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I'm doing some experiments using photodiodes. I raised the temperature to about 1000 degree Celsius for 1 second with melting AuSn near the PIN junction. Then the dark current of photodiode is higher than before. What are the causes? Is the photodiode broken?
PS:
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The way you're checking your circuit is not appropriate. The dark current specified in the datasheet is for 20 V reverse voltage applied. However your circuit does not reverse bias the photodiode.
The datasheet shows typical dark current vs reverse bias:
With no reverse bias, you should expect 0 dark current, as in any other diode with 0 V applied.
You need to test your circuit with some light applied to the photodiode. An input signal on the order of 10 to 100 nW (depending on wavelength) should produce a measurable output signal. Be aware that your photodiode has a strongly directional response, and the input light must be entering from nearly "straight on" to be detected. This will limit your ability to detect diffuse reflections.
I can make a couple of guesses about what might be going on.
You aren't actually measuring the LED current, but instead you are assuming \$I_{LED} = (V_{in} - V_F)/220\$ and assuming \$V_F\$ is a constant. You are operating in a regime where \$V_F\$ will vary significantly (100's of mV), and neglecting this would explain (qualitatively, anyway) the shape of your graph.
The LED has some parasitic conductive path through it that does not cause light emission. For low currents, this path is taking more of the current and resulting in reduced light emissions until some threshold is reached and the proper path begins to dominate.
At very low currents, the optical emission pattern of the LED is changing, causing less of the light to reach the detector and more to be lost into the package.
The way you are hooking up the detector is causing funny behavior. Hooking up the base but not the emitter is not a common way to use a phototransistor (AFAIK), and is not what the device designers would be designing for. Typical phototransistor circuits are shown in an App Note from Sharp starting on page 13.
Best Answer
There are, no doubt, many reasons why the dark current increased, but to answer the question about whether the diode is broken - badly or otherwise - it's just necessary to measure the current with the diode illuminated. If you're using the diode in reverse-biased mode, measuring the reverse current with various levels of illumination will give you some idea of how its response conforms to the data sheet specifications.
If you're using it in photovoltaic mode, then subjecting it to various levels of illumination and load and plotting its response as a load line will also yield salient data.
In the real world, though, if you've exceeded the manufacturer's maximum temperature spec by what... six or seven times? I'd say you've definitely let the magic smoke out.