Notes from Alan Cummings emailed on 07/06/2017

7/6/17 ACC

Why do we say incident energy threshold is 0.5 MeV for LET L1 detector?


V1 LET L1 threshold = 0.226 MeV (noise sigma = 0.027 MeV)

Assume threshold is nominal 0.200 MeV

Range of proton with 0.200 MeV = 2.330 micrometer Si

3 micrometer Al @ 30 degrees inclination angle (FOV = 120 degrees) = 1.155 x 3 =
3.464 micrometer Al.

According to Janni, 0.3 MeV proton has range of 3.07 micrometers in Al and
3.64 micrometers in Si. Ratio = 1.1857.

Thus 3.464 micrometer Al is equivalent to 4.11 micrometer Si.

Total range of incident proton that triggers 0.2 MeV range threshold in L1 is
2.33 micrometers + 4.11 micrometers = 6.44 micrometers Si.

Energy of incident proton with range 6.44 micrometer Si = 0.48 MeV.

----------------

Start over using actual threshold of 0.226 MeV

Range = 2.66 micrometer Si (for 0.226 MeV threshold)
Add     4.11 micrometer for window
       ------
        6.77 micrometer Si

Energy of proton with range of 6.77 micrometer Si = 0.499 MeV

Hency ~0.5 MeV threshold is reasonable.


Additional Note:

All:

I checked what the incident energy of a proton would
be to trigger L1 if the electronic threshold were to be 0.100 MeV.
It is 0.41 MeV. So, >0.5 MeV, would
become >0.41 MeV.

What I have not looked at is the threshold for a normal incident
particle vs one with maximum angle.

Best,
Alan