P
Phil Hobbs
Rene said:That'd be 25mV into 50 Ohms. We're getting far less,
or at least that I measure.
When I'm lucky, I get 3mVpk from 1mW average. Well,
the diode is held into a reflex from a crystal or into the
leakage from a mirror. While the leak of 1mW may be 4mm in
diameter, the diode is not that big. The diode doesn't have
the bandwidth of the pulse and acts as integrator.
Then the dutycycle of the pulses is in the order of 10^-4.
And there are 2m of cable between the sensor and the
amplifier electronics.
Rene,
The responsivity of a good photodiode is on the order of 1A/W, depending on
wavelength--at 1.24 um, the energy of 1 photon is 1 eV, so 1W equals 1A of
photocurrent if the quantum efficiency is 1.0. My 1 mW pulses at 1.55 um
produce about 800 uA of peak photocurrent, because there's no antireflection
coating on the photodiode I'm using. There's definitely some buried treasure
in your setup.
Using a photodiode as an integrator for the pulse is usually a bad idea,
unless the pulse energy is small--the local carrier concentration goes so
high that all the carriers recombine before having a chance to separate, and
the response becomes very nonlinear.
A small integrating sphere is a much better method for stretching the pulse
out--you can get on the order of 50% photon efficiency if you're careful, and
the pulse will be stretched out to about 30*(sphere diameter)/c, which can
easily be several nanoseconds even with a very small sphere.
I've seen people use silicon PDs with long-wavelength lasers e.g.
CO2--running them in forward bias, and detecting the change in forward
voltage with laser power, but that's the only way I know to get sensitivities
as low as you're reporting.
Cheers,
Phil Hobbs