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I bought some laser diodes on ebay recently that were advertised as 5mw which look like the image shown:

enter image description here

To comply with law, I hooked it up to a 9V battery with a 22K resistor.

I made a simple detector as follows with an LM324 op-amp

I connected inverting input to output via 4.7M resistor. Output is connected to cathode of LED via 1K resistor. Non-inverting input is pulled up by 1M resistor and pulled down by the 3mm clear lens photo-diode detector connected in reverse.

The detector works nice if the laser beam shines right at it (which is what I want), but I find that with such a low wattage, I have to be super-precise at shining the beam directly at the detector to get it to respond. If I'm off by even 1mm, the detector won't respond.

I have read about special kind of glass (or something?) that is supposed to scatter laser beam light when hit? but I'm not 100% sure what it is called or if it commonly exists. Can someone shed some light on this? Basically I want my detector to respond if it is in a transparent case and any part of the transparent case that the component side of the circuit is facing is hit with a laser beam.

Can anyone suggest what I should do? and raising power is not an option since laws prohibit lasers more powerful than 5mw.

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    \$\begingroup\$ You can try just taking the smooth surface of the "glass" off the box with fine sandpaper or other rubbing compound. What will happen is the glass will light up. Hopefully, if your detector is biased right, it will be enough for it to detect. You will want to block and OTHER light sources though. \$\endgroup\$ – Trevor_G Mar 27 '17 at 2:27
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    \$\begingroup\$ You're going to have to put a lot more than 0.4mA of current through the laser to get anywhere close to 5mW output! You're probably not even over the threshold current for that diode. I would expect that you need to be somewhere over 10 mA in order to get the rated output. \$\endgroup\$ – Dave Tweed Mar 27 '17 at 2:29
  • \$\begingroup\$ electronics.stackexchange.com/a/69108/81355 \$\endgroup\$ – Ouroborus Mar 27 '17 at 2:38
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    \$\begingroup\$ What you need is to limit the current to the laser. Or, maybe not. There are laser modules that include current limiting, and there are some (many) that don't. The datasheet for the laser module would tell you, but you don't have one. Sometimes it is implicit in the other text from where you bought it. If it says "operating voltage 3 to 12V" or something similar then there's a chance it cotains current limiting circuitry. There's also the possibility that the non-english speaking seller just copied and pasted the text from somewhere else. \$\endgroup\$ – JRE Mar 27 '17 at 5:31
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    \$\begingroup\$ LED brightness is a function of current rather than voltage. Your laser is an LED that s also a laser. To control the brightness, you control the current rather than the voltage. \$\endgroup\$ – JRE Mar 27 '17 at 5:35
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First of all, the relevant regulation requires lasers to emit less than 5mW of radiation. This is not the same as having 5mW electrical power:

21 CFR 1040.11(b) and 1040.11(c), limit surveying, leveling, and alignment, and demonstration laser products to Class IIIa. This means that pointers are limited to 5 milliwatts output power in the visible wavelength range from 400 to 710 nanometers.

LD efficiency is about 30% for the best lasers, the one you have bought probably has about 10%. So you can safely multiply the power you apply by at least 3.

Start by measuring the voltage drop \$V_D\$ on the LD with the resistor you already have, and also measure the current (voltage drop on the resistor multiplied by its resistance: \$I_D=\frac{V_R}R=\frac{9-V_D}R\$). Your target is to have \$P=V_D*I_D=0.005*3=0.015\$. Gradually lower the resistance until you get the right power.

Second, instead of rising the threshold to turn your detector ON, you may consider lowering the threshold for it to go off. Cover your detector with a piece of glass / transparent plastic with a color close to that of your laser beam. It should let most of the laser light through, while rejecting a significant portion of ambient light.

Finally, if you need a bigger detection surface, consider installing a lens of that size, and place the detector in focus behind the lens. In that case, a laser beam hitting anywhere on the lens surface along its optical axe will end up in the detector:

enter image description here

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  • \$\begingroup\$ Ok now I'm curious. How do I calculate how many mw radiation I'm emitting then? I thought I can use a simple ohms law formula ((Battery_voltage - Battery-lazer_voltage_drop)/resistor) * Battery_voltage to calculate wattage. I must be missing something? \$\endgroup\$ – user143136 Mar 27 '17 at 20:46
  • \$\begingroup\$ @Mike See my edit. It's impossible to calculate the right value of R without the datasheet of your LD, but you can measure the power and find the right value by trial & error. \$\endgroup\$ – Dmitry Grigoryev Mar 28 '17 at 12:12
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First, run the laser at its nominal current (refer to datasheet).

If you want the detector to detect over a wider area, then you will need a diffuser, then optics to concentrate the light from the diffuser onto the sensor. For example, a tube with ground glass at one end, and a lens at the other, focused on the detector.

You can extend the length of the tube to the front to make a sunshade.

Then, the smarter thing to do is make your detector more sensitive:

  • Filter out unwanted wavelengths.

You can use a narrow-band detector which is sensitive only to the laser optical wavelength, or use an optical filter (simple colored plastic, or something more high-tech which will better block uncanted wavelengths).

  • Use modulation

By modulating your laser (ie, making it blink at, say, 40kHz) and using a detector with a narrowband, high-Q analog filter tuned to this frequency, you will get rid of constant illumination (sunlight), and other sources of modulated light (flickering LED or CFL lights, etc).

This is how IR receivers for remotes work.

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