I was trying to fix a light sensor module and using a diagnostics tool for it found that 1 of the 4 photodiode sensors was faulty (it was reading 0xFF.) As I don't knowwhich photodiode corresponds to what on the board, I took the following steps.

  1. Put a multimeter in diode measuring mode across each photodiode in both polarities. All 4 photodiodes read 0.8V and 0.45V respectively.

  2. I put a multimeter in current measuring mode across each photodiode and shined a light on it. They measured 0mA. Does this mean it's a phototransistor?

  3. I unsoldered the 4 photodiodes, reversed the order of them (i.e. moved the 1st diode to the 4th, 3rd diode to the 2nd) and resoldered them. I used a soldering iron at around 300C and noticed the chip case slightly melting while unsoldering from the pads.

I reconnected the module and now all 4 photodiodes are showing as faulty (0xFF reading.)

At which step would I have destroyed the photodiodes, and next time how to prevent it from happening?

enter image description here

  • 2
    \$\begingroup\$ Most likely unsoldering them. \$\endgroup\$
    – user16324
    Oct 6, 2020 at 11:45
  • 1
    \$\begingroup\$ noticed the chip case slightly melting while unsoldering from the pads Not willing to state the obvious, but this seems to be the problem. \$\endgroup\$ Oct 6, 2020 at 19:08
  • \$\begingroup\$ All manner of optic diodes are sensitive to heat and may break in case you heat them up several times. There should be a solder temp spec in the datasheet - it might often be a fair bit below 300 dgr C and of course also for a short time. I take it that "case melting" means the pin moving and not the actual silicon case melting. Otherwise, well... you heated it far too long or maybe touched the case with the solder iron directly. \$\endgroup\$
    – Lundin
    Oct 16, 2020 at 8:55

2 Answers 2


I'm not convinced you've destroyed them, especially if that is the 'after' photo. It's fairly hard to invisibly destroy most semiconductors.

The zero current probably was because you need to be on a uA range.

I assume you carefully flipped them to the correct polarity.

As to what is or was actually wrong (you may have introduced new issues), there is a very fried looking ceramic capacitor just above the PDs. Keep in mind that internal diagnostics have their limits. The trace which appears to go to a ferrite bead may also have an open caused by overheating but it's hard to tell from the photo.

My initial diagnosis would be that the cap shorted, causing it to overheat, trace may have overheated and become open/intermittent and the ferrite may also be damaged. The ferrite bead + large value ceramic capacitor could be a filter for the common photodiode bias supply.

  • 1
    \$\begingroup\$ This is a before photo, so the cap was already fried. The only work I did between the before and after was move the PDs around, and doing that caused all 4 sensors to appear faulty. Meter was on uA, but I think these are daylight filtering PDs maybe that's why. \$\endgroup\$
    – tgun926
    Oct 6, 2020 at 11:51
  • \$\begingroup\$ I've been using some similar-looking PDs in a project. The small lead is probably more fragile because it's attached to the wirebond (the fat lead goes under the die). If you actually melt the plastic it would be easy to break the wirebond. Since the die is so huge the wirebond is long (may not be visible with the tinted plastic). \$\endgroup\$ Oct 6, 2020 at 11:58

In Step 3 you destroyed them,

  1. by heating them to such a high temp so that plastic melted.
  2. by connecting them in reverse due to which high current flowed and destroyed the Diode.

set your multimeter in microcurrent measurement mode and check current. it should be as shown below.enter image description here

ref. Vishay Photodiode datasheet

  • \$\begingroup\$ I should've clarified, I didn't reverse the polarity of each individual diode, I changed the location of each diode such that the order was reversed. \$\endgroup\$
    – tgun926
    Oct 6, 2020 at 11:07
  • \$\begingroup\$ so probably it is the heat, but still, you can do one thing set your multimeter to Current mode and read the current. it should be in micro-amps. see the pics I attached in answer \$\endgroup\$
    – Deepak
    Oct 6, 2020 at 11:29

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