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FrozenGate by Avery

Simple Laser Power Meter Using IR Thermometer






No.... it is not sensitive enough to measure a 1mW difference in heat..:cryyy:

Jerry

You can contact us at any time on our Website: J.BAUER Electronics
 
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even if it cant measure to the resolution of 1mw even 5 i think would be okay. just keep testing it and then just take the average.
Would give us a ballpark of how good or bad our lasers are.
I got a 50mw green from the led shop and some of my friends got some too but once mine warms up a bit it seems to be 2x theirs. would love a cheap LPM.
 
Yeah... a thermometer with probe must work as an LPM...
That should definately be able to see a 5mW change in power
of a 50mW green... But what do I know... I could be wrong...


Jerry

You can contact us at any time on our Website: J.BAUER Electronics
 
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Sorry to ask but is that sarcastic or serious?
I seem to have a hard time grasping the difference.
i learned English and Japanese almost at the same time but then forgot how to speak Japanese but didn't forget the grammar so sometimes get confused when reading things.
 
You can't get an accurate wind speed by sticking your finger into
the wind... You will only know if the wind is blowing or not...

You can detect the temperature of a high powered Laser beam
with a digital thermometer... but you need an energy absorption
coating for the wavelength you are using and the entire Laser Beam
Profile will need to be detected/sensed...

You may get a reading.... but what does it mean...:thinking:

To make sense of the readings you will need a Calibrated Laser
Power Meter to compare the Temperature readings to the LPM
wattage readings... while keeping the exact temperature in the T
est area noted...

BTW... have you seen this.....

http://laserpointerforums.com/f64/fs-new-hlpm-ii-200mw-lpm-module-54-99-shipped-46575.html


Jerry

You can contact us at any time on our Website: J.BAUER Electronics
 
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That makes sense. I would need a control setup for that to work.
No i did not see that 55$ one.
I will talk to my friends and see if they want to go in on it, so we can finally find out how powerful our lasers our. or at least in comparison to each other.
 
Is it wrong to use a temp probe from a DMM?

Says 78F. Laser says Output <5mw. Shine for 30 secs...peaks to 83F. 83 - 78 = 5mW..Im guessing no multiplier for this. its a heat probe. painted black.

200mw laser...I focused it to a needle, pointed at the probe for 30 secs.

273 - 78 = 195mW (not new battery)

Not sure if its right to do it that way. Clearly its not 273 X 3.13. Anyone tryed this using a DMM temp probe?
 
Guys, this guide was for using an IR thermometer, exclusively.

The point of an IR thermometer is that it's non-contact, it doesn't interfere with the heating of the foil.

This works by knowing how much energy it takes to raise the temperature of a piece of foil that size by a certain degree, that's how the estimation can be made. All of your temperature probe methods have no way of being related back to real values, all you could use it for is tracking degradation of a laser over a few months, or knowing if one is stronger than another. There's no scale involved.
 
What is ALL the math behind this ? I want to build something like this, but smaller, and I want to use metric system.
 
It should be possible to calculate all of this theoretically, but it would be difficult. The main jewel of this tutorial is the 3.13 magic number, I'd call it the Warske constant or Kw:) With this value he's letting us bypass all the complicated stuff and get straight to the point - relating a change in temperature to power output. Kw rolls into one number multiple values of emissivity and reflectance for several difference surfaces, as well as probably picking up the measurement error in the IR thermometer device. Kw might be true only for aluminum foil targets with one side painted of a size relatively close to 0.25 in^2, so I wouldnt know how this would scale to a patch of 1 m^2...

That said, I have done some online reading and I think I can shed some mathematical light on this. I would really appreciate if some of the smarter people here confirm or deny everything im about to say:)

By shining a laser energy on a target we heat it up. Black paint helps absorb the most energy, it looks black because it does not reflect any colors (wavelengths). A white surface is the opposite, it reflects back all of the light at all wavelengths. I think this property is called "reflectance". Any energy from the light not reflected back still has to go somewhere, so it manifests itself in the form of heat, increasing the temperature of the target. All objects at any temperature emit IR radiation, i think this is called "emissivity". There is a concept of a theoretical object called "blackbody", it emits all of its energy in the form of radiation, and has emissivity e=1. Flat black paint has a lower e, and i dont know what it is (maybe ~0.9), but this difference to e=1 is accounted for with Kw!

Now let's think about our lasers, targets and thermometers. The target is black on one side, very shiny on the other, and the target's temperature is being measured by the thermometer. The black side has a low reflectance, it absorbs most of the energy from the laser. This energy converts into heat and then gets emitted into the thermometer as IR. First, we start with no laser light, just the target floating infront of the sensor - it's at some temperature (room's) T1, and therefore emitts some IR (naturally) at power output P1 (watts). Second, we shoot the laser on the target and raise its temperature to T2 - the target now has much more energy going into it and start emitting at a higher output of P2. We now make an assumption that all of the energy going into the target also gets emitted out from it. This is not 100% true because there is no such thing as a true "blackbody", so in reality we should only get some fraction (80-90% prolly a good guess). We can now relate the power of IR radiated from the target by using the Stefan–Boltzmann law (Stefan, see 3rd eq from top):

P = A * e * s * T^4

where
P is power of the IR output from the target,
e is emissivity of the target's surfaces,
s is the Stefan–Boltzmann constant (0.0000000567) (it's not 's' on Wiki, but i dont know how to type the real lowercase Sigma:)),
T is temperature of the target in K (ex, 63.5F = 17.9C = 290.7K)

Our laser's output power is P(laser) = P2 - P1. A, e, s remain constant, the only change is in the T. Let's do the calculations for the 10F temp change Warske observed with a P(laser) = 31mW = 0.031 W. From the pictures i can see that T2 = 70.3F = 294.43K, so T1 = 60.3F = 288.87K. The square target has side of 0.5in = 1.27cm = 0.0127 m, and area is 0.00016129 m^2; we have TWO sides of the target emitting IR, so A = 0.00032258 m^2. Plug it all in:

P1 = 0.00032258 * 1 * 0.0000000567 * 288.87^4 = 0.1274 W
P2 = 0.00032258 * 1 * 0.0000000567 * 294.43^4 = 0.1374 W
P(laser) = P2 - P1 = 0.01 W = 10 mW

Hmm, something is not right, Warske got 31 mW here... Oh, that's right, we made a ton of assumptions along the way and pretended the target is a "blackbody"! Multiplying by 3.13 * 10 = 31.3 mW, that's why we need Warske's constant:) So the real equation should be:

P(laser) = Kw * A * e * s * (T2^4 - T1^4)
 
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So .. here are some my numbers.

I used target of 1 square cm, which is 0.0001 m^2. With temperatures 27.7C and 46.7C it gives:

A=0.0002
s=0.0000000567
e=1

P1 = 0.0927
P2 = 0.1184

P = 0.025

That is 25mW. Except I was shining 200mw at it. There some fundamental problem in this. If I use lower emissivity, it actually gives lower resulting power ! The paint is far from complete black, but I assume it does not reflect more then 10% of incoming radiation.

Anyway I will do test with larger targets (easier to aim at with the thermometer). Also I think you should paint the target from both sides, because unpainted metal will have lower emissivity (but as I said, overestimated emissivity leads to higher measured laser power, and mine is too low).
 
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hmm, damn, i was really hoping i'd be right about all that:) thanx for another data point!
 
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