Report 2013/04/15. FADC frequency study (cont.) Amplifier gain Problem.
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Transcript of Report 2013/04/15. FADC frequency study (cont.) Amplifier gain Problem.
Report 2013/04/15
• FADC frequency study (cont.)• Amplifier gain• Problem
FADC (cont.)Pedestal
Pulse area
Pulse areas are well-separated from the pedestal
Pulse form with 30MHz
Pulse form with 16.7MHzFrom these plots, we can conclude that 33MHz-FADC can be used
In the cylindrical wire with inner radius a and outer radius b, a pulse amplitude :
€
V t( ) =Q
C
1
Ln b /a( )Ln
2μV0
a2pLn(b /a)t +1
⎛
⎝ ⎜
⎞
⎠ ⎟
1/ 2
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: 10mm.Diameter of 300-mm: 15mm.So, pulse tail of 300-mm is around 2.25 times longer than pulse tail of 30-mm
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
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.
C84 R84
R85
Feedback circuit
Current values:C84=180 pFR84=160 OhmR85=510 Ohm
C84 (pF) R84 (Ohm) R85 (Ohm) Amplitude (mV)
180 160 510 75
180 160 300 0
180 160 400 60
180 160 800 75
180 1 510 400
180 50 510 160
180 300 510 52
300 160 510 78
50 160 510 75
From the above table, we can change the value of R84 and keep the other.So, we can change the output signal around 5 times higher than the current values.
C84=180 pF, R84=1 Ohm, R85=510 Ohm
Overshoot ??
Problem
€
V t( ) =Q
C
1
Ln b /a( )Ln
2μV0
a2pLn(b /a)t +1
⎛
⎝ ⎜
⎞
⎠ ⎟
1/ 2
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 planed
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-mmHow to solve this problem?
€
C =2πεL
Lnd
a+
d2
a2 −1 ⎛
⎝ ⎜ ⎜
⎞
⎠ ⎟ ⎟