Doubt about the capacitor in a equivalent circuit (ion chambers)

Question

I'm delving into the topic of ionization chambers, but as someone without a background in electrical engineering, I'm finding the equivalent circuits a bit challenging to comprehend. Specifically, I'm puzzled by the placement of the chamber's capacitance in the equivalent circuit. I thought if the capacitor symbolizes the ion chamber's capacitance, it should be on the left of the voltage generator. But in the picture, it's on the right. Why is it positioned like that?

another image, in which \$V_{ch}\$ is the voltage across the chamber, is the following

Source of the pictures is the book Glenn F. Knoll-Radiation Detection and Measurement

Answer 1

The voltage source is a supernode, meaning it doesn't matter which side certain components are connected to it. Capacitors and ideal current sources are such components.

Put another way: consider a change in voltage at the terminals. What current flows? Move the capacitor to the left side. What current flows? Results are identical, therefore the circuits are equivalent.

Illustrating the capacitor here, with dotted lines, I think is intended to remind the reader that chamber capacitance is effectively here and so will affect output bandwidth, independent of the load resistance connected (indeed, inversely proportional to be exact, thus trading off with sensitivity).

Answer 2

The illustrations are misleading. The ionizing chamber is already drawn using the capacitor symbol and has a presumed capacitance. The additional C on the right side of the Vo source could only add parasitics outside of the chamber.

And you're right, if the chamber was drawn to look not like a capacitor, then the equivalent elements should be right next to it - on the left side of the picture, just to communicate things well.

As others have mentioned, the circuits are equivalent no matter whether the C is on the left or right side of Vo. But the purpose of the illustration is to communicate ideas, not to be "clever" and obfuscate the purpose of the illustration.

If I were drawing the thing, I'd draw the ion chamber as a "black box" - a two-terminal device, with equivalent devices shown explicitly:

^{simulate this circuit – Schematic created using CircuitLab}

IMHO, the diagram below would do a much better job at communicating the central idea here:

^{simulate this circuit}

Answer 3

Ionisation chamber shown in your image is not a capacitor or can't be modelled as a capacitor/capacitance. It can be modelled as a current source instead.

^{simulate this circuit – Schematic created using CircuitLab}

But, as every other practical electrical components, it exhibits a parallel capacitance, \$C_i\$, which is a part of the chamber:

^{simulate this circuit}

So the capacitance \$C\$ in the image represents that parallel capacitance, \$Ci\$, plus everything else (wires, components, measuring devices, etc):

^{simulate this circuit}

Answer 4

I thought if the capacitor symbolizes the ion chamber's capacitance, it should be on the left of the voltage generator.

The large capacitor on the left is representing the capacitance of the ionization chamber.

But in the picture, it's on the right. Why is it positioned like that? Here C represents the capacitance of the chamber plus any parallel capacitance

The capacitor on the right is representing the parasitic capacitance of cables/connections etc.. Hence why we show it with dotted lines. This implies an unpredictable value.

It needs to be taken into account when measuring pulses across the resistor and, is shown in your diagram as a reminder.