412 Chapter 6. Scintillation Detectors and Photodetectors
Example:
An APD having a gain of 20 and responsivity of 8A/W is subject to a
photon beam of intensity 6. 5 × 109 s−^1. Calculate the quantum efficiency of
the APD at 1.6μm. Also compute the average photocurrent.Solution:
The efficiency can be calculated from equation 6.5.67.ξ =R
〈G〉
hc
eλ=8
20
(
6. 63 × 10 −^34
)(
2. 99 × 108
)
(1. 602 × 10 −^19 )(1. 6 × 10 −^6 )
=0. 31
To compute the average photocurrentIγwe use the basic definition of respon-
sivity, that is
R=Iout
P
UsingIout=〈G〉Iγthis can be transformed intoIγ =〈P〉R
G
=
Rφhc
〈G〉λwhereφ=Pλ/hcis the incident photon intensity. Substituting the given
values in the above equation yieldsIγ =(8)
(
6. 5 × 109
)(
6. 63 × 10 −^34
)(
2. 99 × 108
)
(20) (1. 6 × 10 −^6 )
=3. 2 × 10 −^10 A
=0. 32 nA.C.4 ModesofOperation
Just like PMTs, APDs can also be operated in analog or digital modes. However
for APDs they are generally referred to aslinearandGeigermodes.
We saw in the chapter on gas filled detectors that operating a detector in Geiger
region means applying a bias voltage high enough to cause breakdown in the gas.
This is also true for APDs operating in the Geiger mode. If a high enough reverse bias
is applied to the APD, it will cause a large current pulse whenever a photon produces
charge pairs in the bulk of the material. Hence in this way the detector can be used
as a photon counting device just like a PMT. The problem, however, is the dark
current in the bulk of the material, which is significant even at room temperatures.
The trick, therefore, is to operate the detector at very low temperatures. Silicon
APDs have been found to have very low leakage or dark currents and therefore high