Report 2013/04/15. FADC frequency study (cont.) Amplifier gain Problem.

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Report 2013/04/15

Transcript of Report 2013/04/15. FADC frequency study (cont.) Amplifier gain Problem.

Page 1: Report 2013/04/15. FADC frequency study (cont.) Amplifier gain Problem.

Report 2013/04/15

Page 2: Report 2013/04/15. FADC frequency study (cont.) Amplifier gain Problem.

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

Page 3: Report 2013/04/15. FADC frequency study (cont.) Amplifier gain Problem.

FADC (cont.)Pedestal

Pulse area

Pulse areas are well-separated from the pedestal

Page 4: Report 2013/04/15. FADC frequency study (cont.) Amplifier gain Problem.

Pulse form with 30MHz

Pulse form with 16.7MHzFrom these plots, we can conclude that 33MHz-FADC can be used

Page 5: Report 2013/04/15. FADC frequency study (cont.) Amplifier gain Problem.

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

Page 6: Report 2013/04/15. FADC frequency study (cont.) Amplifier gain Problem.

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

Page 7: Report 2013/04/15. FADC frequency study (cont.) Amplifier gain Problem.

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.

Page 8: Report 2013/04/15. FADC frequency study (cont.) Amplifier gain Problem.

C84=180 pF, R84=1 Ohm, R85=510 Ohm

Overshoot ??

Page 9: Report 2013/04/15. FADC frequency study (cont.) Amplifier gain Problem.

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 ⎛

⎝ ⎜ ⎜

⎠ ⎟ ⎟