Welcome to Laser Pointer Forums - discuss green laser pointers, blue laser pointers, and all types of lasers



Basic questions for thermal LPM

Georgi96

New member
Joined
Jan 3, 2022
Messages
5
Points
1
Hello, community,

I have several basic questions about thermal LPM for powers between 100-500W.

So, the basic measurement concept is measuring the temperature absorbed by a metal plate and then calculating the resulting power. As I know the most companies use thermopiles as a sensor.

But I do not have an answer to the following three questions:
  1. What kind of thermopiles are used?
    • thermopile sensors, which can measure from a distance (like an IR camera)
    • or do they use thermopile that is attached to the metal plate?
  2. Why just simple senor type K or J or PT1000 is not used? It actually does the same work.
  3. Most LPM for measuring range 100-500W have a cooling fan too. But what does it cool?
    • Is it the ambient temperature for the electronics or not?
    • Cooling the metal block(absorber) makes no sense for me, as we want to measure its temperature.
Many thanks for considering my request!
 



bostjan

Active member
Joined
Dec 29, 2011
Messages
202
Points
43
I know nothing about the large LPM's, but, I think I can partially answer just based on general LPM measurement.

You do not want to measure the temperature of the hot spot (T). You want to measure the heat (Q), which is usually done either by measuring the thermal gradient along the surface of the target material (dT/dr) or by measuring the cooling power required to maintain a constant temperature, if the material is a good enough thermal conductor. This can be done by using an array of thermocouples as you mentioned, but they would be situated at different distances away from the target spot.

Remember that power is a measurement of the flow of energy, and temperature is the measurement of thermal equilibrium. Heat is the flow of thermal energy from high temperature to low temperature. If we want to measure the optical power of the laser, we have to convert the optical energy into thermal energy and then measure how much of it there is. If I shoot 100W of heat into an object, how much does the temperature of the object increase? Well, no one can answer unless provided with the type of material (for example: Aluminum vs Gold), the mass of the material (for example 0.1 g vs 15 kg), and, if we want to get precise measurements, the conditions (for example, is the material clamped down or free, what is the temperature, etc.). And, realistically, since we are talking about a point source of the heat, it's a little more complicated than that, since we don't want to wait several minutes for things to reach equilibrium.
 

Georgi96

New member
Joined
Jan 3, 2022
Messages
5
Points
1
Thank you for the answer, bostjan. My question appeared to me, after reading the post ,which can be found in the attached document.

Moreover, he has mentioned in his post cooling water, but I can not see what he cools in the picture, but it is not the aluminum block.

By measuring temperature, I meant actually the change of temperature over time, e.g., 60 seconds. So, for example, I know the start temperature(ambient temperature), and I measure for 60 seconds, and then i know how much the temperature has increased.

A simple example:
I have a 400 W Laser that shoots for 60 seconds in the middle of an aluminum block. The parameters of the block are 10*10*0.5cm. Using the density of aluminum, I get a mass of 0.135 kg.
Using the following equation Q=m*c*ΔT where Q is the energy (Q = 400W*60s = 24000 Joule), the mass is 0.135kg, and c is the heat capacity of aluminum I get a heat gradient of 197,5 K. that means that the temperature has increased by 197,5 K.

It is definitely not the most precise solution, but it should generally work, doesn't it?
 

Attachments

  • Thermal measurement of laser power.pdf
    541.4 KB · Views: 1

Anthony P

Well-known member
LPF Site Supporter
Joined
Oct 7, 2018
Messages
519
Points
63
You are correct. Your equations work to an extent, under ideal conditions. You also have to factor in ambient loss, purity of your sample, etc. A commercial power meter already has calibration for all relevant factors.
 

Georgi96

New member
Joined
Jan 3, 2022
Messages
5
Points
1
You are correct. Your equations work to an extent, under ideal conditions. You also have to factor in ambient loss, purity of your sample, etc. A commercial power meter already has calibration for all relevant factors.
Thank you a lot for the answer, Anthony.

Yeah, I know that commercial product is calibrated.
For me was just important if the equation even works and if the method shown in the blog is, to an extent, correct.
 




Top