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Newby question here ;) I have pulled apart a kids toy containing a small laser. The laser part doesn't have any partnumber on it, so I did some measurements on the actual device. Please note that I took these measurements with my multimeter, which probably isn't the most accurate because of all the losses in the long leads that I used.

When laser is at full brightness these are the characteristics:

  • 5.0V supply voltage (no load)
  • 1.0V voltage after drop (with load, so the voltage drop is 4.0V)
  • 220 mA current

When laser is at minimal brightness these are the characteristics:

  • 4.1V supply voltage (no load)
  • 0.7V voltage after drop (with load, so the voltage drop is 3.4V)
  • 85 mA current

Now my idea is to drive this laser myself using a Raspberry Pico. The Pico has a 5.0V pin which is directly connected to the input USB, delivering the power. I don't think the pin has any current limiting protection on it.

Question:

How to protect the laser, so it can safely operate within the boundaries as specified above? I think I need to add a current limiting resistor, since the USB input can have various input currents. If this is indeed what I need to add, please let me know how to properly calculate the correct resistance, since there is a difference in voltage and current in both situations (full brightness vs dimmed brightness), resulting in different ohms.

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  • \$\begingroup\$ Please be more explicit about the three bulleted measurements you made and explain what dropping circuit was used. \$\endgroup\$
    – Andy aka
    Commented Dec 3, 2023 at 12:54
  • \$\begingroup\$ I measured the voltage directly in the input of the component. So when the toy was switched on at full power, I measured 5.0V on the laser input when the laser was disconnected. I did that by holding the multimeter leads to the 2 pins of the JST connector box. When I connected the laser, I measured 1.0V. Hope this makes sense? \$\endgroup\$
    – Mar Tijn
    Commented Dec 3, 2023 at 13:56

1 Answer 1

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If I'm interpreting your measurements correctly, you see (at max power setting) 1 V across the laser diode with 220 mA flowing. To replicate that off a 5V supply, you need a series resistor with value R=(Vsupply - Vlaser)/Ilaser = 4/0.22 = 18.2 ohms. I'd use a 20 ohm resistor to be safe with power supply and resistor tolerances. To run at lower laser power, use a larger resistor. To be clear, this resistor should be placed in series between the 5V supply and the laser. The voltages you measured with the laser disconnected don't matter.

A couple caveats: (1) This resistor needs to be rated for more than 1W (calculate Ilaser^2 * R). That's a lot bigger than your typical kit resistor.

(2) USB is limited to 500 mA, and you're consuming roughly half of that in this laser. If you're also loading the RPi with other components you should be careful about your total power consumption.

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  • \$\begingroup\$ Thanks a lot for this answer. I still have a couple of questions. I noticed on the board, just above the laser JST connector there is a resistor labeled 6R8. My guess is that that is the resistor limiting the input current of the laser. Any idea why they would use a 6.8 Ohm resistor as opposed to the calculated 18.2 Ohms? I noticed that that resistor gets really hot when the laser is operating though. And you mentioned the USB is limited to 0.5A. However the manufacturer mentioned on the package that you need at least a 5V/1A adapter to use the toy. So I guess this USB isn't limited to 0.5A? \$\endgroup\$
    – Mar Tijn
    Commented Dec 5, 2023 at 10:50
  • \$\begingroup\$ For your first question: it's really hard to tell without knowing the schematic of the toy. It's likely that they are using some upstream voltage or current regulator. Doing the entire current limiting with a single resistor like we're discussing is a bit quick and dirty. \$\endgroup\$
    – chrisp
    Commented Dec 5, 2023 at 14:37
  • \$\begingroup\$ For USB: 0.5 A is the old standard USB 1.1/2.0 limit, like if you were powering this from a computer. Lots of dedicated chargers can supply more than that, as can some computer ports. Now that we've decided to make USB effectively a universal charging standard the options are endless and confusing. But many wall plug chargers will happily just act like a dumb 5V power supply up to their max labeled current, if you don't need use the USB for data. \$\endgroup\$
    – chrisp
    Commented Dec 5, 2023 at 14:43
  • \$\begingroup\$ Thanks again Chris! I'll go investigate a bit more and let you know if I need more info! \$\endgroup\$
    – Mar Tijn
    Commented Dec 6, 2023 at 9:58

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