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Heater and black paint for Peltier cell based power meter

a12b34c56

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It looks like quite a few people have built Peltier cell based laser power meters. I’ve seen different approaches to the heater for calibration and black paint.

One approach to the heater is to attach a couple of resistors to the front side of the cell. The problem with this approach is that only the part of resistor that is in direct contact with the cell transfers the heat directly to the cell and the rest of the resistor dissipates the heat via radiation in all directions so god knows what percentage 60, 80… get actually transferred to the cell. Another approach is to attach looped resistive wire which probably transfer more heat to the cell. However the biggest question what to use to glue the wire. Any type of glue will be a heat barrier unless the glue is heavily loaded with good heat conductive fillers silver, cooper… but still even with this approach the efficiency of heat transfer will be less than 100% because the heat conductivity coefficient of best TIM materials on the market is less than 15 W/m*K and that with more than 95% loading rate of the filler. So with the silver powder and 60% loading rate it will be less than 5W/m*K and probably close to 2 W/m*K.

Another issue is the black paint. I’ve seen people using markers and spray can paint which can create a problem as both paints use dies that usually absorb more light in certain wavelength range. Some people are using graphite discs cut from battery’s electrodes glued to the cell. The problem is the same as with the wire heat loses in the glue and heat conductivity of carbon is pretty low. Some people are making paint in which they use activated carbon die in some sort of glue. This seems more valid and I’d like to try this approach but I’m not sure what type of glue to use. In industrial thermal pads the most common polymer is silicon but it is very thick for the paint even if it is highly diluted.

I wander how people on this forum solved the problems?
 



farbe2

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Hey,

i build a Peltier sensor myself so i have a few tips.

0. Use the smallest tec you can get, this will help to reduce the effect of uneven heat distribution on its surface. After all the alumina ceramic is not a super good thermal conductor. So keep the area as small as you can (and need) to have a low thermal differential across the surface.

1. I used small resistor (SMD 0603) for the calibration heaters. Not too small to have low contact area (more drop across the TIM). I used normal thermal epoxy to glue them down, i soldered afterwards to guarantee that the resistor would be flat on the tec. Dont use just one resistor, spreading the thermal load across multiple resistor will help reducing the effect of the thermal resistance by the TIM. I used 4 resistors and calibrated with 100mW, this is only 25mW for each resistor and thus the TIM will only see a few mK drop across it.

Your thoughts are not completely correct. The radiation that will be emitted by the resistors is do to them being hot, thats the same as a laser beam hitting them and heating the resistors up. So the laserbeam will also produce (the same) thermal radiation that will be radiated and will not go trough the tec.
You just need to have the resistors and the laser surface evenly covered in your absorbing paint.
So combining low individual resistor load and same emissivity on resistor and tec will get you very close. Think about it, if you drop 25mW across a resistor, even the worst thermal compound only produce a few mK of temperature raise. This means only a a very very small fraction more radiation from the very slightly hotter resistor than from the surrounding sensing surface -> insignificant.

2. I used lamp black / soot. I applied it by lighting a gas lighter under the cold tec. The will deposit a even layer of soot thats a quite broad absorber. Use only gas flames for soot, candle flame deposits wax residue thats reflective.
Its easy to do, however it is not adhered like paint would, so no touchy touchy. If you got your sensing surface damaged by laser or touch, just wipe it and reapply the soot coating.

I had good results with this, i actually compared it with a real laser power meter and it was closer than 5%!
 

paul1598419

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Most TEC based laser power meters don't add resistors to the heat sink or around the TEC itself. Consider that in making yours. I have used Rustoleum black paint in a spray can with acceptable results. Graphite is likely better though. Good luck to you.
 

farbe2

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Yes, the resistors are only needed if you want to calibrate the instrument without having a known stable power laser.
If you have, say a 155mW Laser module thats stable and has been measured recently, go ahead without resistors. So you can just use the Laser as a transfer standard.

If you dont have one, you need the resistor (not on the heatsink, on the tec sensing surface) to calibrate the meter for known power. Resistors are easy because P=U*I will give you very accurate results with just a simple DMM.
 

a12b34c56

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Hey farbe2,

Thank you very much for very detailed and encouraging answer. 5% accuracy is very impressive! I’ve hoped to get at least 20% accuracy.
Where on the cell did you glue the resistors near the perimeter of the cell?
Do you use multi meter to read the voltage directly or op amp to amplify the signal and what min power can you measure?
Do you remember the model of Peltier cell that you've used? I’m finding only 40X40mm cells. I’d think smaller cell should allow to measure lower power that the large one.
What brand of thermal epoxy have you used? My colleague recently tested a bunch of thermal pastes and greases from Amazon and most of them did not meet their spec by a very significant margin.

Recently I bought 660 nm line 300 mW laser pointer from ebay that did not look to me like 300 mW compared to ebay 100mW and 200 mW 655nm dot pointers. I have Extech Easy View 33 light meter that I use for comparative measurements which has a limited range and really not a proper tool. Anyway it gave me roughly 100 mW for so called 300 mW pointer compared to 100mW and 200 mW ones. When I talked to the seller he told me … off because my measurements were not correct and one can’t use light meter to measure laser power and also the power output can be only measured without lenses which by the way are not removable.

As 300 mW pointer produces the line I needed a pretty large area to capture all output from 5 cm distance. I picked up TEC1 12706 from Amazon yesterday and coat one side with thermal grease leaving roughly 3 mm on each side clean to glue it to heat think with super glue. I used a candle to blacken the other side of the cell. It took me less than 30 min to make a working prototype. 100mW pointer gave me 8 mV reading and 200mW gave me 16 mV reading and 300 mW gave me 6 mV. I bought 100 mW and 200 mW pointers from different sellers and at different time so it is very unlikely that they are both off but produce the same ½ voltage ratio as their power ratio.
 

a12b34c56

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Most TEC based laser power meters don't add resistors to the heat sink or around the TEC itself. Consider that in making yours. I have used Rustoleum black paint in a spray can with acceptable results. Graphite is likely better though. Good luck to you.
I don't have a calibrated light source so I need somehow to calibrate it.
 

paul1598419

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Yes, the resistors are only needed if you want to calibrate the instrument without having a known stable power laser.
If you have, say a 155mW Laser module thats stable and has been measured recently, go ahead without resistors. So you can just use the Laser as a transfer standard.

If you dont have one, you need the resistor (not on the heatsink, on the tec sensing surface) to calibrate the meter for known power. Resistors are easy because P=U*I will give you very accurate results with just a simple DMM.

I would think that if you are going to calibrate your LPM this way, you need only use the resistor temporarily. LPMs are usually calibrated using a single wavelength laser and that wavelength depends on what you intend on measuring with it. Many are calibrated using an IR laser, but it really depends on what you are going to measure.
 

a12b34c56

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I would think that if you are going to calibrate your LPM this way, you need only use the resistor temporarily. LPMs are usually calibrated using a single wavelength laser and that wavelength depends on what you intend on measuring with it. Many are calibrated using an IR laser, but it really depends on what you are going to measure.
I'm using cheap Chinese mutimiter so the calibration needs to be done periodically but I think even expensive one are not very stable.
I'm mostly interested in red 630-660nm and IR 830-904 nm ranges.
What are the dimensions of your cell?
 

farbe2

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I used a tec that I salvaged out of a old DPSS 532nm laser. But I think it was 15x15mm.

I made a OP circuit that had offset and gain adjustment. I used a cheap 200mV panel meter and amplified the signal to have 0-2W power meter with 1mW resolution. I would not count on 1mW accuracy. Even good power meters are very touchy at these ranges with a high power (2W/3W) measurement head.

As for epoxy: I used almost the same method as planters / tech ingredients, I mixed fine copper powder with standard epoxy. This gives great results and is cheap. If I don't need much, I use simple arctic silver thermal epoxy.

My resistors where glued in a circle around my intended measurement spot.

To attach the resistors temporarily would not work, this will change the radiation load after removing the resistors, so the calibration would need to be done without the resistors if you want to measure without resistors. -> Impossible.

I would go with resistors as calibration, thats the most accurate/easy (for home use). Otherwise, make sure to use a high quality tec, your voltage values seem quite low for around 100mW. If I remember correctly, I got way more voltage out of my tec.
I have also experienced big big differences on cooling power with 12706 tec from random eBay sellers vs good quality ones.
 

a12b34c56

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I used a tec that I salvaged out of a old DPSS 532nm laser. But I think it was 15x15mm.

I made a OP circuit that had offset and gain adjustment. I used a cheap 200mV panel meter and amplified the signal to have 0-2W power meter with 1mW resolution. I would not count on 1mW accuracy. Even good power meters are very touchy at these ranges with a high power (2W/3W) measurement head.

As for epoxy: I used almost the same method as planters / tech ingredients, I mixed fine copper powder with standard epoxy. This gives great results and is cheap. If I don't need much, I use simple arctic silver thermal epoxy.

My resistors where glued in a circle around my intended measurement spot.

To attach the resistors temporarily would not work, this will change the radiation load after removing the resistors, so the calibration would need to be done without the resistors if you want to measure without resistors. -> Impossible.

I would go with resistors as calibration, thats the most accurate/easy (for home use). Otherwise, make sure to use a high quality tec, your voltage values seem quite low for around 100mW. If I remember correctly, I got way more voltage out of my tec.
I have also experienced big big differences on cooling power with 12706 tec from random eBay sellers vs good quality ones.
Do you have any idea of D50 of Cu powder that you've used?

I have tried different coatings: candle, black marker and graphite( which I just rubbed in) and got pretty much the same voltage readings for 100 &200mW 650nm and200Mw 830 nm pointers so I think low voltage output is indeed related to the efficacy of TEC that I got. 200 mW 650 nm was giving me 16mV and 200mW 830nm 20mV. In theory the coating should absorb 650nm better and 830 nm will be reflected more.
I tried to play with different thickness of graphite coating but got pretty much the same result. I will try lamp black.

I need resistors as I don't have an access to calibrated LD and when I was looking for calibrate LEDs I checked a number of suppliers that claimed that they sell them and found out that they don't really have them. I was reading that somebody was covering resistors to reduce heat loss but what you have said about radiation pattern makes sense because probably 20-40% of light is reflected so reducing resistor’s heat loses does not make much sense.

I found 15X15 cell produced by UK based European Thermodynamics that claim that they manufacture in UK . It has Vdc =15.5V and they also make 10X12 cell with Vdc=8.4V. It seems to me that 15X15 will generate higher voltage and therefore will be more sensitive.
 

a12b34c56

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Do you have any idea of D50 of Cu powder that you've used?

I have tried different coatings: candle, black marker and graphite( which I just rubbed in) and got pretty much the same voltage readings for 100 &200mW 650nm and200Mw 830 nm pointers so I think low voltage output is indeed related to the efficacy of TEC that I got. 200 mW 650 nm was giving me 16mV and 200mW 830nm 20mV. In theory the coating should absorb 650nm better and 830 nm will be reflected more.
I tried to play with different thickness of graphite coating but got pretty much the same result. I will try lamp black.

I need resistors as I don't have an access to calibrated LD and when I was looking for calibrate LEDs I checked a number of suppliers that claimed that they sell them and found out that they don't really have them. I was reading that somebody was covering resistors to reduce heat loss but what you have said about radiation pattern makes sense because probably 20-40% of light is reflected so reducing resistor’s heat loses does not make much sense.

I found 15X15 cell produced by UK based European Thermodynamics that claim that they manufacture in UK . It has Vdc =15.5V and they also make 10X12 cell with Vdc=8.4V. It seems to me that 15X15 will generate higher voltage and therefore will be more sensitive.
 

paul1598419

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None of my meters are cheap and therefore are not bound to wander.
 

farbe2

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... radiation pattern makes sense because probably 20-40% of light is reflected ...
No no, you did not understand that right.

The energy produced in the resistors will fully convert to heat, this will heat them up. (Thats clear..)
The Laserbeam will strike your sensing surface and will be either reflected or converted to heat e.g. heating the sensing surface. Ideally you would have a coating thats 100% absorbent, meaning 100% of the lasers energy will be converted to heat energy -> 0% reflected.
Thats why i recommended coating the surface with lamp black e.g. burning a sooty gas flame underneath the cold sensing surface thus depositing a thin layer of lamp black / carbon black. This will absorb almost 99% of the incoming light and convert it to heat.

To calibrate the response of you sensor (mV per mW), you will need to apply a known amount of power to the surface.
By gluing resistors to the surface before you blacken the surface, you have an easy mean of applying any power you want.
If you apply the power to the resistor you can expect to have losses (power that does not get to your sensor, thus skewing the calibration) that are two fold:
1. convective losses
2. radiative losses

Convective is essentially that your resistors will get hot, this will make the air around them hot, therefor the air is less dens and raises thus carrying away heat.

Radiative losses are similar: the resistors get hot and therefor emit infrared light (spectrum temperature depended), this will also lower the power at your sensing element.

HOWEVER! and this is big.

Imagine a laser hitting the sensor (with the resistors attached), the coating will absorb the light and heat up. This will also heat up (small tec area required) the resistors, which in turn will show the same losses as they would if they where heated by the current flowing though them.

Example: Hitting the surface with the laser (assume 100mW) will heat the sensor surface to 10k above ambient. The resistors would be glued to the tec surface and will heat up to the same 10k above ambient. (actually just little less because TIM resistance) So the resistors would produce Convective and radiative losses with 10k temperature difference, reducing the power at the sensing element to 90mW.

Powering the resistors with 100mW will also heat the sensor surface to 10k above ambient, the resistors will also be at around 10k above ambient. (just a little more because TIM resistance) Thus producing the same convection and radiative losses. This results in 90mW at the sensing element.

So in the calibration example the resistors have almost the same temperature as in the laser example. Only a tiny fraction difference because of the TIM. This does not matter because its thermal resistance. Its the same as electrical resistance: only a high flow through the resistance will cause significant drop. As the power inside the resistors will only be a few mW, the drop across the tim resistance is negligible. The other way around e.g. resistors get heated by the laser hitting the sensing area near the resistors: now only the radiative and convection losses are flowing through the tim, even less impact.

So your resistors only calibrate for your sensing element, you still need to make sure to absorb all of the light on the sensing element.
If you only absorb 80% of the light, your calibrated result will be 20% off. (As you make 20% less heat than calibrated for)

Pure carbon black / soot has known absorption curves, so you could even calibrate for that by using a multiplier.
If you know that soot will absorb 95% of light at your desired wavelength, you could calibrate with 95mW resistive and adjust your meter that it shows 100mW.

As for your tec: i would go for the higher voltage one, it should be good.
 

a12b34c56

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I understand the theory quite well as in my current line of work I involved with the design of fillers for TIM and EMI shielding. I started my career working in optoelectronics industry but I have not touched lasers for decades so I completely forgot that a human eye is one of the most sensitive instruments that can detect only three photons so I overestimated the magnitude of reflection. I did a pretty lousy candle coating job as I was afraid to bust the only TEC that I got so it was not absorbing well.
I coated some Al scrap with graphite, black paint and candle and measured the reflection from each. With 100mW 650nm laser I got only 22% from Al, 17% from graphite 5% from black paint and 0.5% from the candle coating. I have not tried carbon black because I don’t have it but I will get some canola oil tomorrow and will try it. As I don’t have an optical bench and proper tools the measurements are semi- quantitative. In this case covering resistors with Al foil glued to Styrofoam makes sense as the surface area of resistors is larger than the one of the laser spot and hence thermal losses are larger and are not known. In theory they can be calculated.
 




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