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.