# Required math for setting up a laser diode

I'm new to electronics and come from a software background. Recently I got a laser diode that I would like to use with my arduino. The manufacturer says it is 5mW and draws 2.8-5.2V.

I'd like to learn how to decide which components I need for a project (i.e. their specific ratings). I know I need a diode and a capacitor for this "circuit." What maths would I need to do to figure out the rate of the capacitor and the diode as well as any resistors that I would possibly need.

Edit: The manufacturer says === the specs on the web site. Also as far as the math, I'm asking more so for the functions needed/ algorithms to compute which parts need to be what strength.

• "The manufacturer says" means you asked them over the phone? Or is there an actual datasheet? Also devices usually draw current. The maths you need are usually addition, subtraction, multiplication and division. – PlasmaHH Mar 25 '15 at 14:41
• @PlasmaHH sorry, I will edit my post to clarify. When I ordered the product that's what was listed in the specifications. Also I guess I more-so meant the different algorithms/ functions that would be needed for the computations to figure out what each component would need. – Sleep Deprived Bulbasaur Mar 25 '15 at 14:42
• Ωhms law will usually suffice. – PlasmaHH Mar 25 '15 at 14:45
• I could imagine that the laser diode connected to a 5.2v supply may go "pop". I think you need to link the data sheet. Why do you need a diode and capacitor for "this" circuit? It's much more likely you need a resistor. – Andy aka Mar 25 '15 at 15:16
• Ah it's not just a laser diode but an integrated laser and driver packaged up. Still no data sheet. – Andy aka Mar 25 '15 at 15:24

For capacitor ratings all you would need to concern yourself with is the voltage rating which will be listed with the component, so if the maximum voltage you are going to operating at is 5V say, then you'll need to get a capacitor that can handle more than this.
Here is an example of a thru-hole components marking:

With resistors you'll have to look out for the power rating.

$$Power = I \times V$$

And you can easily work these values out with Ohm's Law:

$$V = I \times R$$

With these two equations you only need to know two of the values (which you will, likely the voltage will be known - if not, use a maximum so as not to fall short, and the resistance needed or maximum current will be known making it easy to get the power rating)

You can find a lot of information about the ratings of different packaged resistors on the web, if you're using surface mount that package name hold a bit of information on it, as to thru-hole components, I'm not so sure but you'll easily be able to find out when purchasing the components. It will also be easy to find out if you have a part number for the resistors you are using.

Diodes will tell you on the datasheet what their maximum current ratings are, both continuous and pulsed currents so you don't fry these components. Always check the datasheet for information like this as they are really useful and you can quite easily determine if they are the component for you - so long as you know what to look for!

You don't need a diode and you don't really need a cap.

All the math you need is this:

1. Is the voltage source I intend to use within the voltage range the laser needs?

2. Can that voltage source provide enough current? That is, is the amount of current that voltage source can provide equal to or greater than the amount of current the laser needs (also known as how much it "draws")?

For some devices, such as plain LEDs without a driver, you do have to limit the amount of current, but this device has a circuit built into it that keeps it from "accidentally" drawing too much current, as long as the voltage applied to it is less than the maximum allowed.