Presentation is loading. Please wait.

Presentation is loading. Please wait.

Michael Lupberger Dorothea Pfeiffer

Similar presentations


Presentation on theme: "Michael Lupberger Dorothea Pfeiffer"— Presentation transcript:

1 Michael Lupberger Dorothea Pfeiffer
VMM2 Studies Michael Lupberger Dorothea Pfeiffer RD51 miniweek This work was funded by the EU Horizon 2020 framework, BrightnESS project

2 Schematic of Setup Pulser Oscilloscope CH1 CH2 PC with Georges VMM DCS
GND VMM Hybrid 10 kW VMM2 50 W Spartan FPGA GND VMM2 50 W 10 kW VMM 1 Ch 11 VMM 1 digital data SRS NIM for SRS trg 1.7 pF M0 analog mon

3 VMM2 Logic

4 VMM2 hybrid connections

5 Probe at capacitor connected
Overview Zoom on falling fast falling edge (negative charge) Pulser CH2 Voltage before capacitor M0: VMM analog output NIM Trigger for SRS

6 Probe at capacitor disconnected
Zoom on falling fast falling edge (negative charge) Zoom on falling fast falling edge (negative charge) Pulser CH2 Pulser CH1 M0: VMM analog output NIM Trigger for SRS

7 Single pulse from pulser CH 1
Voltage before capacitor NIM Trigger for SRS M0: VMM analog output Increasing input charge

8 Pulses from pulser CH1 and CH2
Delay CH1 – CH2: 200 ns Peaking time 25 ns Delay CH1 – CH2: 500 ns Peaking time 50 ns Delay CH1 – CH2: 200 ns Peaking time 50 ns Delay CH1 – CH2: 500 ns Peaking time 200 ns

9 VMM2 digital output narrow pulses
Question: What is the minimum delay, such that two pulses are resolved by the VMM logic (in the analog part, as we have seen, they are clearly separated at e.g. 200 ns and 25 ns peaking time) -> Answer: (by just looking at the frame size of the SRS packages): ≈500 ns More quantitatively: events with double pulses - over threshold flag of measured hits Dt = 600 ns Dt = 500 ns Dt = 480 ns 2x60000 pulses seen ≈ 2x60000 pulses seen But some are not over threshold (tail caught) ≈ 1x60000 pulses seen

10 VMM2 digital output narrow pulses
Question: What is the minimum delay, such that two pulses are resolved by the VMM logic (in the analog part, as we have seen, they are clearly separated at e.g. 200 ns and 25 ns peaking time) -> Answer: (by just looking at the frame size of the SRS packages): ≈500 ns More quantitatively: events with double pulses – amplitude of measured hits Dt = 600 ns Dt = 500 ns Dt = 480 ns 2x60000 pulses seen ≈ 2x60000 pulses seen But some are not over threshold (tail caught) ≈ 1x60000 pulses seen

11 VMM2 digital output narrow pulses
Question: Can we reconstruct the time difference between the two pulses? -> Answer: Yes we can! Dt = 600 ns Dt = 500 ns ≈ 2x60000 pulses seen But some are not over threshold (tail caught) 2x60000 pulses seen

12 VMM2 digital output Summary narrow pulses
BC clock 40 MHz TAC slope 125 ns Gain 1 mV/fc Pulse frequency 1 kHz Shaping time 25 ns Time of threshold crossing For a time delay >= 600 ns, both pulses are recorded with the correct ADC and delay For delays between 600 ns and 480 ns, the second pulse is detected less often, with a lower ADC, and the reconstructed time delay is too small For delays < 480 ns the second pulse is lost

13 VMM2 digital output Wide pulses, comparison APV25
Digitized full waveform ADC Time bin [25 ns]

14 VMM2 digital output Wide overlapping pulses
VMM2 detects first peak with correct ADC and time VMM2 detects second peak only when threshold is very high, incorrect ADC (peak minus trough) ADC Time bin [25 ns]

15 VMM2 digital output Summary wide overlapping pulses
BC clock 40 MHz TAC slope 125 ns Gain 1 mV/fc Pulse frequency 1 kHz Shaping time 25 ns or 200 ns (overlapping) Time of threshold crossing or peak time The second peak of overlapping pulse is still partially resolved with a time delay of 600 ns, but only if the discriminator threshold is high


Download ppt "Michael Lupberger Dorothea Pfeiffer"

Similar presentations


Ads by Google