SCT Detector shielding and grounding  Tony Smith Jan 00

 

Magnetic field susceptibility of the detector and front end

 

The front ends of Silicon strip detector modules are relatively immune to magnetic fields in their passband due to the very small loop area encompassed by the strip, the backplane metalisation, the fanin structure and the chip input and reference (analog power ground) connection. Depending on the angle of incidence of the magnetic field there are a number of identifiable loops.  Any magnetic coupling into these loops will generate a noise voltage in the loop, and hence a current determined by the loop impedance.  The coupling is directly proportional to frequency, current in the offending conductor, inversely proportional to the distance.from the culprit conductor and proportional to the cosine of the angle of incidence. 

 

Let us examine the six cases of a magnetic field cutting the detector, these are

a)      A culprit radiating wire parallel to the strips and above or below the detector  

            b) A culprit radiating wire parallel to the strips in line with the detector plane at either side of the detector

            c) A culprit radiating wire orthogonal to the strips and above or below the detector

            d) A culprit radiating wire orthogonal to the strips and in the detector plane at either end of the detector

            e) A culprit radiating wire normal to the detector plane and centrally at the side of the strips

            f) A culprit radiating wire normal to the detector plane and at the end of the strips

   

For the above cases there are different coupling loops formed by the strips and backplane, between strips and any loop area enclosed by traces to the backplane decoupling capacitor and from the capacitor back to the amplifier ground.

  

Cases a) and b) are clearly the worst cases for loops where the strips and backplane form the greater part of the loop.

They also cut any loop in the backplane decoupling capacitor traces.

These cover the case of pipework or cabling in the barrel and some of the pipework and cabling in the forward

Figure 1

 

Cases c) and d) do not contribute any coupling to loops formed by the strips and backplane as the field is in the plane of the loop

They do couple into loops formed between adjacent strips, the inter-strip capacitance, and the amplifiers. For strips in the centre of the detector the opposing current in the neighbouring strip will cancel the effect. For strips at the edge the compensation is not balanced and some net effect may be seen. They also couple into any loop formed by the decoupling capacitor traces. These cover the case of some parts of the pipework and cabling in the forward detectors.

Figure 2

 

Case e) Field lines cut the strip and backplane loop tin both directions so the net flux is zero in the strip and backplane loop. There will be no contribution to a loop formed by connections to the decoupling capacitor, as these traces will be in the same plane as the field.

 

Case f) The field lines cut the strip backplane loop and will couple into it but the field is decreasing by 1/r along the length of the loop

Again, There will be no contribution to a loop formed by connections to the decoupling capacitor, as these traces will be in the same plane as the field.

 

From the above it can be seen that:

1) The largest coupling to the strip backplane loop will be from a current carrying conductor running parallel to and above or below the strips

2) Any loop formed by the conductors to the backplane decoupling capacitor can couple signals into the input loop.   

 

Differences between barrel and forward detectors with respect to electromagnetic coupling

 

Figures 1 and 2 essentially represent the geometry of a forward module with amplifiers placed at the ends of the strips. In a barrel module the amplifiers are placed at the centre of the strips and the loop geometry is rather different. Here the strip and backplane loop is essentially split into two halves

 

 

 

Example calculation for order of magnitude of the effects in the strip backplane loop

 

Let us take the loop area enclosed by a 12 cm strip and its backplane on a 300 um thick detector as an example and calculate the current required to flow in a single wire spaced 1 cm from the loop to generate a 10 MHz sinusoidal signal of an amplitude equal to that necessary to give a signal at the amplifier due to a single ionising particle in the detector.

 

The loop impedance will determine the current generated in the loop by the induced voltage so it is necessary to estimate the impedance. The impedances in the loop are those of the amplifier input impedance, the strip resistance, the backplane resistance, the impedance of the strip to backplane capacitance. and the return path from the backplane to the amplifier ground. The issue of the return path is one where there is some debate. Figure 1 shows two possible paths one via the decoupling capacitor and one via the virtual earth inputs of the other strip amplifiers. In the case where a single strip is stimulated by the charge from a single particle it could be argued that the decoupling capacitor and the amplifier inputs are in parallel and there will be current sharing between them in the ratio of their impedances. Here however we must assume that all the strips are equally influenced by the impinging field so all the amplifier inputs will have their input voltage raised equally and the only remaining decoupling path is via the backplane decoupling capacitor. Further to this there will be net flow from all the strips via this capacitor therefore its effective capacitance for a single strip loop calculation will be reduced by  a factor equal to the number of strips decoupled by it.

 

Loop impedance calculation

                        Amplifier input impedance – will be complex but assume low resistance at 10 MHz --- say 30 ohms           

                        Strip resistance                                                                                                             say 100 ohms

                        Strip to backplane capacitance 0.2 pF / cm * 12 cm                                                     2.4 pF

                        Backplane decoupling capacitance (effective) 10 nF / 1536 strips                                  7.25 pF

 

The capacitive elements of the loop will add reciprocally and the effective capacitance will be 1.8 pF

                       

Then the reactance at 10 MHz will be 2.86K ohms

 

 And this being the dominant impedance in the loop, most of the induced voltage will appear between the strip and the backplane.  

 

(Note that the strip to backplane capacitance and the strip resistance are distributed parameters but are lumped in this calculation. If the strip resistance were the dominant impedance then this would need to be accounted for as field lins cutting the loop at the near end to the amplifier would see a lower impedance and produce more current from the induced noise voltage. If the strip to backplane capacitance is dominant as above, then this is not an issue)

 

 Given that the dimensions of the loop are 12 cm by 300um for the strip and backplane plus a further 300

 

.   .   The noise voltage in the loop Vn  = 2* pi * f  * B * A cos theta     ----- 1

Where, f = 10e7 , B is the flux density, A is the loop area and theta is the angle of the loop to the field (assume cos theta =1 for ideal orientation of the conductor axis being in the plane of the loop.

 

The flux density B at a radius r from a conductor carrying a current I, is given by

 

                                                B = u * I /2* pi * r                               ----- 2

 Where u is the permeability of free space = 4* pi *10-7 Henries/metre and r is greater than the conductor diameter.

 

  The current required to give 22000 electrons in 25 nS                         I = q/t = 22000 * 1.6 * 10e-19 / 25e-9

                                                                                                = 1.4 uA

 

To induce this current in the loop impedance of 100 ohms we would need Vn to be 1.4 uA * 100 ohm = 140 uV

 

Substituting 2 into 1 for B and rearranging we get I = Vn * 2*pi * r  / (2*pi* f * u * A * cos theta)

 

                                                                   = 1.4e-4 * 1e-2 / 1e7 *4pi *1e-7 * 3.6e-5

 

                                                                   = 11 mA  @ 10MHz @ 1cm

 The coupling at different frequencies, ignoring the shaper response would be

                                                                        1 mA @ 100MHz

                                                                        100mA@ 1MHz

                                                                        1 A     @ 100KHz

                                                                        10A    @ 10KHz

 

Although these frequencies are outside the shaper passband it should be remembered that the front end amplifier has a much wider bandwidth than the shaper and therefore could suffer from being driven outside its operating range if large signals were applied 

 

This would seem to be a quite large effect for quite a small current but it must be remembered that the calculation is done for a single conductor where the return current is flowing in a return conductor, which is remote and does not affect the magnetic field. In practice this will not be the case for power and signal conductors, which will be arranged so that the return current travels in close proximity to the outgoing current and contributes a nearly equal and opposite field. In this case the coupling will be several orders of magnitude smaller.  There is however an exception to this balanced situation with field cancellation and this is where current flows through shielding foils and the cooling tubes due to external loops and common mode currents. Fortunately for the overall external shield, if the current flow in it is uniform, then there will be no magnetic field inside it due to current flow in the shield.

 

 In the case of the cooling tubes within the shield a net magnetic field will exist if current is allowed to flow through the tubes. This raises the question of trying to demonstrate the effect and deciding if it is likely to be significant in the design of the grounding and shielding scheme for the detector.  

 

Experimental verification

 

Barrel system test

To test if the above calculation is correct to an order of magnitude I suggest that a test be made with a current being passed along the cooling tube with the return current part of the loop at a reasonably large distance (say 30 cm)  so that it subtracts only a very small fraction from the field generated by the tube. For this test it will be necessary to isolate the tube at one end so that the current cannot flow by any other route than the tube. Figure 1 shows the arrangement of the test setup required

Note the position of the 50 ohm load  which terminates the signal generator output. Its position is important as if it were placed anywhere else then some portion of the loop would also radiate electrically as well as magnetically. Note also that the detector analog ground should be shorted to the tube so that there is no electric field coupling.

Figure 3

 

Method

The setup is constructed as in figure 3. The tinned copper wire loop should be of a substantial cross section (say 16 gauge or larger or possibly made from a self-supporting piece of aluminium angle or tube)

A signal generator is connected via short coax cable. It should be powered, set to 10MHz sinewave and the output voltage set to zero.

A calibration run is now done to establish a baseline noise for the setup.

The output voltage of the signal generator voltage should now be adjusted so that the voltage measured differentially across the 50 ohm load is 0.5 volt, which will drive 10 mA of current around the loop. Note that it is important to measure the size of the signal on each side of the 50 ohm resistor and subtract the two signals because at higher frequencies the inductance of the loop will become significant, and a simple measurement of the generator output voltage will overestimate the current. (Note.  I estimate the reactive impedance of this 3 metre loop is around 100 ohms at 10MHz  so the voltage required at the signal generator output to provide

0.5 volts across the resistor will be around 1,2 volts Current leads voltage by 90 degrees in the inductance so total drive voltage is the quadrature addition of the voltage across the inductor and the voltage across the resistor) 

A noise scan should now be done

Depending on the result of this noise scan, further scans should be taken after adjusting the output level and /or the frequency

 

I suggest frequency should be scanned in decades from 10KHz to 50MHz in half decade steps. The output voltage of the generator should be adjusted each time the frequency is changed so there is always the same voltage across the resistor and hence the same current in the loop. If there is a marked peak in the response versus frequency then a finer scan may be useful.

 

Non-sinusoidal stimulus

 

It would be interesting to see what the response is with square wave signals and with varying mark space ratios, however it is difficult to see how to interpret the results as the inductance will limit the current of the high frequency components of the signal so the current waveform will not be the same as the applied voltage waveform. This should be evident from the voltage waveform seen across the resistor.

 

What we might expect to see

 

The calculation has been done assuming that the magnetic field cuts the loop area of the detector at right angles (cos theta = 1) however in this setup the cooling tube is very nearly in the same plane as the detector so the field will cut at a small angle so the effect will be considerably smaller. The positive side to this is that the angle of incidence will be nearly constant with distance from the tube so, if the effect is measurable, it should be possible to see a difference in the noise on strips depending on the strip proximity to the tube. Note that the 1/r dependence will be convoluted with the quadrature addition of noise so the falloff of noise with distance will be more rapid than the 1/r relationship alone would give.

 

Potential inaccuracies

The strips are sensitive to capacitively coupled charge

 

 

What if we don’t see an effect (or even if we do).

 

Should it prove impossible to see an influence on the noise from current in the tube (or even if it is possible) it would be useful to check that it is possible to see the effect from a wire 1cm above the detector, Figure 2 . This would involve almost the same setup except that instead of the cooling tube carrying the loop current, the loop is closed by a tinned copper wire suspended 1cm above the detector and preferably not centrally but offset to one side.

 

Figure 4

   

 NEED RESULTS HERE TO DRAW CONCLUSIONS

 

ALSO NEED SETUP WITH OVERALL FOIL SCREEN AND TUBE SHORTED TO IT AT BOTH ENDS TO SEE IF CURRENT IS DIVERTED TO THE SCREEN AND THE EFFECT REDUCES 

 

Susceptibility of detector and front ends to electric fields

 

Strip detectors are particularly sensitive to changes in electric fields due to their charge sensitive front ends and large area strips, which have significant parasitic capacitive coupling to conductors or other detectors. 

 

Example calculation for order of magnitude of the effects

 

Take a simple case of two modules overlapping along the full length of 12 cm with a 3 mm gap between the strips on one module and the strips on another module. (as in the Atlas SCT forward region )

The amplifier inputs are virtual earths so the strips on each detector will be forced to the AC potential of the amplifier ground reference point on each of the modules.  Therefore any potential difference between the ground reference points on the two modules will appear as a potential difference between the two silicon detector faces. From the point of view of a single strip on one of the detectors in the overlap region, the opposing face of the superposed detector will appear as an equipotential second plate of a parallel plate capacitor with a capacitance determined by the area of the strip and the separation of the strip and plate.

For a 12 cm strip of width 100 um (assume all field lines in the width of one strip pitch terminate on the strip) and separated by 3 mm from the overlying detector then:

 

                        C = 12e-2 * 1.0e-4  * 8.85e-12

                               ------------------------------------               = 35.4 fF

                                                3e-3

 

This parasitic capacitance is connected directly to the input of the strip amplifier and consequently any voltage change across it will inject a signal in the same way as the on chip calibrate capacitor is used to inject a known calibration charge.

 

In order to get a feel for the problem which may be caused by this parasitic capacitance let us calculate the magnitude of a voltage step which would be required across this parasitic capacitance to inject a charge equal to 1 mip.

 

=                                    22500 * 1.6 e-19

                                                ------------------------       =100 mV for 1 mip

 

                                                            3.54e-14

 .

 

Now this is for a 1 mip signal injection and clearly such a large parasitic input would completely swamp real signals- what we really need to know is the maximum noise voltage which we can allow between the detector grounds before the performance of the detector is compromised.

 

 For this purpose let us now assume that we are dealing with non-correlated random noise where the noise contributions are RMS values and add in quadrature.

 

The noise in RMS electrons of the unirradiated module is expected to be around 1500e-. If we now say we can allow an additional 100 electrons total noise so that the total noise will become 1600e then we can allow a contribution from the injected signal of 556 electrons. - dividing this by 22500e and multiplying by 100mV gives us an allowed noise voltage between the two detector grounds of 2.5 mV RMS. 

 

It is clear that one way to minimise voltage differences between module grounds, and hence minimise injected noise, is to short the sensor reference grounds together by a low impedance path..

 

It may be possible to use the cooling tube as a common reference conductor and also as a common (safety)  ground point in the detector.