Let QLED(i) be the charge deposited in the silicon by the LED on
channel i on a given side of the detector.
Assume
QLED(i) = A(i) * DeltaTime + C
where
A(i) depends on the shape of the LED beam
DeltaTime is the length of the LED pulse
C is a constant
Then, summing over all channels
Sum{ A(i) } = B (1)
where B is some constant, WHICH IS THE SAME FOR BOTH THE n AND p SIDES
OF THE DETECTOR (conservation of charge...).
The voltage level at the output of the shaper for the ith channel is
V(i) = eff * g * QLED(i) + Offset(i)
where
eff = charge collection efficiency
g = gain of the AToM chip
Offset(i) = offset of ith channel amplifier
and where we assumed that "eff" and "g" are constants (on a given side of
the detector, n or p) independent of channel
Then
V(i) = eff * g * A(i) * DeltaTime + eff * g * C + Offset(i)
V(i) = eff * g * A(i) * DeltaTime + K(i)
where
K(i) = eff * g * C + Offset(i)
and K(i) is independent of DeltaTime.
Just as in internal charge injection, the 50% point of the threshold
turn on curve, T(i), measured in THR DAC counts, is simply related to
V(i) as
T(i) = d * V(i) + doff
where
"d" and "doff" are constant which may have small chip to
chip variations.
Therefore
T(i) = eff * g * d * A(i) * DeltaTime + d * K(i) + doff
Fitting for
T(i) = slope(i) * DeltaTime + Constant(i)
gives
slope(i) = eff * g * d * A(i)
summing over all channels and using equation (1)
Sum{ slope(i) } = eff * g * d * sum{ A(i) }
Sum{ slope(i) } = B * d * (eff * g)
Therefore the sum of slopes for all channels is a (relative) measure
of (eff * g), the product of charge collection efficiency and gain.
This assumed that "eff" and "g" are constants for all channels illuminated
by the LED. If they are not, then the sum of slopes is a measure of
the charge weighted product of efficiency and gain.
When comparing relative (eff * g) for channels on different chips, there
is an uncertainty due to the chip-to-chip variation in the parameter "d",
which is the conversion factor between Voltage and THR DAC counts.
The data from Honeywell suggests that this is a 2% uncertainty.