Scientech 362 arrived!!

[SideFX]

Finally!!!!
My scientech 362 powermeter I was talking about arrived!!
It was in the best shape one could imagine.. not a single scratch!

I hooked it up right away and started measuring my lasers.. I couldn't resist sharing this on PL

First with my projector currently only containing a 150mw (as how it was sold to me) greeny:



After all optics (dichro, x/y) .. only 70mw!!
I removed the scanners.. then I measured 90mw after only the dichro.
Final stage:

the greeny without optics measured at 130mw.. well close to 150mw don't you think?

Now ofcourse I expect that this meter needs some calibration.. but i can at least determine some relatives..
30% loss on the dichro? Is that normal?

Then I measured my 635 red which was sold to me as a 300mw..

A whopping 60mw!!
Maybe it's calibration is way off.. but considering it was doing well with the greeny i think i got conned bigtime on this..
Too late to ask for a refund i guess..

Someone here who has experience with calibrating these babies? I read somewhere the 2 white connectors should be used to power up an internal resistor of 40 Ohm.. you should be able to measure the exact amount of mW's put in to the resistor..
is that correct?

[daedal]

Well, congrats on the new purchase A meter that is well worth it indeed!

Can you take a picture of the sensor's inside. Is it matte or reflective (thermal or silicon)?

If it's a silicon sensor, there's going to have to be a conversion factor for the wavelength...

Enjoy your new toy

--DDL

[SideFX]

Well, congrats on the new purchase A meter that is well worth it indeed!

Can you take a picture of the sensor's inside. Is it matte or reflective (thermal or silicon)?

If it's a silicon sensor, there's going to have to be a conversion factor for the wavelength...

Enjoy your new toy

--DDL

Hey DDL.. Thanks!

I was under the impression that wavelength conversion was not an issue with thermophiles.. but i could be mistaking..

Picture of the inside (full size here):



There's some damage there.. could that possibly affect the readings?

[daedal]

Good picture-taking skills

That looks like a thermal head to me... and as such should most probably NOT be wavelength dependent.

The damage will affect your readings, but it looks like you have enough there to be able to avoid that spot

Sux about the lasers then...

Where'd you get them?

--DDL

[mixedgas]

That is a thermal head. Call scientech with the serial numbers and get the cal procedure and constants for your head. You then need a variable power supply and a really good voltmeter,

The damage wont hurt you much. Zero being off will however.

proper zeroing is to hold a black metal object in front of the detector.

Steve

[SideFX]

That is a thermal head. Call scientech with the serial numbers and get the cal procedure and constants for your head. You then need a variable power supply and a really good voltmeter,

The damage wont hurt you much. Zero being off will however.

proper zeroing is to hold a black metal object in front of the detector.

Thanks for the comments!

I applied 2V to the internal resistor. According to physics this should result in: P = U^2 / R = 2^2 / 40 = 100mW.
And this is exactly what the meter showed.. spot on!



I checked again with 4V <=> 400mW.. same thing.
I'll try contacting Scientech for the 'official' procedure... but if I can trust these measurements then I guess this baby is in pretty good shape

For anyone interested BTW, I purchased this meter from RDY-sales (ebay store).

Sux about the lasers then...

Where'd you get them?

China

Well I think I'd better get my facts straight (= calibrating power meter, second meter) before bashing a manufacturer..

Cheers!

[trwalters001]

Hi SideFX,

Check this out:

http://www.photonlexicon.com/forums/...ead.php?t=1160

Also, call Scientech at 800-525-0522 and talk to Bob Lee. He's been there forever and knows these meters very well.

[SideFX]

Hi SideFX,

Check this out:

http://www.photonlexicon.com/forums/...ead.php?t=1160

Also, call Scientech at 800-525-0522 and talk to Bob Lee. He's been there forever and knows these meters very well.

Wow.. thank you for this.. all the information I need was right here on PL! must've missed it.

This formula looks a little funny to me though:
V = (Rc x C x W)1/2

shouldn't that 1/2 be SQRT()?

[colouredmirrorball]

Maybe they meant:
V = (Rc x C x W)^1/2 = Sqrt (Rc x C x W)

[SideFX]

Maybe they meant:
V = (Rc x C x W)^1/2 = Sqrt (Rc x C x W)

ok that makes sense
the ^ probably had fallen off