RPC Signal Fraction

H. Wieman
9/8/04

When a charge moves across the gas gap in an RPC it induces charge flow from one conductor pad to the other. The distance that the charge moves, the dielectric thickness and the dielectric permittivity affects the amount of induced charge flow. A simple one dimensional analysis can provide the correct function. This analysis considers only AC effects and should not be confused with the DC picture. In the DC picture due to the finite conductivity of the dielectric there is no field across the dielectric; the full field appears across the gas gap. For AC signal induction, however there are fields in the dielectric as determined by the dielectric permittivity and the charges created in the gas by ionization.

Consider a sheet of charge, s, at position x in the gas gap. The following analysis using Gausses law with Gaussian pill boxes, the dielectric field matching condition and the AC potential across the detector calculates the surface charge on one electrode. The change in this charge as the sheet charge is moved across the gap provides the value of the induced signal that is measured. The analysis could be done by separating the problem into the positive ion motion and the electron motion, but here we in the end move a single charge sheet across the full gas gap which gives the same total as one gets by moving the positive ion component part way and the electron component the rest of the way. One could of course also use Ramo's theorem to solve this problem, but this simple approach is adequate for getting the fraction of induced charge compared to ionization charge.

The case under consideration is shown in figure 1. There are two conductive plates connected with a 0 impedance charge integrator (AC picture, 0 volts between the plates ). There is a gas gap with a sheet charge s at location x and a dielectric to the right of the gas gap. Also shown are the electric fields in the various regions which we will evaluate. The rectangle shows the boundary of the Gaussian pill box used to get equation 3. A similar pill box is used to get equation 4 by moving the right boundary to the left of the sheet charge s.

Figure 1

The potential drop from one conductor to the other is 0, i.e.

gives the following equation:

Eq. 1

The field changes going across the dielectric boundary due to polarization charge and is given by the following equation

Eq. 2

Using gausses law and the Gaussian pill box shown in figure 1 gives the following equation

Eq. 3

is the surface charge on the left hand conductor

is the sheet charge of one sign in the gap created by ionization

Use a Gaussian pill box again, but this time include the surface charge only to get:

Eq. 4

Solve equations 1-4 to get an expression for the conductor surface charge as a function of x

Eq. 5

Get he total induced charge passing through the integrator when the charge sheet moves the full distance across the gas gap:

using Eq. 5

Total induced charge

Consider the case where the dielectric thickness is 5 times the gas gap and the relative permittivity constant is 4

close to SiO2 value

The fraction of the ionization charge that is detected at the amplifier