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I am reading "Techniques for Nuclear and Particle Physics Experiments" by William R. Leo. I have some questions about the ionization detectors chapter (chapter 6). In chapter 6, there is a figure like this:

 The basic configuration (Fig. 6.1) consists of a container, which we will take to be a cylinder for simplicity, with conducting walls and a thin end window. Along its axis is suspended a conducting wire to which a positive voltage, + Vo, relative to the walls is applied.

I understand the basic concept of how the signal is registered (radiation penetrates the cylinder causing the gas to become ionized. The electrons then moves toward the wire causing the constant voltage to drop. The power supply then must do some work to bring the voltage back to $V_0$, and the work is what being registered as signal...please correct me if I am wrong. I am not taking a class on this. This is just independent study). However, I do not understand the electronics part of this. For example, what are the functions of the two resistors and the capacitor? I also would like some more details about how the signal is registered.

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  • $\begingroup$ You should ask only one question at a time—taking 'one question pretty broadly, but the two you have here are very different in character. If you have multiple distinct question they should go in multiple distinct posts. I'm going to suggest that you remove the electric field question from this post and stick to the electronics. $\endgroup$ – dmckee Sep 26 '17 at 18:16
  • $\begingroup$ Now, you should be able to get that field as a good approximation for most of the volume of the tube (and essentially all the volume near the wire) if you've had an upper-division course in electromagnetism. If you have not had such a course there are going to be many parts of Leo that you have to take on faith. But the book is still worth you time. $\endgroup$ – dmckee Sep 26 '17 at 18:16
  • $\begingroup$ The question is edited. $\endgroup$ – Kane Billiot Sep 26 '17 at 18:31
  • $\begingroup$ One can watch the signal on an oscilloscope. Shape of the pulses depends on the RC-time. $\endgroup$ – Pieter Sep 26 '17 at 18:37
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    $\begingroup$ @KaneBilliot - to be blunt, it would appear that you need to establish some basic background in analog circuits, like what 'capacitive coupling' means for ac signals. Otherwise you are in way over your head. $\endgroup$ – Jon Custer Sep 26 '17 at 19:09
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The device that you would connected to Signal is also connected to ground. Without the resistor near Signal, the reading would always be $0$.

$V_0$ is the maximum voltage applied to the detector. Without the resistor, it could provide infinite current so that the everything to the left of the capacitor was at a constant voltage. With the resistor, only a finite amount of current can be delivered. This resistor and $V_0$ are an idealization of a real voltage source with internal resistance (though the detector also has its own internal resistance), which can only provide a finite amount of current.

The capacitor acts as a high pass filter. If it weren't there, there would always be a current between ground and $V_0$. This creates a voltage at Signal of $V_0\frac{R_{V_\text{Signal}}}{R_{V_0}+R_\text{Signal}}$. This decreases the effective range of voltages you can measure and creates a possibly destructive current that can damage your measuring device. It also reduces the effects of low frequency noise for similar reasons.

Here's a timeline of events:

  1. When there is no radiation, the capacitor is fully changed. There is no current flowing because the left plate is at the potential of the voltage source and the right plate is at ground potential. Because there is no current flowing, the voltage drop across the resistors is $0$.
  2. When some radiation ionizes the medium between the outside of the tube and the wire, it creates a conductive path with little resistance.
  3. A current forms, discharging the capacitor (since there is now a circuit formed between the two plates of the capacitor that doesn't go through a voltage source), bringing the potential of the left side down and the right side up. This registers as a spike in Signal. The current only has to pass through one resistor: the one near Signal. Ideally, the resistance of this resistor is small to allow the capacitor to discharge as much as possible while the conducting path exists.
  4. Once the conducting path between the tube and wire no longer exists, the current from $V_0$ can recharge the capacitor, bringing it's left side back up to $V_0$ and its right side back down to $0$. Ideally, the resistor near the voltage source is large, limiting the current that can flow. This keeps the voltage source safe by limiting the amount of current that it can produce, keeps the effect of the current from the voltage source to a minimum while the conducting path exists, and allows the capacitor to charge slow enough for the signal to be more easily visible by your measuring device.
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The electronics there implement a simple high-pass filter that prevent flow of the DC high-voltage supply to the DAQ (or ground) but allows the (rapidly changing) signal pulse to pass almost unaffected.

The very simple form used here is acceptable because there is suck a stark difference between the signal you want to keep (time scale of a few tens of nano-seconds meaning frequencies around 100 MHz) and the one you want to exclude (the DC high-voltage).

The feature to notice is that there is no wire that connects the high-voltage to ground anywhere (because a capacitor has a gap, right?). At the same time as charge move on and off the capacitor due to things happening on the tube side of the circuit it causes a detectable change in potential (voltage) on the DAQ side of the circuit.

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  • $\begingroup$ Thank you. What about the resistor near the input? $\endgroup$ – Kane Billiot Sep 26 '17 at 19:27
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    $\begingroup$ It creates an input impedence and prevents the free-flow of a lot of current in the case of something going wrong. It is usually chosen to be pretty large, because you want to limit the damage the HV can do. 1000+ volts is scary. $\endgroup$ – dmckee Sep 26 '17 at 19:29

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