Models 2151 & 2153 Operation • 9
will have an output voltage of approximately
(0.1 nW)·(0.5 A/W)·(2×1010 V/A) = 1 V.
The maximum optical power that can be detected by
the photoreceiver is determined by the input optical
power at which the transimpedance gain stage
saturates. We can calculate the saturation power at
900 nm for the Model 2151 assuming a maximum
output voltage of 5 volts. (The output can typically
generate greater than 5 volts, but to be conservative we
assume a maximum output of 5 volts.)
Using the expression 5 V = Psat·R·G, the Model 2151 has
a saturation power of 0.5 nW for the DC Low and
AC Low gain settings and 0.05 nW for the AC High
gain setting. At other wavelengths where the
responsivity is lower, the saturation power increases
inversely with responsivity.
Shielding the Photodetector
Since the femtowatt photoreceivers have extremely
high sensitivity, you should to be careful to shield the
photodetector from any unwanted light sources. The
simplest technique is to use baffling or other physical
barriers such as black cloth or opaque beam tubes to
block stray light. Use of the Model 1280 1" filter holder
(see page 19) with appropriate optical bandpass filters
is also highly recommended.
To illustrate the problems that can be caused by even a
low power point source far from the photoreceiver,
take a look at the following calculation:
where ris the photodiode radius, Ris the distance from
a point source to the photodiode, Pois the power
emitted from a point source, and Preceived is the power
which will be incident on the photodiode from that
point source at that distance.
Preceived Poπr2
4πR2
-------------1
4
---Por
R
---
20.25 1mW 0.5mm
1m
----------------
20.06nW
˙
=⋅⋅== =
215x revC.fm Page 9 Tuesday, May 21, 2002 11:04 AM
