How to mesure a laser diode's power ?

[sam1902]

Hello,
I'm making a laser triggered electronics project and I'm using one of those Arduino laser diode module to emit the laser (c.f. images). The specifications says it works on 5V and has a power of 5 MW (though I think they meant mW or else I'll have other problems). They also indicate that it has a power of 650nm and an OD of 6mm.
I don't know why they said OD was 6 mm as it doesn't seem to be a mesure of distance.
Upon powering the laser with my USB powered Arduino Uno, I thought the laser was quite powerful (compare to a simple presentation laser pointer) so I tried mesuring the tension and intensity it was using. I also measured it when powered through my Arduino Nano over USB, and to the same Arduino Nano on a 9V battery. I got the following reading off it:

Source	Voltage (tension)	Ampereage (intensity)	Watts (power)
Arduino Uno on USB	4.89 V	28.35 mA	138 mW
Arduino Nano on USB	4.34 V	23.1 mA	100 mW
Arduino Nano on 9V bat	4.2 V	15 mA	63 mW

From these result, I concluded that I had a Class 3B laser (5 to 500 mW) and after reading a bit about it, it seems I need to wear laser goggle protection so I ordered a pair of these which are rated for 190-470 and 610-760nm at OD 4+ (google for MCWlaser Lunettes de Protection Laser Safty Lunettes 190-470 & 610-760 nm).
This laser diode being a standard mass produce arduino module, I was a bit astonished by getting the above readings. So my questions are the following:

Are my readings realistic and is that the proper way to know the "power" of a laser ?
I'm only reading the electrical input to the diode, not the beam power itself because I only have a multimeter, not a spectroscope or anything fancy.

Is that protection adapted for that laser ? It's my first project involving laser ever so I have no idea if I'm buying the correct goggles.

Thanks for taking the time to help me figure this out

PS: I have to add, as it seems relevant, that I'll be powering the laser constantly, not for just short pulses.

 

[Deleted member 57455]

Hello,
I'm making a laser triggered electronics project and I'm using one of those Arduino laser diode module to emit the laser (c.f. images). The specifications says it works on 5V and has a power of 5 MW (though I think they meant mW or else I'll have other problems). They also indicate that it has a power of 650nm and an OD of 6mm.
I don't know why they said OD was 6 mm as it doesn't seem to be a mesure of distance.
Upon powering the laser with my USB powered Arduino Uno, I thought the laser was quite powerful (compare to a simple presentation laser pointer) so I tried mesuring the tension and intensity it was using. I also measured it when powered through my Arduino Nano over USB, and to the same Arduino Nano on a 9V battery. I got the following reading off it:

Source	Voltage (tension)	Ampereage (intensity)	Watts (power)
Arduino Uno on USB	4.89 V	28.35 mA	138 mW
Arduino Nano on USB	4.34 V	23.1 mA	100 mW
Arduino Nano on 9V bat	4.2 V	15 mA	63 mW

From these result, I concluded that I had a Class 3B laser (5 to 500 mW) and after reading a bit about it, it seems I need to wear laser goggle protection so I ordered a pair of these which are rated for 190-470 and 610-760nm at OD 4+ (google for MCWlaser Lunettes de Protection Laser Safty Lunettes 190-470 & 610-760 nm).
This laser diode being a standard mass produce arduino module, I was a bit astonished by getting the above readings. So my questions are the following:

Are my readings realistic and is that the proper way to know the "power" of a laser ?
I'm only reading the electrical input to the diode, not the beam power itself because I only have a multimeter, not a spectroscope or anything fancy.

Is that protection adapted for that laser ? It's my first project involving laser ever so I have no idea if I'm buying the correct goggles.

Thanks for taking the time to help me figure this out

PS: I have to add, as it seems relevant, that I'll be powering the laser constantly, not for just short pulses.

The laser of yours is no more than 5-10mw else it would have already blown up, And most of the power is wasted across the resistor in series with the diode. As you can also see when the voltage rises the current does too because the resistor stays the same in terms of resistance but voltage increases. I = V:R. The diode itself only needs about 20mA at 2.2v.

 

[loreadarkshade]

I think you can directly hook up a thermistor to a heatsink and measure it's resistance (temperature).
It can give you a rough estimate based on heat output.
I think this is how cheap laser power meters work.

Note: This probably is really not ideal for low power lasers, since even a small air current cooling the heatsink can lower your reading down to 0.

 

[Deleted member 57455]

I think you can directly hook up a thermistor to a heatsink and measure it's resistance (temperature).
It can give you a rough estimate based on heat output.
I think this is how cheap laser power meters work.

Note: This probably is really not ideal for low power lasers, since even a small air current cooling the heatsink can lower your reading down to 0.

No, and to give you the TLDR, Diode efficiencies vary vastly even with the same diodes, heatsink mass, material quality, and general surface area vary vastly even with the copper modules. And a lot of other things... The best option is to use a photodiode, then TEC for higher powers, and Idk why I haven't seen anybody do this but SHIELD THE CABLES TO THE TEC AND BACK PPL WE ARE MEASURING MILLIVOLTS HERE.

 

[loreadarkshade]

No, and to give you the TLDR, Diode efficiencies vary vastly even with the same diodes, heatsink mass, material quality, and general surface area vary vastly even with the copper modules. And a lot of other things... The best option is to use a photodiode, then TEC for higher powers, and Idk why I haven't seen anybody do this but SHIELD THE CABLES TO THE TEC AND BACK PPL WE ARE MEASURING MILLIVOLTS HERE.

See here

davidegironi . blogspot . com/2014/07/a-cheap-and-simple-laser-power-meter.html#.X95ahNhKiUk

It's a cool device!

 

[sam1902]

The laser of yours is no more than 5-10mw else it would have already blown up, And most of the power is wasted across the resistor in series with the diode. As you can also see when the voltage rises the current does too because the resistor stays the same in terms of resistance but voltage increases. I = V:R. The diode itself only needs about 20mA at 2.2v.

You're right ! I did notice resistors on there but they were smd components so I had trouble reading them. With the help of several magnifying glass, I have a 103 resistor and 910 resistor, which should correspond to 10K ohms and 91 ohms. However it still doesn't match the readings: 28.35 * 1e-3 * 10*1e3 = 283.5 V != 4.89 V and 28.35 * 1e-3 * 91 = 2.58 V != 4.89 V.
At least I'm glad it's not that powerful.

It may be a silly question, since you're talking about ThermoElectric Cooling module (TEC, it's not easy to google this) and op amps, but why not use a photoresistor ? Also I assume a TEC is basically a small and precise version of a Peletier module right ? I wonder if by by using a divergent lense it would be possible to cover more of its surface and have a better reading.
Photoresistors should be more precise than heat based methods because it uses the photoelectric effect and not a proxy effect like temperature which can be affected by wind like @loreadarkshade mentionned. I don't know if this method is viable for more powerful lasers but that's apparently not a concern here.

 

[Deleted member 57455]

You're right ! I did notice resistors on there but they were smd components so I had trouble reading them. With the help of several magnifying glass, I have a 103 resistor and 910 resistor, which should correspond to 10K ohms and 91 ohms. However it still doesn't match the readings: 28.35 * 1e-3 * 10*1e3 = 283.5 V != 4.89 V and 28.35 * 1e-3 * 91 = 2.58 V != 4.89 V.
At least I'm glad it's not that powerful.

It may be a silly question, since you're talking about ThermoElectric Cooling module (TEC, it's not easy to google this) and op amps, but why not use a photoresistor ? Also I assume a TEC is basically a small and precise version of a Peletier module right ? I wonder if by by using a divergent lense it would be possible to cover more of its surface and have a better reading.
Photoresistors should be more precise than heat based methods because it uses the photoelectric effect and not a proxy effect like temperature which can be affected by wind like @loreadarkshade mentionned. I don't know if this method is viable for more powerful lasers but that's apparently not a concern here.

You absolutely shouldn't use photoresistors, Because both they and photodiodes use the photoelectric effect but photodiodes are much much more precise and FYI don't use a photodiode for lasers above 100mw. Then use a TEC and coat it with the flattest spectrum response coating you can find(carbon-based very matt paint, vantablack would work well.)