View Full Version : Minolta CS-200


TomHuffman
07-14-09, 01:57 PM
What type of instrument is this?

Minolta's literature implies that it is some type of hybrid between a tristimulus colorimeter and a spectroradiometer. It apparently reads spectrographic data, which makes it sounds like a spectroradiometer. However, it is quite fast and reads very low light levels, which makes it sound like a colorimeter.

It seems like an unconventional tristimulus colorimeter, but I am not really sure what that means.

Can anyone explain the operating principles behind this instrument?

ghibliss
07-14-09, 04:04 PM
http://www.konicaminolta.com/sensingusa/products/display/luminance-color-meters/cs200/specifications.html

Tom,

This piece is not particularly good at measuring low light levels as it has a specified accuracy to 1 cd/m2 which is approximately .2919 fL with color accuracy specified to be within (x,y) of .007 and "repeatability" at this light level to only +/- 1 digit. If you check out the spec sheet you will find that they do not even provide accuracy values when the light level is below .5 cd/m2 at which point the "accuracy" is quoted to be good to only within +/- 1 digit. The instrument uses a photo diode array which utilizes a spectral fit method to derive chromaticity values. The piece has a spectral bandwidth which is basically the same as an i1Pro which is 10 nm/pixel. This is why the instrument can on certain display types provide inaccurate data.

A true spectral device provides a far greater degree of certainty for a broader range of applications then this piece allows. The price point which is about $12K is quite expensive for the degree of accuracy which it offers.

krasmuzik
07-14-09, 09:36 PM
At it's heart it is a colorimeter in principle much like the Spyder - calibrating multiple filter differences to come up with the weighted passbands used to form the CIE integration curves. The whole point is to make it a suitable field instrument designed to be calibrated with the lab spectroradiometer on your light sources. It differs from a conventional colorimeter in that you don't use XYZ optical filters that match CIE, XYZ are derived quantities much like in a spectroradiometer - but it also is not sampling the spectra.

The only truly accurate device is whatever ghbliss happens to be selling this year - anyone that reads this forum knows this by now...

TomHuffman
07-15-09, 12:15 AM
It differs from a conventional colorimeter in that you don't use XYZ optical filters that match CIE, XYZ are derived quantities much like in a spectroradiometer - but it also is not sampling the spectra.This is consistent with what Minolta's literature says, but I must say that my eyes glaze over when I try to make sense of this. It is like a tristimulus colorimeter, except it doesn't use optical filters. It is like a spectroradiometer, except it doesn't sample the spectra. I am officially clueless.

Michael TLV
07-15-09, 01:10 AM
Greetings

It is considered a Hybrid of the two ... not just a tri stim ... and not a spectro.

Pluses are Portability in that it runs on AA's and you really don't need to connect it to anything else to work. Take readings and read off the side screen. Never any uncertainty about what you are aiming at. Put dot on the place you want the reading from.

Also fast ...

Of course it is compatible with Calman and Colorfacts ...

Minus ... it is big to carry around and having a spare set of batteries is always a good idea. (But you don't need a tripod and a laptop)

I use it when doing calibration jobs for installers when no fancy graphs are needed. Pop in ... shot the display ... tweak it ... and away I go. Also great for my store front work at the various stores.

Regards

krasmuzik
07-15-09, 09:10 AM
Try an audio analogy - do you understand the difference between a dBA filter, an RTA 3rd octave filter set, and an FFT?

Bear5k
07-15-09, 10:56 AM
This is consistent with what Minolta's literature says, but I must say that my eyes glaze over when I try to make sense of this. It is like a tristimulus colorimeter, except it doesn't use optical filters. It is like a spectroradiometer, except it doesn't sample the spectra. I am officially clueless.
As near as I can tell, it uses a grating to divide the spectrum into measurable quanta, but then uses photodiodes like a colorimeter to sense the amount of light present from a given "slice". Since the photodiodes tend to be a lot larger than the CCD arrays used in many "true" spectro designs, this probably complicates the light path inside the unit, driving up the cost. However, I'm not about to disassemble one to find out. :)

ghibliss
07-15-09, 12:53 PM
krasmuzik

The only truly accurate device is whatever ghbliss happens to be selling this year - anyone that reads this forum knows this by now...

What information in my reply do you find so offensive to warrant your snide comment? The specifications which I list are taken directly from the Minolta websites specifications. The only comment that I made was that for the price a true spectroradiometer would be a better choice. How does this comment have anything to do with what I do? I also happen to support this instrument in my software package just as my competitor does for those that happen to use one.

Gregg Loewen
07-17-09, 12:56 PM
hi guys

Tom / Michael.... Michael are you still wanting to sell yours ??

Michael TLV
07-17-09, 12:57 PM
Greetings

Yes ...

regards

TomHuffman
07-17-09, 01:02 PM
Tom / Michael.... Michael are you still wanting to sell yours ??Gregg: I am not an owner. I was just curious to understand how this thing works. It seems like a unique type of instrument.

TomHuffman
07-17-09, 01:03 PM
As near as I can tell, it uses a grating to divide the spectrum into measurable quanta, but then uses photodiodes like a colorimeter to sense the amount of light present from a given "slice". Since the photodiodes tend to be a lot larger than the CCD arrays used in many "true" spectro designs, this probably complicates the light path inside the unit, driving up the cost. However, I'm not about to disassemble one to find out. :)Thanks. That's the best explanation I have heard so far.