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I want to hook up a 9v battery to a laser diode(linked to the ones I will get), but have seen laser drivers scattered across the internet and vaguely have heard of them. Do I need one to use the laser?

The laser will not be constantly on for very long, just whenever I'm pressing a limit switch in short bursts. I checked out this question which recommended a driver, but since I'm only using one, do I need a driver?

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    \$\begingroup\$ "Working Voltage: 5v Operating Current: must less than 20 mA". How are you planning to ensure these limits without the use of a driver? Is the module self regulating? \$\endgroup\$
    – AJN
    Commented Sep 11, 2020 at 13:58
  • \$\begingroup\$ Those ones say "Working voltage 5V" so they may have the driver built in (IF you trust the advert). What 9V will do to it is anybody's guess, but probably nothing good. \$\endgroup\$
    – user16324
    Commented Sep 11, 2020 at 13:59
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    \$\begingroup\$ Simplest solution: use a resistor to drop: 9 V - 5 V = 4 V, 4 V / 20 mA = 200 ohm (220 Ohm will do as well) in series with the laser diode. Probably that will just work. \$\endgroup\$ Commented Sep 11, 2020 at 14:01
  • \$\begingroup\$ @Bimpelrekkie, 200 would be a red-black-brown or 220 is a red-red-brown? i cant quite get the hang of reading resistors. and then the last band doesnt matter as much, right? \$\endgroup\$ Commented Sep 11, 2020 at 14:13
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    \$\begingroup\$ @Andyaka I believe the 5 mW is meant to be the optical power. The rest is spent as heat. \$\endgroup\$ Commented Sep 11, 2020 at 14:45

2 Answers 2

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Looks like you don't need a driver. You will, however, need to deal with the extra voltage.

From the first Q&A on your link:

"There is a 92 Ohm tiny resistor (I did measured 92 Ohm with my Ohmmeter) at the back of the Led, in series with the actual Led device. So with a 5V tension source connected on both wires, you get 20mA through the circuit, spliting as about 2V at the resistor and 3V at the led (2.7V is the expected value). What this means is they are safe to connect directly to a 5V source as they are, as the included resistor will limit the current."

Since you want 20mA through the diode, and you have a 9 V source...

We will need to drop: $$9\ V - 2.7\ V= 6.3\ V$$ With resistors.

Using Ohm's Law, we know R=V/I: $$\frac{6.3\ V}{0.02\ A}= 315\ Ω$$

And we know we already have 92 Ω: $$315\ Ω-92\ Ω=223\ Ω$$ So a 220 Ω resistor may work, but you might want to use a 270 Ω or even 330 Ω resistor to be safe.

This resistor should be in series with the diode.

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    \$\begingroup\$ But as OP said that she/he wants to use a 9V battery, would not it be adviceable to put another resistor for cope with 9 V to 5 V? \$\endgroup\$
    – mguima
    Commented Sep 11, 2020 at 14:33
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    \$\begingroup\$ @mguima Thank you, please see my edit. \$\endgroup\$ Commented Sep 11, 2020 at 14:44
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Your best bet is to use a linear regulator such as LM78L05 to get 5V from the 9V battery.

There are much better regulators available but that one is cheap and has low requirements for external components.

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