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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?

Ceramicmrno0b
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    "*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? – AJN Sep 11 '20 at 13:58
  • 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. –  Sep 11 '20 at 13:59
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    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. – Bimpelrekkie Sep 11 '20 at 14:01
  • @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? – Ceramicmrno0b Sep 11 '20 at 14:13
  • [What to check for when buying an electronic component or module](https://electronics.stackexchange.com/questions/504044/what-to-check-for-when-buying-an-electronic-component-or-module). No data sheet link means high risk. And that diode spec on shamazon is not making sense to me. 5 mW means 1 mA current but, what do you expect when buying from disreputable sources who don't understand what they sell. – Andy aka Sep 11 '20 at 14:36
  • @Ceramicmrno0b 200 is red-black-brown but they are very rare. 220 is red-red-brown and they are easy to find. – mguima Sep 11 '20 at 14:37
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    @Andyaka I believe the 5 mW is meant to be the optical power. The rest is spent as heat. – evildemonic Sep 11 '20 at 14:45

2 Answers2

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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.

evildemonic
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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.

Spehro Pefhany
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