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enter image description hereI opened a charge sensitive amplifier (CSA) board which is used in radiation detection.

I am trying to figure out how it works.

It is totally confusing me that one BNC is used for two signals. How is it possible?

The operating voltage of the CSA is -20V, and the output of the CSA is square pulses. One BNC (left side in the pic) is used to supply -20v and to take CSA output (square pulses.)

How is it possible?

Similarly, on the right side, just one BNC is used to supply high voltage to the detector and to take the output of the detector (charge pulses- which then go to CSA.) Again one BNC for two purposes.

How does it work? Any suggestions?

CSA

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That's simply a matter of adding a high-pass filter in line: your DC supply won't pass, but high-frequency pulses are let through. In the power line, add a low-pass filter, so that noise generated on the supply rail doesn't bleed back into your signal line:

--PWR+Signal---+----|HPF|---<signal----
               |
               \----|LPF|-->DC---->

This is a pretty commonly employed technique in things like RF amplifiers: technically, their core is biased transistors, and you can imprint the biasing from the supply, using pretty much the inverse to the above (called a bias-T, often for the fact that it's a T-junction that adds a bias):

--PWR+Signal---+----|HPF|---<signal----
               |
               \----|LPF|---<DC----<power supply
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  • \$\begingroup\$ When you say high pass filter in line, does it mean, there must be a filter in the CSA board? But still, the cable, which I have to connect to this BNC connector should carry two signals in the opposite direction ???? Or the device that supplies this DC (-20v) voltage and analyzing the signal (square pulse) would have another High pass Filter? \$\endgroup\$ May 7, 2020 at 8:16
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    \$\begingroup\$ that's all answered in my answer: yes, on the Board. Signals don't care about the direction of other signals. \$\endgroup\$ May 7, 2020 at 8:29
  • \$\begingroup\$ I added a block diagram to my post based on your explanation. I hope it is right. Please have a look. Thank you \$\endgroup\$ May 7, 2020 at 9:06

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