Calculating mw from ma.

[newbie101]

Hello, I've resantly built a red 650nm laser using the DDL driver. I have 4xAA powering it and am using 3.3ohm's of resistance on it and was wondering if it would be possible to calculate the power output. I've searched through this site many times looking for a formula to convert ma's to mw's but was unsuccessful. I'd greatly appreciate any help and tips that you can give.

 

[chido]

There is no formula, you need to get a laser power meter and measure the output of your laser.

 

[rkcstr]

You might be able to guestimate from looking at power output vs current graphs around the forum for whatever diode you have.

 

[rog8811]

The best tip I can give you is that you need more volts....

4 X AA = 6v the regulator needs 3v the LD needs 3v so you have zero overhead which means your mw will reduce rapidly as the batteries sag. All you need is one more AA in there and you will be burning quicker and longer.

Regards rog8811

 

[GooeyGus]

Like others have said, there is no direct formula that you can use because they are two totally separate units. All you can base your power on is other people's experiences at similar current levels.

 

[SenKat_Stonetek]

Do a fast search for Hemlock Mike's power graphs - that will give yo ua rough estimate (fairly accurate, though !!!) of the output power you are getting - for example, it has been proven that the newer PHR-803T diodes will output about 100mw at about 100ma For the red diodes I sell, you will get 240mw+ at 400ma.

 

[newbie101]

The best tip I can give you is that you need more volts....

4 X AA = 6v the regulator needs 3v the LD needs 3v so you have zero overhead which means your mw will reduce rapidly as the batteries sag. All you need is one more AA in there and you will be burning quicker and longer.

Regards rog8811

Wouldn't adding another AA overload the LD and burn it out or no? Also would it be possible for me to get a link to buy a reliable and accurate laser power meter. And thanks alot for all your advise.

 

[rog8811]

wouldn't adding another AA overload the LD and burn it out or no?

No, the LM317 can work with an input voltage of up to around 34v, what it doesn't use it turns into heat...adding another 1.5v to the input will be fine. As long as the regulator is working the LD will not see anything more in the way of power than it does now but it will see it for longer

Regards rog8811

 

[newbie101]

wouldn't adding another AA overload the LD and burn it out or no?

No, the LM317 can work with an input voltage of up to around 34v, what it doesn't use it turns into heat...adding another 1.5v to the input will be fine. As long as the regulator is working the LD will not see anything more in the way of power than it does now but it will see it for longer

Regards rog8811

Oh, O.K. thanks. I'm gonna have to mod it to fit in another AA...but thanks a lot for the tip I'm definataly going to have to try this out.

 

[newbie101]

Hey rog8811, would it be possible for me to just hook up a 9v batt?

 

[rog8811]

A 9v battery would last no time at all I am afraid, it has too low a ma/hr output.
Stick to AA's or for a slightly smaller package go for 2 x RCR123a's (rechargable 3.6v)

Regards rog8811

 

[Xylicon]

So if you hooked up two DMMs and measured the current and operating voltage and did Ohm's law E*I its not what the laser is outputting?

I built my own driver based off information from this site (LM317 based) and I measured a voltage of 2.53 with 42.6mA and calculated 107.778mW. So is this what the beam output is or what the laser diode itself is dissipating as heat? (or am I totally off here)

*Just signed up*

 

[chido]

You're totally off, like we said, there's no formula, you need to get a laser power meter and measure the output.

 

[GooeyGus]

So if you hooked up two DMMs and measured the current and operating voltage and did Ohm's law E*I its not what the laser is outputting?

I built my own driver based off information from this site (LM317 based) and I measured a voltage of 2.53 with 42.6mA and calculated 107.778mW. So is this what the beam output is or what the laser diode itself is dissipating as heat? (or am I totally off here)

*Just signed up*

I believe this formula just gives you how many watts of power the diode is using (basically heat). I guess if a diode was 100% efficient it may work, unfortunately thats not the case ;D