontal displacement of kicker from axis: 338cm*10mrad ≈ 3.5cm • Length: 25cm - given • Gap: 1cm - reasonable given small beampipe diameter • Impedance: 120ohms - typical for an efficient stripline geometry • Orientation: kicker assumed parallel to the magnet axis • - for ease of analysis • - would actually be possible... • - forces little different if aligned to beam C.Perry - Oxford - 21 April 2010

  3. 3: The Kicker Geometry: Type: stripline with strips in vacuum Operating mode: shorted at far end (ie no electrostatic deflection) Treat kicker as a single-turn rectangular coil in the vertical plane Connections made at one point: can disregard feed currents Short sides would be split to clear beam: little effect on forces Deflection: Assumed kicker current: 5A Kicker L per unit length: Zo/c = 120/3E8 = 0.4uH/m Coupled flux: 5A*0.4uH/m = 2uWb/m For 1cm gap kicker, field: 2uWb/m / 0.01m = 200uT For a uniform kicker field, this is also B near to the axis. From : p = 300 * B * r [MeV/c, T, m] Curvature for 1.75TeV: 300 * 2E-4 / 1.75E6 = 3.4E-8 m-1 Angle for 0.25m kicker: 0.25 * 3.4E-8 = 8.5nrad Displacement at IP: 3.4m * 8.5nrad = 29nm ==>So 5A drive is reasonable but generous C.Perry - Oxford - 21 April 2010

  4. 4: Forces - 1 For 5A: 29nm at IP Br = 72mT Br mean = 58mT Bz = 2.29T z = 350cm Bz = 3.12T Br = 45mT 3.5cm z = 325cm NOT TO SCALE! (Bz>>Br) C.Perry - Oxford - 21 April 2010

  5. 5: Forces - 2 For 5A: 29nm at IP Br = 72mT 73mN Br mean = 58mT Bz = 2.29T 156mN 112mN z = 350cm Bz = 3.12T Br = 45mT 73mN 3.5cm z = 325cm C.Perry - Oxford - 21 April 2010

  6. 6: Forces - 3 For 5A: 29nm at IP Br = 72mT 73mN Br mean = 58mT Bz = 2.29T Transverse force = 21mN 156mN 112mN 10mN z = 350cm 10mN Bz = 3.12T Br = 45mT 73mN 3.5cm z = 325cm C.Perry - Oxford - 21 April 2010

  7. 7: Forces - 4 For 5A: 29nm at IP Torque = 68mNm Br = 72mT 73mN Br mean = 58mT Bz = 2.29T Transverse force = 21mN 156mN 112mN 10mN z = 350cm 10mN Bz = 3.12T Br = 45mT 73mN 3.5cm z = 325cm Calculation from coupled flux: force = work/ds = I*(dΦ/ds) Mean dB/dr: 167 gauss/cm = 1.67T/m Horizontal transverse force: dBr/dr*I*A = 0.021N Vertical transverse force zero by symmetry Axial forces (on short sides) are negligble C.Perry - Oxford - 21 April 2010

  8. 8: Motion For single pulse: If current is maximum for full 250ns, impulse: 0.021N * 250ns = 5.2nNs Assume effective mass 100kg for kicker, QD0 and support structure. Velocity: 5.2nNs / 100kg = 0.05nm/s If resonant at 50Hz, amplitude: 0.05nm/s / (2π * 50Hz) = 0.15pm For repetition at 50Hz: Worst case: full impulse given & stays in phase with structural resonance. If damping timeconstant 200ms, amplitude: 200ms * 50Hz * 0.15pm = 1.5pm On more reasonable assumptions, under ~0.5pm peak. Note: Predicted motion is horizontal - but much could be converted to vertical. C.Perry - Oxford - 21 April 2010

  9. 9: Frequency Domain View System response is linear, so we can consider it in the frequency domain. Only the low frequency (~50Hz) components of the drive current are significant. Worst case: Drive: 5A 0.25us pulses repeated at 50Hz Spectrum roughly flat up to 2MHz Amplitude of 50Hz component: 5A*50Hz/2MHz = 125uA Force component at 50Hz: 0.021N*125uA/5A = 0.5uN On a free 100kg mass, amplitude: 0.5uN/100kg*(2π*50Hz)**2 = 0.05pm Resonance at 50Hz increases response If Q=30 amplitude 0.05pm*30 = 1.5pm ==>*very* rough - but agrees with previous calculation At LF induced currents in eg cable screens are insignificantly small - so feed connections should be routed together. We can remove LF components from drive: greatly reduces mechanical effects. C.Perry - Oxford - 21 April 2010

  10. 10: Driving the Kicker Example output stage performances possible with available MOSFETs: 1: Simple:+/-8A drive, +/-50nm, 7ns or better risetime 2: Maximum Kick:+/-30A to each segment, +/-160nm, 6ns or better risetime Kicker divided into 4 segments each with own driver Electronics plugged directly onto the kicker 3: Maximum Speed:+/-8A drive, +/-50nm, 3.5ns or better risetime As 2, but with lower voltage, faster MOSFETs Notes: - no information on rad-hardness of devices - magnetic field prohibits ferrite cores in amplifier: examples allow for this - there are forces from currents in amplifier: can be kept negligible with care C.Perry - Oxford - 21 April 2010

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