Understanding The Detector "Ear"

Using Ramo/Green's Reciprocity Theorem

MIGUEL DAAL

NADER MIRABOLFATHI

 

This study tests the idea that the ear shape, seen in the graph qo/prc versus qi/prc, is due to charges stuck on the cylindrical detector crystal surface, which escape collection at the electrodes.

It follows a method for calculating induced charges/currents proposed by S. Ramo [S. Ramo, Proc. IRE, Sept. 1939, pgs. 584 & 585]. This method is equivalent to Greens Reciprocity Theorem...

The Ramo Theorem

Consider a charge, q, in the presence of any number of grounded conductors (below left).

We want to calculate the charge induced on conductor A due to the motion of q. Surround the charge by a tiny equipotential sphere, Sq.

Let V be the electrostatic potential due to of the charge in the presence of the grounded conductors. Then,

in the region between conductors. Call Vq the value of this potential on this Sq. Note that

where the derivative is in the direction of the outward normal of S. Now, remove the charge and set the potential of conductor A to 1. Let V' be the resulting electrostatic potential. Note that in the space between the conductors

Call V'q the  value of V' at the place where the charge used to be.Then, we have Green's Theorem,

where R is the volume of space between the conductors and Sq. Nabla R denotes the surfaces,

The left side of Green's Equality is zero becauce

The right side of Green's Equality can be divided into three integrals: (1) One over the surfaces of all the other conductors except A's. This integral is zero becasue V=V'=0 on these surfaces. (2) One over the surface of A. This reduces to

since V' and V equal zero on these surfaces. (3) One over the surface of the sphere, Sq, which becomes

The second integral is zero due to Gauss' Law. (There is no charge in V'.) We obtain,

where QA is the induced charge on conductor A. Along the trajectory of the charge, we may write,

If we know where the charge begins and ends its trajectory, we may integrate,

This expression is key to understanding the Detector "Ear".

Our hypothesis is that the "ear" shape is due to the charges that orginate on Qo and end up on the cylindrical surface of the detector...

Suppose charges, q, originate on Qo, travel via electric field lines to the cylindrical surface of the detector, and end up stuck on the cylindrical outer surface of the detector.

The induce charge on Qo will originate on Qo and then end up in the cylindrical surface. The Ramo theorem dictates that,

The induced charge on Qi will be due to the same charge originating on Qo and ending up on the cylindrical surface. Hence,

Using the Matlab PDE toolbox, We simulate the potential inside the detector. The simulation uses the correct germanium dielectric constant and places the crystal in a grounded case.

We generate the Ramo potentials for  Qi and Qo. (This means I set Qx = 1V and Qy = 0V.)

Then, we plot the values of these Ramo potentials along a vertical line close to the edge of the crystal. The line happens to be x = -3.74.

Finally, here is the "ear" from the above figure plotted on top of data from UCB Run 296, for instance. To compensate for the bad normalization of Qo, we multiply the y-values by 1.4. Otherwise, no other transformations are used. [Thank you, Jeff, for the data plots!]

The "ear" that results from the simulation should go all the way to Qi =0 since there is nothing stopping charge from collecting on the 1mm of bare cyrstal at the top and bottom edges of the detector. Extending this simulation as such...

 

The 1in Detector

The Matlab Simulation Code (.tar)