How to calculate the current/voltage that need to be applied on the laser diode

[Skokovich]

I've been searching the forums tutorials on how to make a laser, but most of them are made for people who don't care for laser fundumentals, i.e people who just want to buy parts and build a powerful laser.

I want to learn more than that. So, what i actually want is of mathematical matter; for ex., if i buy a 2w laser diode, what's the max current it can take without blowing up, i.e what's the max current i need to apply to get the "maximum" out of laser diode?

The laser i will build requires a 1.25 amp laser diode driver.

I guess i need to use W=V*I formula, but i still want a profesional opinion.

Sorry if thread already exists or if the question sounds stupid. Feel free to redirect me to some good tutorial where i can learn this.

 

[Speedy78]

1.25A? To get the highest peak output of any particular diode you have to power it with dc supply and adjust the current until yoy reach the point where more current doesnt equal more output. They are all slightly different in efficiency and will vary in output.

 

[Blaster]

1.25A? To get the highest peak output of any particular diode you have to power it with dc supply and adjust the current until yoy reach the point where more current doesnt equal more output. They are all slightly different in efficiency and will vary in output.

been doing electronics as a job for 30 years and your formula is true, but as speedy 78 said as far as laser diodes go is spot on. Get to that current point then back off a bit 5% at least for safety margin and use that current, and heatsink the diode really well biggest killer of laser diodes is heat leading to thermal runaway and cooked diode.

 

[Skokovich]

been doing electronics as a job for 30 years and your formula is true, but as speedy 78 said as far as laser diodes go is spot on. Get to that current point then back off a bit 5% at least for safety margin and use that current, and heatsink the diode really well biggest killer of laser diodes is heat leading to thermal runaway and cooked diode.

I'm familiar with basics such as heatsinking.
I've read that it is, even though it is not mandatory, good to use thermal paste to ensure a good thermal contact.

 

[NKO29]

And with LDs it's only the current you need to regulate. But if you use a linear regulator the more "excess" voltage the more heat the driver will produce.

 

[DrSid]

You need so called current voltage graph for the diode. For any voltage applied on the diode, some current will flow through it. This voltage multiplied by this current will give input power. This is not the same as output optical power. Output power is more or less linear function of input power, there are just some losses, so it may be like 30% of input power or so. It usually differs for individual pieces of the same diode type, you can find diodes twice as powerful at the same current as average diodes.
So .. back to the graph .. for every voltage there is current .. or vice versa .. for every current there is voltage. We usually use current driver, as this dependency of current on voltage does change with temperature, and with constant voltage drivers your current can get too high and kill the diode.

Check this image (taken from this thread: http://laserpointerforums.com/f65/3x-445nm-ld-piv-plot-53927.html)

Look at the red line .. you see the bend ? That's where the diode starts to produce light. Let's say at 100mA, or 4.5V. First non-coherent only.
At 200mA, or 4.8V you can see the output power rise from zero .. that means coherent light is being produced. This is called 'lasing threshold'.
After that the curves are more or less linear, both current to voltage, and power to current.
What can't be seen in this graph, that after some current (or voltage) the output power stops rising, and diode will either stop lasing, or will burn (that's why most people don't push it).
You also can't see dependency on temperature. As the diode gets hotter, it will pass more current with the same voltage applied. It's really hard to measure, and not important, if you use current driver. That's why we mostly talk about current and power, and not voltage. But this graph applies to basically all electronic parts, especially semiconductors, and often the voltage is used as primary value. Not so with laser diodes.

There is also some 'sweet spot' .. some current at which the diode does what you want, with reasonable live time. When this graph was made, these diodes were usually run at 1W. Which means current 1A, voltage 4.5V. Now here where the voltage is important. There are different current drivers .. and some can only decrease voltage of the battery (buck driver).. and some can only increase it (boost driver).. and there are even some which can do both. So if you have 3V battery .. and you see the diode need 4.5V .. you need boost driver, able to input 3V and output 4.5V.

Anyway .. you need those graphs. It is different for each diode type. Especially for different colors .. for shorter wavelength you need higher voltages, but it also depends on construction of the diode and some other things.

 

[Svenvbins]

I hope it's ok to revive this thread cause I have a similar question that doesn't need a whole new thread, I think.
I have a red diode, but absolutely no idea what its specs are (sold as <5mW... ). I have access to a laser power meter, and am just planning to up the current bit by bit until I find this "flattening" in the current-power curve. However, as DrSid writes:

What can't be seen in this graph, that after some current (or voltage) the output power stops rising, and diode will either stop lasing, or will burn (that's why most people don't push it).

My question: Will there be enough 'current difference' between the flattening and the burning to stop before I break stuff, or am I risking breaking the diode? It was cheap, but I'd rather not wait another two weeks for stuff to arrive from china

 

[DrSid]

There might be no warning at all .. that's why we use torture test .. we destroy few diodes of popular types to find out where the limits are.