From: Phil Hobbs
Subject: Re: Anyone Need A Laser Power Meter?
Date: Tue, 08 Oct 2002 11:20:03 -0400
Organization: IBM T. J. Watson Research Center
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NNTP-Posting-Date: 8 Oct 2002 15:20:06 GMT
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Ed Edmondson wrote:
> I need a power and wavelength meter to use with my laser collection. However, I
> am not an engineer, just a technical person.
Chuck Simmons wrote:
> The power meter is simple enough. Either a silicon solar cell or a large
> area photodiode will do fine. The problem is calibration and
> compensation for responsively variation with wavelength.
Right. Given that the OP is a hobbyist and not a professional, the
performance requirements aren't as stiff and the cost constraints are
much tighter, so here's a suggestion.
Except at very long or very short wavelengths (<400 nm or >950 nm),
silicon photodiodes have pretty flat quantum efficiencies with respect
to wavelength. The main problem is the front surface reflection. One
good way to solve this (and to eliminate polarization sensitivity) is to
wire three photodiodes in parallel, physically arranged so as to require
the beam to make five reflections before it can exit again. The
arrangement is as follows:
1. Imagine the light coming in vertically downwards.
2. Hits a 45 degree surface that sends it horizontally due north.
3. Hits a 45 degree surface that sends it horizontally due east.
4. Hits a 0 degree surface that sends it back due west.
That way it has to hit the first two diodes twice, making five bounces
in all. Because of the particular orientations, the polarization
dependence of the front-surface reflection goes away, and because of the
5 reflections, the power lost is less than 0.2% even if the photodiodes
are not antireflection coated. Because of the high effective quantum
efficiency of this arrangement, you can assume that you'll get one
photoelectron per incident photon, and so the photocurrent will be
I = P_optical*lambda*e/(h*c).
where e is the charge on the electron (1.602x10**-19 coulombs), c is the
speed of light (3x10**8 metres/second), and h is Planck's constant
(6.626x10**-34 joule seconds). The magic factor of e*lambda/(hc) is the
charge (e) on each electron divided by the energy (hc/lambda) in each
photon, so it converts optical power to electric current. If lambda is
in microns, the factor e/(hc) is 0.806 amps per watt.
This trick also greatly reduces the ripples in the sensitivity caused by
interference fringes inside the photodiode windows, which are
surprisingly serious (10% peak to peak for a single longitudinal mode
For a hobbyist, silicon photodiodes are more nearly constant with
wavelength than anything else you can easily buy or make. If you happen
to have a pulsed laser, a pyroelectric detector based on some PVDF
plastic film (as used in automatic front porch lights), painted flat
black, will be flatter with wavelength but very much harder to
Use real photodiodes or high efficiency solar cells for this job--'twere
it me, I'd use 1-inch square solar cells, about $5 each--but don't use
low-efficiency cells such as thin-film or amorphous silicon.