Photodiode depletion region confusion

[Tandy]

In my quest to better understand how photodiodes work I have had to look closer at the physics of semiconductors.

As I understand it a photon hitting the semiconductor with the correct energy will shift an electron from the valiance band into the conduction band. If the electron is released in the depletion layer then the electrons and holes will migrate to the terminals of the diode. If the electrons are displaced outside the depletion zone they will usually just recombine as therefore not adding to the current produced.

So the thing that is confusing me is that it is generally accepted that the depletion zone shrinks as the temperature rises and yet a photodiode becomes more sensitive as temperature rises, something isn't adding up?

[Andy Watson]

Heat contributes to the thermally generated electron-hole pairs - think of it as another form of light radiation.

[Tandy]

I'm not sure that heat explains the sensitivity issue as if the depletion layer shrinks then I would expect to see less sensitivity to longer wavelengths but the opposite is true.

[John Coloccia]

I'm not a photodiode expert, but if I think of everything I know about them, I'd probably come to these conclusions:

as it heats up:

1) response favors longer wavelength (i.e. less energetic photons) simply because there's already excess energy available to knock electrons about

2) dark current goes up...so efficiency goes down

3) the depletion region shrinks and more current will come from diffusion....therefore efficiency goes down

So you're partly right, I think, but you're confusing sensitivity with efficiency.  In this case, it's important to separate the two, whereas in most other instances if sensitivity goes up it's often because effeciency went up.

[TimFox]

The photodiode has a depletion region at zero bias, but it becomes thicker with applied reverse bias.  With reverse bias, there is more volume to stop photons (more important at higher energy than visible), the capacitance decreases, but the leakage current increases (especially at high temperature).  At zero bias, the device looks like a resistance and capacitance in parallel with the photocurrent generator (the resistance is the slope of the I-V curve through the origin), and the resistance is a noise generator.  However, the dark current (and its noise) goes away.
In reverse bias, with a high-impedance load, the capacitance charges up until the voltage reaches the bias and the device conducts further current away.  This is common in arrays, but the capacitance is non-linear.  If you discharge the device into a charge-sensitive amplifier, the output is more linear than if you only read the voltage.
The device can be run into a virtual ground with zero or reverse bias to measure photocurrent.

[Tandy]

My conundrum is that I want to mount a photodiode in a permanent position that can measure sunlight exposure. Clearly in the winter the temperature can go very low and when direct sun hits the sensor enclosure at the hight of summer it can get very hot.

From my understanding, if for arguments sake a 400nm light wave hits the photo diode, as long as it can pass through the upper layer of the semiconductor and reach the depletion zone the short wavelength will mean a lot of energy is absorbed and chances are very little energy will be left if it reaches deep enough to pass out of the other side of the semiconductor. A longer wavelength like 800nm however may still have significant energy that was not absorbed.

So to widen the range of wavelengths absorbed by a photodiode we have a PIN diode with an added intrinsic layer to widen the depletion zone. This is why I am puzzled by the behaviour under increased temperature.

I would have expected that as the depletion region shrinks at higher temperatures less of the long wave energy would be absorbed. I had hoped that I could perhaps control the width of the depletion zone by changing the reverse bias in-line with temperature.

My main concern is the wavelength sensitivity, I can deal with extra current due to temperature just by adjusting results according to temperature but ideally I don't want to have to deal with the influence of extra wavelengths as that is not predictable. For example the temperature could be 50degC now making the photodiode sensitive to 900nm where it wasn't before. There is no way of knowing if the 20% increase in measurement is due to 20% increase in light in the visible spectrum or the IR spectrum.

I'm sure I am missing something somewhere about how the photodiode is working, I have that frustrating feeling that there is just one hole in my knowledge on the subject that is just stopping me from quite piecing it together to answer my question. The problem is all the material I can find is either aimed at complete novices so doesn't give anything but a simplified model or is theoretical physics that launches straight into complex mathematics. I don't seem to be able to find something in-between.

[Tandy]

Well I finally cracked it after reading a 1958 book on semiconductors. Where I was going wrong is focusing to hard on the depletion zone.

The situation is that as the temperature rises there are two changes in the diode.
1. The depletion zone shrinks.
This means that the light must penetrate deeper into the diode in order to release an electron/hole pair. This does make an overall marginal reduction in efficiency as less energy is absorbed in the narrower depletion zone. This is somewhat offset by the heat energy making it less obvious than you might expect. So by using a monochromator to isolate a narrow (10nm wide) spectrum of light on to the photodiode and heating it up it shows a reduction in current produced in photovoltaic mode.

2. The band gap decreases.
The energy gap in silicon at room temperature is 1.1eV meaning that light wavelength must be under 1127nm to produce the required 1.1eV from the photon to push the electron out of the valance band into the conduction band. As the temperature increases the energy gap decreases meaning that for example if the gab becomes 1eV then wavelengths as high as 1239nm that previously could not release electrons can now do so.

Ultimately what happens is that the photodiodes sensitivity to lower wavelengths at the blue end begins to roll-off while the peak sensitivity moves further towards the red spectrum while at the same time extending further into the red spectrum.

The bottom line for me is that controlling the depletion zone width with reverse biasing won't solve the issue. It could help widen the sensitivity into the blue spectrum but not stop the increase into the red. So if I want constant readings I have to either control the temperature of the sensor or optically filter the light.