1. F D F November 8, 2006Alessandro G. Ruggiero1 of 8 1.0-GeV 10-MWatt Proton Driver Target 200-MeV DTL 1.0-GeV FFAG H – Stripping Foil Injection Energy, U i 200 MeV Extraction Energy, U f 1.0 GeV Beam Ave. Power, P = I U f 10.0 MWatt Repetition Rate, F 1.0 kHz Repet. Period,      1.0 ms Beam Ave. Current, I = Ne F10.0 mA Total No. Protons, N6.25 x 10 13 DTL Peak CurrentI L Revol. Freq.f = c  inj / C Chopping Ratio  Revol. PeriodT = 1 / f FFAG CircumferenceCNo. Protons / TurnN P =  I L T / e Injection  inj No. Injected Turnsn = N / N P = Ne /  I L T Acceleration PeriodT acc Rep. Period  = T inj + T acc Injection Period T inj = nT = Ne /  I L --> not dependent on C and  inj  = 0.5 I L = 60 mA -->T inj = 0.333 ms & T acc = 0.667 ms Filling Acceleration

2. F D F November 8, 2006Alessandro G. Ruggiero2 of 8 General Beam- RF Considerations Average Energy Gain / TurnW No. Revol. Acceleration Periodm = (U f – U i ) / W Average Revol. PeriodT ave = C /  ave c Acceleration PeriodT acc = m T ave = T ave (1 – U i / U f ) (1 +  ) U f / W  extra acceleration time dilation factor (1 – U i / U f ) (1 +  ) ~ 1 U f = P  / N e Acceleration PeriodT acc = T ave P  / N e W Cycle Period  = T inj + T acc = T inj / (1 – P / P B ) Average Beam Accel. PowerP B = W N e / T ave P B = (3 / 2) P = 15 MWatt Average RF PowerP RF = P B + P cavity = 2 P B = 30 Mwatt FFAG Cycle EfficiencyP / P RF = 33%or P / P AC = 25% Energy Gain / Unit LengthW / C = (5.0 /  ave ) keV /m

3. F D F November 8, 2006Alessandro G. Ruggiero3 of 8 More on RF Acceleration System FFAG Circumference C = 201.8 m  ave = 0.75  Energy GainW = 1.35 MeV / Turn  RF Peak VoltageV RF = 1.8 Mvolt  Harmonic Numberh = 36  No. Empty Buckets9 out of 36  Protons / Bunch2.4 x 10 12  No. of RF Cavities40  No. of Gaps / Cavity1  Cavity Length1 m  Peak RF Voltage / Gap45 kVolt  Power Amplifier / Cavity0.8 MWatt  Cavity Inter. Diameter10 cm Energy Range, MeV2001,000  0.5660.875 Rev. Frequency, MHz0.8411.300 Revolution Period, µ s1.1890.769 RF Frequency, MHz30.2846.80 Peak Current, Amp12.6519.55 Peak Beam Power, MW15.223.5 Issue # 1 Can the RF ferrite be swept in 2/3 ms ? 25 MHZ / ms

4. F D F November 8, 2006Alessandro G. Ruggiero4 of 8 Period Layout (68 Cells) Side View Diagnostic & Steering Boxes D-Sector Magnet F-Sector Magnets Flanges & Bellows Vacuum Pump 10 cm Top View D-Sector Magnet F-Sector Magnets Flanges & Bellows Vacuum Pump 20 cm Diagnostic & Steering Boxes RF Cavity Diagnostic & Steering Boxes D-Sector Magnet F-Sector Magnets Vacuum Pump 100 k$ 600 k$ 0 m 1.0 m 2.0 m 3.0 m

5. F D F November 8, 2006Alessandro G. Ruggiero5 of 8 Multi-Turn Injection (H – ) B1 B2 C1 Foil C2 From DTL Injection Orbit Bump Orbit 20 x 20 mm Foil Injected Beam 200 MeV 1.0 GeV Circulating Beam 10 cm x 20 cm Vacuum Chamber Linac Peak Current60 mA Revolution Period1.89 µs No. of Protons / FFAG pulse6.25 x 10 13 Chopping Ratio0.50 Chopping Frequency30.283 MHz Single Pulse Length0.333 ms No. of Turns Injected / pulse165 Linac/FFAG Rep. Rate1.0 kHz Linac Duty Cycle0.33 % Linac Beam Emittance, rms norm.1 π mm-mrad Final Beam Emittance, full norm.150 π mm-mrad Bunching Factor3 Space-Charge Tune-Shift0.35

6. F D F November 8, 2006Alessandro G. Ruggiero6 of 8 Single-Turn Extraction Revolution Period 1.89 µs Beam Gap 330 ns Kicker Magnet, Length1.0 m Field1 kG Rise-Time< 300 ns Septum Magnet, Length1.0 m Field10 kG Repetition Rate1.0 kHz The Kicker field remains constant for the duration of the beam pulse (about 1.6 µs), and it is finally reset to zero-value in about 1.0 ms, to be fired again the next cycle. Kicker Septum F D F F D F F D F

7. F D F November 8, 2006Alessandro G. Ruggiero7 of 8 Rule #4 for the FFAG Design Make FFAG Circumference as large as possible Chose a number as large as possible of Periods Avoid the Curvature effect at Low Energy small Rings x  + h 2 x / (1 +  ) + h 2 n y / (1 +  ) = h  / (1 +  ) y  – h 2 n y / (1 +  ) = 0 To keep small Tune Variation with the AFP minimize the Edge Effects 200 MeV 1 GeV x, cm s, m Half Period h i B i (x co ) B i (1 +  ) (tan x co ') y  y' = – x co ' x co F D Vertical == Focusing Horizontal == Defocusing

8. F D F November 8, 2006Alessandro G. Ruggiero8 of 8 Conclusions FFAG Proton Accelerators are a very promising alternative to other Accelerator Architectures (Super-Conducting Linacs, Cyclotrons, Rapid Cycling Synchrotrons), especially in view of the recent progress in beam dynamics and of new proposed design approaches. FFAG’s rely on conventional Magnet Technology, thus appealing to many centers of research. They are supposed to be less expensive: a crude estimate of the Proton Driver described here shows a cost of about 50M$ (excluding DTL and Tunnels). In the case of Protons, the Path Length variation with Momentum is not a concern if there is at any time only a single beam pulse circulating. A 10 MWatt beam power requires a AC source of about 40-50 MWatt. It is then imperative to demonstrate methods to create energy with the use of sub-critical nuclear material. Target 200-MeV DTL 1.0-GeV FFAG H – Stripping Foil Sub-critical Fissionable material Experimental Areas: Spallation Neutrons Waste Transmutation Tritium Production Radio-Isothops Production Exotic Elements Production, ….. AC Power The next step is to resolve the 2 main Issues