If you go to this link you see a nice laser driver for a small laser...didn't he mean to put that diode in series with the + and - or is it supposed to be parallel and I"m just not understanding for some reason?

enter image description here http://www.rog8811.com/LM317%20components01.jpg?0.35312663627278806

also his reasoning there: http://www.rog8811.com/laserdriver.htm

Plus isn't the diode possibly unnecessary since the laser is a diode and assuming it can handle the voltage?

  • \$\begingroup\$ Which diagram are you referring to? If you can't include images, choose the direct link to the image you are referring to and add it to the bottom of your post, and one of us will edit it in. \$\endgroup\$ – Jay Greco Jun 11 '13 at 22:40

The 1N4001 diode is included for reverse-power protection. As the linked site says, the diode "protects LD if batteries are inserted the wrong way round".

In normal operation, the diode is reverse biased and will have very little effect on the circuit operation.

But if you accidentally miswired something, or hooked up your power backwards so that you were trying to push current the wrong way through the laser diode, the 1N4001 diode could save the laser from being destroyed (assuming your power source has enough internal resistance that it doesn't just blow up the 1N4001 and the laser).

Laser diodes are generally optimized for efficient light output rather than ability to withstand high reverse voltages, so protection circuits like this are often needed to improve the reliability of laser diode circuits.

  • \$\begingroup\$ I think I understand so the laser diode and the 1N4001 diode must be in parallel and with oposite orientation(cathode of laser to anode of 1N4001). So wouldn't a tad better design be to the add a resistor after the diode in series with 1N4001 so if it is used it doesn't have a large current...asuming it can be that large. \$\endgroup\$ – itb Jun 11 '13 at 23:55
  • \$\begingroup\$ @itb, in a reverse power situation you want absolutely no current through the laser. So only an infinite resistor value (an open circuit) will protect you. But that would make the forward operation of the circuit "somewhat" less useful. \$\endgroup\$ – The Photon Jun 12 '13 at 0:01
  • \$\begingroup\$ If you're talking about adding a "smallish" series resistor to avoid the situation I alluded to where the 1N4001 is blown up, I think it's not there because the original article is talking about getting power from flashlight batteries. And he's relying on the internal resistance of the batteries being enough to not blow out the 1N4001. \$\endgroup\$ – The Photon Jun 12 '13 at 0:03
  • \$\begingroup\$ ahhh ic so that small resistor would be usefull assuming not very large internal resitance of a battery? And yes I was trying to fix the problem where you said the 1N4001 might blow. \$\endgroup\$ – itb Jun 12 '13 at 0:04
  • 2
    \$\begingroup\$ If I was me, I'd put the 1N4001 before the LM317, with a fuse before the whole thing, so that, if the supply polarity was reversed, the current through the 1N4001 would blow the fuse, protecting the laser diode. \$\endgroup\$ – Peter Bennett Jun 12 '13 at 1:25

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.