1. Report 2013/04/15

2. FADC frequency study (cont.) Amplifier gain Problem

3. FADC (cont.) Pedestal Pulse area Pulse areas are well-separated from the pedestal

4. Pulse form with 30MHz Pulse form with 16.7MHz From these plots, we can conclude that 33MHz-FADC can be used

5. In the cylindrical wire with inner radius a and outer radius b, a pulse amplitude : From the above formulae, the pulse tail is dependent of inner radius a. In our case, the pulse tail is proportional to square of wire radius. Diameter of 30-mm: 10  m. Diameter of 300-mm: 15  m. So, pulse tail of 300-  m is around 2.25 times longer than pulse tail of 30-  m Output signal after being amplified The pulse tails are almost same for both cases. So, we can use 33MHz-FADC for both 30-mm anode wire and 300-mm anode wire. 30-mm anode wire 300-mm anode wire

6. Amplifier gain: In the post-amp circuit, feedback circuit exists. So, we can change some elements in the feedback circuit in order to change the amplifier gain. The amplitude in this case is around 75 mV. C 84 R 84 R 85 Feedback circuit Current values: C 84 =180 pF R 84 =160 Ohm R 85 =510 Ohm

7. C 84 (pF)R 84 (Ohm)R 85 (Ohm)Amplitude (mV) 18016051075 1801603000 18016040060 18016080075 1801510400 18050510160 18030051052 30016051078 5016051075 From the above table, we can change the value of R 84 and keep the other. So, we can change the output signal around 5 times higher than the current values.

8. C 84 =180 pF, R 84 =1 Ohm, R 85 =510 Ohm Overshoot ??

9. Problem The signal from anode wire can be expressed as the following equation So, an amplitude of signal is dependent on capacitor of wire C. Cathode plane d L wire Length of anode wire of the next prototype is 300-mm, 10 times longer than the current one  signal of 300-mm will be 10 times smaller than signal of 30-mm  How to solve this problem?