Physics and Engineering of Radiation Detection

6.5. Photodetectors 399

Example:

If the temperature of a PMT based detector working at 200Kis elevated

to the room temperature of 300K, what would be the relative change in its

Johnson noise.

Solution:

Since all other parameters are supposed to remain constant at the two tem-

peratures, we can write equation 6.5.30 as

σamp = K

√

T

where K =

√

4 FampkBB

Reqv

.

The percentage relative change in Johnson noise can be obtained by writing the

above equation at the two temperatures and taking their relative difference.

σamp =

σamp, 1 −σamp, 2

σamp, 1

× 100

=

√

T 1 −

√

T 2

√

T 1

× 100

=

√

200 −

√

300

√

200

× 100

=22.5%. (6.5.32)

All of the noise components described above contribute to the overall signal-to-
noise ratio. If we know all the individual currents, we can compute the signal-to-noise
ratio from

S/N =

Iγ

[

σ^2 st+σ^2 bk+σ^2 d+σamp^2

] 1 / 2

=

Iγ

[2eμ^2 FB{Ipe+Ibg+Id}+4FampkBTB/Reqv]^1 /^2

. (6.5.33)

However, practically it is difficult to separate the individual currents from the mea-
surements. The easiest and most practical thing to do is to take two sets of mea-
surements, one without and the other with incident light. The shot noise associated
with the measurement without incident light is given by

σbg,d,amp=

√

σ^2 bk+σ^2 d+σ^2 amp, (6.5.34)

while the total shot noise after shining the photocathode with incident light is

σtot=

√

σ^2 st+σ^2 bk+σ^2 d+σ^2 amp. (6.5.35)

Now, the currents corresponding to these noise levels can be subtracted to get the
current due to light only, that is

Ipe=Itot−Ibg,d,amp. (6.5.36)