burning laser diode?

Discussion in 'The Projects Forum' started by Gadersd, Jul 9, 2013.

  1. Gadersd

    Thread Starter Member

    Dec 8, 2012
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    I found this diagram on a burning laser tutorial. I understand that the capacitor is there to smoothen the voltage and that the diode is there to prevent voltage spikes. What I don't understand is how the diode prevents voltage spikes. Wouldn't the diode create a short circuit? The tutorial said that if some one removes the laser diode and then puts it back, the diode will prevent the capacite'rs current from burning out the laser diode. How is this possible? Why will the capaciter output current through the diode and not the laser diode?
     
  2. wayneh

    Expert

    Sep 9, 2010
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    Something is wrong with your attachment. It's too tiny to see.

    But I think the diode is there to protect the laser against placing the battery in backwards. That might also ruin the LM317 but the laser is more expensive. That circuit needs a diode to divert current from the capacitor around the LM317 in the event the battery is removed. Otherwise, the energy in the capacitor could ruin the LM317, when the output is at a higher voltage than the input.
     
    Last edited: Jul 9, 2013
  3. Gadersd

    Thread Starter Member

    Dec 8, 2012
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    The image should become fairly clear if you click on it. The tutorials says this:The diode is there to soak up voltage spikes. If your led were to come disconnected while the circuit was on and then reconnected the stored charge in the capacitor would be sent right into your laser diode. It would kill it instantly. You would be sending a charge of around 6 volts to it. How does the diode not create a short circuit? Does the laser diode not need a resistor? I see that the regulator is outputting voltage directly to laser diode.
     
  4. wayneh

    Expert

    Sep 9, 2010
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    It's not clear - too small.

    I'll speculate that the LM317 is in a constant current configuration and that the diode is a zener that will conduct in reverse at its specified breakdown voltage. That would be consistent with your description.

    With the laser removed, the LM317 will raise the voltage to the max in a failed attempt to maintain current, putting a max charge on the capacitor. Reconnecting the laser would destroy it.

    The diode caps the voltage that the cap can be charged to. At that voltage, it will begin to conduct the current and act as a dummy load if the laser is missing. For this to work, the zener voltage should be very close to the laser's forward voltage drop.

    It seems to me that removing the capacitor or cutting its size would make more sense, but don't quote me.
     
  5. Gadersd

    Thread Starter Member

    Dec 8, 2012
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    I just took another look at the tutorial and it turns out that the diode he used was a 1N4001 rectifier diode not a zener.
     
  6. wayneh

    Expert

    Sep 9, 2010
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    Well then I'm baffled. :confused:
    Other than hooking the battery up wrong, I can't see how that diode would ever conduct.

    Well, maybe if the laser is more conductive than the diode, the laser will take the current preferentially unless the laser is removed, and then the diode takes all of it. It needs to be forward biased to work this way, but acts as a dead short for any voltage over ~0.65V in that orientation. Which way is it oriented?

    Again, a better picture or a link would help a lot.
     
    Last edited: Jul 9, 2013
  7. Gadersd

    Thread Starter Member

    Dec 8, 2012
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  8. wayneh

    Expert

    Sep 9, 2010
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    That's a constant current driver. The diode is somewhat optional, to protect the laser from reverse polarity. That's all.
     
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  9. Gadersd

    Thread Starter Member

    Dec 8, 2012
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    Actually I made a mistake. He said that the capacitor protects the laser diode from voltage spikes. Wouldn't the capacitor cause a short circuit when the battery is removed since there is no resistance between the cap and the laser diode?
     
    Last edited: Jul 9, 2013
  10. wayneh

    Expert

    Sep 9, 2010
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    No, because it will only be at the voltage it was when running, when the laser was happy, and will drop quickly from that.

    The only spike it might absorb is a transient that gets thru the regulator, perhaps as the circuit starts up. I'm not convinced it's really needed but it doesn't do any harm. It would absorb some current and get the regulator working before the laser turns on, so that could be good.

    BUT, it does risk blowing the laser if you disconnect and re-connect it. As I said before, the cap will charge to max voltage without the laser in place, and then deliver a whopping jolt to the laser when it's replaced.
     
  11. Gadersd

    Thread Starter Member

    Dec 8, 2012
    98
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    I don't understand. I thought if a voltage is applied to a diode directly it will burn out. So if I connected a battery to it without any resistance it will blow, but if I connect a capacitor to it it won't blow.
     
    Last edited: Jul 9, 2013
  12. wayneh

    Expert

    Sep 9, 2010
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    A charged capacitor is just like a low-capacity battery at the same voltage. If your LED would be destroyed at that voltage, touching the leads to the capacitor will likely destroy it.

    Now, depending on the capacitance value of the capacitor, the voltage may drop quickly and the total energy stored in the capacitor might not be enough to do damage. But some capacitors, like one in a TV power supply, are quite large and can melt tools that are placed across their poles. Been there.
     
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