"Bradley1234" writes:

"Sam Goldwasser" wrote in message
...
"Bradley1234" writes:

"Sam Goldwasser" wrote in message
...

The current is proportional to optical power and is more or less
independent of applied voltage. That's not what is normally
considered to be a resistor.

The current? so we agree that current changes. Im saying its because
the
resistance changes, not because power is created inside a diode that is
adding to the circuit. Are you saying optical power is transferred
right
out of the diode? It comes in as light and is converted to current?

So how do you explain that a photodiode can operate in photovoltaic mode
with no bias and generate power?

I dont deny this, its the effect silicon has which is "special" there is
current flow from the conversion, it doesnt need bias, but if there is bias,
the converted energy, as I understand it, is typically used to control the
size of the depletion region, which is the resistance

The photodiode response has nothing to do with the size of the depletion
region. Changing the amount of light on a photodiode doesn't affect the
depletion region significantly. Light generates charge carriers which
pass through the depletion region.

Im saying light affects the depletion region that causes a change in
resistance, its proportional to incident optical power but assuredly not
linearly proportional

But the current in a photodiode is very linearly related to optical power
until the device saturates.

Okay, here we are both describing a graph in general terms. If there is a
point where the device saturates, its no longer linear after that. I call
this nonlinear, if a typical photodiodes response contains a nearly linear
region? Then its sometimes very linearly related to optical power

Yes, over several orders of magnitude. Even a resistor is nonlinear once
it starts smoking.

There is a fundamental difference between a device which is non-linear
and one which has a linear operating region with some limits where it
becomes non-linear.

Each photon generates one or more electron-hole pairs. That is where the
current comes from.

Well this is the ideal response, but not the actual response. To get to a
point to be able to get 1 to 1 reaction? This is the photon counter
problem. To be able count photons takes a very precise, temperature
controlled, and expensive photodiode, like APD (avalanche photodiode) which
isnt perfect. But in a general photodiode, its not necessarily efficient
where each photon gets converted.

I can go to Digikey and buy a $2 photodiode which will demonstrate linearity
over several orders of magnitude. Is it perfect? No, but darn close. Is
it a photon counter? No. But on average, it's doing a decent job of
counting large numbers of photons.

The conversion causes current, but in typical sensor circuits Im guessing
the application is usually to control a bias current in a known linear
region, so it would be in the input to an opamp, with some bias and by the
photodiode varying resistance would provide varying current in response to
incident light, that output would feed an a/d converter and to be accurate,
it would require calibration

Go do an experiment. Take a photodiode, almost any one will do, Rip it
out of your PC mouse if you like. Put a reverse bias on it of a few V
and a mA meter in the circuit. Now get your calibrated light source and
try it at various intensities. The current will be very nearly linear
with respect to optical power up a point. That point is where secondary
effects come into play. But for a typical 1x1mm active area, a couple
mW should be fine. You can try a laser pointer or LED on it.

Except for a fixed dark current (may be a uA or so), the current will
be linear with optical power.

Now remove the bias but leave the mA meter in place. Repeat the experiment.

In some applications, the photodiode feeds into the summing junction of
an op-amp, between the + and - inputs, with a feedback resistor to set
the transimpedance gain. There is no voltage across the photodiode.
(OK, there is a uV or nV across it depending on the open loop gain of
the op-amp but an ideal op-amp with infinite gain and exactly 0 V
across the photodiode would work as well.)

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