4 ms·

Oh neat, I just reproed it with a Pi 2 and a Canon Speedlight flash. I'll put my scope on the power lines and see what's happening when you flash the board. S

by tdicola 12y ago

Oh neat, I just reproed it with a Pi 2 and a Canon Speedlight flash. I'll put my scope on the power lines and see what's happening when you flash the board. Sounds like from the thread one of the power ICs is photo sensitive. edit: Wow yeah, here's a look at the 3.3V power line when you flash the board, it drops almost down to 0V and then wildly fluctuates for about 100 nanoseconds: http://imgur.com/hG86pRy http://imgur.com/hG86pRy edit 2: Another interesting measurement, with the board _totally unplugged_ and flashing it you can see a big voltage spike on the 3.3V rail. Up to 6-7 volts or so for a few nanoseconds: http://imgur.com/td262QK http://imgur.com/td262QK I guess not only can you learn about electronics but also Einstein's photoelectric effect with the Pi 2!

noonespecial 12y ago
Are you sure that's not induced current from the power spike required to operate the flash or the spark-gap inside the xenon tube? You've got the gear set up... would you mind terribly repeating the experiment but blocking the light with cardboard or something?
- robkix 12y ago
  If I understand correctly the Speedlite is battery powered, so as such shouldn't cause any voltage spike on mains.
  - nitrogen 12y ago
    It's not a mains spike, but an EM discharge caused by the rapid voltage and current changes in the flash, that the parent comment is considering (IIUC).
    - noonespecial 12y ago
      Yes thankyou. This is what I'm getting at. The momentary power flow in a xenon flash is very large. A great big capacitor charges up and then discharges very rapidly to create the bright flash of light. There is also a high voltage exciter coil that fires to kick the xenon tube into conduction at the beginning of the flash cycle. Both of these could be very disruptive to nearby digital electronics. I thought the original poster would be in a very good spot to help us determine if the effect if photoelectric or the result of EM because he's got everything set up. All he has to do is repeat the experiment while blocking the light from hitting the Pi.
      - tdicola 12y ago
        I'll put it in a box and try in a little bit, sorry have to do a few things first. I'm 99% sure it's light dependent though, I had to aim the flash at the board and keep it somewhat close to get it to lock up (like 6" away or so).
      - karlshea 12y ago
        From the original thread it does appear to be a photoelectric effect, since blocking a specific chip on the board causes it to survive the flash.
- jayrhynas 12y ago
  The most recent post [1] seems to indicate that covering one of the chips with blu-tac prevents the freeze [1] http://www.raspberrypi.org/forums/viewtopic.php?f=28&t=99042#p687959 http://www.raspberrypi.org/forums/viewtopic.php?f=28&t=99042...
  - analog31 12y ago
    Not knowing what the IC is, one thought is that the blu-tac could be blocking light, or providing a wee bit of extra capacitance to tame a wicked electrical signal.
  - bluehex 12y ago
    I'm thinking of trying a more permanent fix with some Sugru[1] when my Pi 2 arrives. Their about says it's insulating and okay for electric repairs under 24 volts. [1] http://sugru.com/about http://sugru.com/about
    - bluehex 12y ago
      Maybe not such a good idea according to: http://sugru.com/questions/can-i-use-sugru-to-secure-shock-mount-a-sensitive-pcb http://sugru.com/questions/can-i-use-sugru-to-secure-shock-m...
- tdicola 12y ago
  Alright I put the Pi 2 in a cardboard box and held the flash at the same spot (but the box obscures the light). It keeps running just fine and doesn't have the nasty spike. Definitely is something light related.
  - noonespecial 12y ago
    Thanks! Well done. Photoelectric it is. Now I wonder what color its most susceptible to...
    - jakobegger 12y ago
      Since the effect seems to be caused by a silicon component that isn't properly shielded from light, anything with a photon energy higher than the bandgap should do. The bandgap of silicon is 1.1eV, which means the effect should start in near-infrared.
      - psgbg 12y ago
        What about glass, then that should work, right?
        jakobegger 12y ago
        Not sure I understand you correctly. The problem should happen with near-infrared, visible, and UV-light. Glass is transparent in the visible range, and depending on type, also in IR / UV, so glass won't work as a shield. The easiest, reliable way to shield would be wrapping in aluminum foil (also helps against RF-interference).
        psgbg 12y ago
        Yeah you are right. I was assuming that the chips would block visible light but I forgot about UV.
    - tomkinstinch 12y ago
fnordfnordfnord 12y ago
I want to see this with a bare regulator. I might have some time Monday.
swamp40 12y ago
My money is still on an electromagnetically coupled pulse onto either the 3.3V or ground rail. I have seen that happen plenty of times, but I have never seen a light flash disrupt a power rail like that. I'm not saying it couldn't happen, just playing the odds...
- crb3 12y ago
  It can and did happen, as the afflicted chip was the SMPS controller. At last, a use for all those write-protect stickers we collected for floppies...
scythe 12y ago
The historical importance of the photoelectric effect is to demonstrate the relevance of the work function Ω. It works like this: V ~ E_0 - Ω where hf is the energy per photon and Ω is the work function of the target. If Ω > E_0, no voltage is produced. This provided the basis for Einstein's demonstration that energy is proportional to frequency (E_0 = hf), and won him the Nobel prize. What does this have to do with xenon? Xenon lamps produce not only visible light, but also ultraviolet light, in fact, xenon lamps extend further into the UV than natural sunlight: http://en.wikipedia.org/wiki/Xenon_arc_lamp#mediaviewer/File:Xenon_arc_lamp_profile.png http://en.wikipedia.org/wiki/Xenon_arc_lamp#mediaviewer/File... LEDs by contrast produce almost no UV light: http://en.wikipedia.org/wiki/Light-emitting_diode#mediaviewer/File:White_LED.png http://en.wikipedia.org/wiki/Light-emitting_diode#mediaviewe... With this data we can determine that the work function of silicon (in the Raspberry Pi 2) is somewhere in the range of hc/(400 nm) < Ω < hc/(250 nm) or 3.1 eV < Ω < 5 eV. Using data from http://journals.aps.org/pr/abstract/10.1103/PhysRev.127.150 http://journals.aps.org/pr/abstract/10.1103/PhysRev.127.150 http://www.sciencedirect.com/science/article/pii/0022369759902914 http://www.sciencedirect.com/science/article/pii/00223697599... we see that the work function of silicon is around 4.7-4.9 eV, which agrees well with our observations. draws a little box
- jakobegger 12y ago
  The problem with your theory is that, as others have noted, the effect seems to happen with a red laser pointer as well. To interfere with the electronics, the wavelength only has to as large as the band gap in silicon (as someone else has noted, around 1.1 eV). This means that IR light (and all visible light) can cause the interference too. A strong enough light source will lift a lot of electrons from the valence band across the band gap into the conduction band, effectively turning a transistor into a conductor for the duration of the flash. Since the sensitive part seems to be related to the power supply, it's quite plausible that this leads to the voltage fluctuations on the power rail.

by tdicola 12y ago

noonespecial 12y ago

Are you sure that's not induced current from the power spike required to operate the flash or the spark-gap inside the xenon tube? You've got the gear set up... would you mind terribly repeating the experiment but blocking the light with cardboard or something?

robkix 12y ago
If I understand correctly the Speedlite is battery powered, so as such shouldn't cause any voltage spike on mains.
- nitrogen 12y ago
  It's not a mains spike, but an EM discharge caused by the rapid voltage and current changes in the flash, that the parent comment is considering (IIUC).
  - noonespecial 12y ago
    Yes thankyou. This is what I'm getting at. The momentary power flow in a xenon flash is very large. A great big capacitor charges up and then discharges very rapidly to create the bright flash of light. There is also a high voltage exciter coil that fires to kick the xenon tube into conduction at the beginning of the flash cycle. Both of these could be very disruptive to nearby digital electronics. I thought the original poster would be in a very good spot to help us determine if the effect if photoelectric or the result of EM because he's got everything set up. All he has to do is repeat the experiment while blocking the light from hitting the Pi.
    - tdicola 12y ago
      I'll put it in a box and try in a little bit, sorry have to do a few things first. I'm 99% sure it's light dependent though, I had to aim the flash at the board and keep it somewhat close to get it to lock up (like 6" away or so).
    - karlshea 12y ago
      From the original thread it does appear to be a photoelectric effect, since blocking a specific chip on the board causes it to survive the flash.
jayrhynas 12y ago
The most recent post [1] seems to indicate that covering one of the chips with blu-tac prevents the freeze [1] http://www.raspberrypi.org/forums/viewtopic.php?f=28&t=99042#p687959 http://www.raspberrypi.org/forums/viewtopic.php?f=28&t=99042...
- analog31 12y ago
  Not knowing what the IC is, one thought is that the blu-tac could be blocking light, or providing a wee bit of extra capacitance to tame a wicked electrical signal.
- bluehex 12y ago
  I'm thinking of trying a more permanent fix with some Sugru[1] when my Pi 2 arrives. Their about says it's insulating and okay for electric repairs under 24 volts. [1] http://sugru.com/about http://sugru.com/about
  - bluehex 12y ago
    Maybe not such a good idea according to: http://sugru.com/questions/can-i-use-sugru-to-secure-shock-mount-a-sensitive-pcb http://sugru.com/questions/can-i-use-sugru-to-secure-shock-m...
tdicola 12y ago
Alright I put the Pi 2 in a cardboard box and held the flash at the same spot (but the box obscures the light). It keeps running just fine and doesn't have the nasty spike. Definitely is something light related.
- noonespecial 12y ago
  Thanks! Well done. Photoelectric it is. Now I wonder what color its most susceptible to...
  - jakobegger 12y ago
    Since the effect seems to be caused by a silicon component that isn't properly shielded from light, anything with a photon energy higher than the bandgap should do. The bandgap of silicon is 1.1eV, which means the effect should start in near-infrared.
    - psgbg 12y ago
      What about glass, then that should work, right?
      - jakobegger 12y ago
        Not sure I understand you correctly. The problem should happen with near-infrared, visible, and UV-light. Glass is transparent in the visible range, and depending on type, also in IR / UV, so glass won't work as a shield. The easiest, reliable way to shield would be wrapping in aluminum foil (also helps against RF-interference).
        psgbg 12y ago
        Yeah you are right. I was assuming that the chips would block visible light but I forgot about UV.
  - tomkinstinch 12y ago

fnordfnordfnord 12y ago

I want to see this with a bare regulator. I might have some time Monday.

swamp40 12y ago

My money is still on an electromagnetically coupled pulse onto either the 3.3V or ground rail. I have seen that happen plenty of times, but I have never seen a light flash disrupt a power rail like that. I'm not saying it couldn't happen, just playing the odds...

crb3 12y ago
It can and did happen, as the afflicted chip was the SMPS controller. At last, a use for all those write-protect stickers we collected for floppies...

scythe 12y ago

The historical importance of the photoelectric effect is to demonstrate the relevance of the work function Ω. It works like this: V ~ E_0 - Ω where hf is the energy per photon and Ω is the work function of the target. If Ω > E_0, no voltage is produced. This provided the basis for Einstein's demonstration that energy is proportional to frequency (E_0 = hf), and won him the Nobel prize. What does this have to do with xenon? Xenon lamps produce not only visible light, but also ultraviolet light, in fact, xenon lamps extend further into the UV than natural sunlight: http://en.wikipedia.org/wiki/Xenon_arc_lamp#mediaviewer/File:Xenon_arc_lamp_profile.png http://en.wikipedia.org/wiki/Xenon_arc_lamp#mediaviewer/File... LEDs by contrast produce almost no UV light: http://en.wikipedia.org/wiki/Light-emitting_diode#mediaviewer/File:White_LED.png http://en.wikipedia.org/wiki/Light-emitting_diode#mediaviewe... With this data we can determine that the work function of silicon (in the Raspberry Pi 2) is somewhere in the range of hc/(400 nm) < Ω < hc/(250 nm) or 3.1 eV < Ω < 5 eV. Using data from http://journals.aps.org/pr/abstract/10.1103/PhysRev.127.150 http://journals.aps.org/pr/abstract/10.1103/PhysRev.127.150 http://www.sciencedirect.com/science/article/pii/0022369759902914 http://www.sciencedirect.com/science/article/pii/00223697599... we see that the work function of silicon is around 4.7-4.9 eV, which agrees well with our observations. draws a little box

jakobegger 12y ago
The problem with your theory is that, as others have noted, the effect seems to happen with a red laser pointer as well. To interfere with the electronics, the wavelength only has to as large as the band gap in silicon (as someone else has noted, around 1.1 eV). This means that IR light (and all visible light) can cause the interference too. A strong enough light source will lift a lot of electrons from the valence band across the band gap into the conduction band, effectively turning a transistor into a conductor for the duration of the flash. Since the sensitive part seems to be related to the power supply, it's quite plausible that this leads to the voltage fluctuations on the power rail.