Laser test stand

SLAC system, for reference

SLAC laser test stand. The SLAC system has a large motorized stage instead of our small hand-driven one, but the optics package is similar.

Laser power calculations

ALPIDE is 50 um thick. Mean energy loss for a MIP is 388 eV/um, so 19.4 keV. (Most probable energy loss is lower by a factor of ~2.) The number of electron-hole pairs you get is (energy deposit)/(ionization energy), ionization energy is 3.67 eV in silicon (PDG 34.7.1).

This calculator gives transmission coefficients of 0.077460 for 850 nm and 0.93796 for 1056 nm. The number of electron-hole pairs you get is just the number of photons that are absorbed.

How much laser energy do we need to simulate a certain MIP energy deposition? The amount of energy we need per laser pulse is (hc/(wavelength)) * ((MIP energy deposit)/(ionization energy)) / (1-transmission).

For 850 nm: (h*c/(850 nm)) * ((19.4e3 ev)/(3.67 ev)) / (1-0.077460) = 1.34e-15 J.

At nominal power 20 mW and pulse width 4 ns, a pulse would be 80e-12 J. Need 6e4 attenuation, about 48 dB or OD 4.8.

For 1064 nm: (h*c/(1064 nm)) * ((19.4e3 ev)/(3.67 ev)) / (1-0.93796) = 15.9e-15 J. At nominal power 10 mW and pulse width 4 ns, a pulse would be 40e-12 J. Need 2.5e3 attenuation, about 34 dB or OD 3.4.

For normal operation let's run with an OD 4.0 filter in the 850 nm lens tube, and OD 3.0 in the 1064 nm lens tube.

Laser power calculations (new, use most probable deposition and epitaxial layer thickness)

Epitaxial layer of ALPIDE is 18-30 um thick (process variation?). Mean energy loss for a MIP is 388 eV/um, so 19.4 keV. (Most probable energy loss is lower by a factor of ~2.) The number of electron-hole pairs you get is (energy deposit)/(ionization energy), ionization energy is 3.67 eV in silicon (PDG 34.7.1).

This calculator gives transmission coefficients of 0.077460 for 850 nm and 0.93796 for 1056 nm. The number of electron-hole pairs you get is just the number of photons that are absorbed.

How much laser energy do we need to simulate a certain MIP energy deposition? The amount of energy we need per laser pulse is (hc/(wavelength)) * ((MIP energy deposit)/(ionization energy)) / (1-transmission).

For 850 nm: (h*c/(850 nm)) * ((19.4e3 ev)/(3.67 ev)) / (1-0.077460) = 1.34e-15 J.

At nominal power 20 mW and pulse width 4 ns, a pulse would be 80e-12 J. Need 6e4 attenuation, about 48 dB or OD 4.8.

For 1064 nm: (h*c/(1064 nm)) * ((19.4e3 ev)/(3.67 ev)) / (1-0.93796) = 15.9e-15 J. At nominal power 10 mW and pulse width 4 ns, a pulse would be 40e-12 J. Need 2.5e3 attenuation, about 34 dB or OD 3.4.

For normal operation let's run with an OD 4.0 filter in the 850 nm lens tube, and OD 3.0 in the 1064 nm lens tube.

Laser focus

The beam diameter after the collimator is 2.4 mm for both collimators. The focal length is 25 mm for both lenses.

The diffraction-limited spot size is 4*wavelength*focal length / (pi * beam diameter). This is the 1/e^2 diameter, and 86% of the power is inside this diameter.

For 850 nm: 4*(850 nm)*(25 mm)/(pi * 2.4 mm) = 11.3 um.

For 1064 nm: 4*(1064 nm)*(25 mm)/(pi * 2.4 mm) = 14.1 um.

Both numbers are significantly smaller than the MAPS pixel dimensions: a circle with diameter 14.1 um fits very comfortably inside a square with sides 28 um. So we should be able to hit a single pixel. We could do math to see how much energy hits adjacent pixels.

The lenses are not perfect, so there will be some broadening. Also, the beam only comes to a perfect focus at one Z, so the profile of the beam as it passes through the sensor is more like a cone. Not a big effect with a 50 um sensor. The 850 nm beam is mostly absorbed in the shallow part of the sensor, the 1064 nm beam illuminates the sensor uniformly.

The "working distance" of our lenses is 19.6 mm: see figure to understand what this means. This is the required distance from the end of the lens housing to the ALPIDE.

Calibration

We have two photodiodes for calibration: one Si photodiode for 850 nm, and one InGaAs photodiode for 1064 nm. The Si photodiode is fast enough (1 ns rise and fall times) to resolve a short pulse. The InGaAs photodiode is slower (10 ns rise and fall) but it should still be possible to integrate to measure the pulse power.

Si photodiode: about 0.35 A/W at 850 nm according to spec sheet, or 17.5 V/W into a 50-ohm load. A MIP pulse (80e-12 J from above) should make a voltage pulse of integral 1.4 e-9 V-s.

Pat's LED source

From above, the minimum pulse power you need is around 1.34e-15 J (that's for 850 nm, which is fully absorbed in the silicon). It's not clear how short a pulse needs to be to realistically stimulate the ALPIDE. Assume 100 ns, much slower than a real MIP deposition but (I think) faster than the analong frontend of the ALPIDE. The CW power you need is then 13.4 nW.

This power needs to be coupled into the single-mode fiber that goes to the optics. This fiber (Corning HI780) has a mode field diameter approx. 5 um. Assuming (very bad assumption!) you have 100\% coupling inside a circle of 5 um diameter, the power density you need is 13.4 nW/(pi/4*(5 um)^2) = 0.68 mW/mm^2.

Pat's LED pulser boards use Avago LED transmitters (AFBR-1629Z?): 650 nm, -2 dbm peak power into 1 mm plastic optical fiber. -2 dbm is 0.63 mW.

RF attenuators

Previous experience with AVTech pulsers is that the pulse shape is cleaner when you drive the max amplitude (10 V). So, run the pulser at max and use a fixed RF attenuator to bring the pulse height down to the nominal level for the diode. This also makes it impossible to destroy the diode by overdriving the pulser.

For the 850 nm diode, nominal operating current is 45 mA, and nominal operating voltage is 1.9 V (max is 50 mA). So the nominal pulser voltage is (45 mA)*(50 ohm) + 1.9 V = 4.15 V. The necessary attenuation is 20*log10((4.15 V)/(10 V)) = 7.64 dB. Round up to be safe: we will use the 8 dB attenuator for 850 nm.

For the 1064 nm diode, nominal operating current is 50 mA, and nominal operating voltage is 1.33 V (max current is 55 mA). So the nominal pulser voltage is (50 mA)*(50 ohm) + 1.33 V = 3.83 V. The necessary attenuation is 20*log10((3.83 V)/(10 V)) = 8.34 dB. Round up to be safe: we will use the 9 dB attenuator for 1064 nm.


Sho Uemura
Last modified: Thu Jan 25 17:19:55 MST 2018