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Gold-coated DMD mirrors

Chris Burak
Prodigy 40 points
Other Parts Discussed in Thread: DLP9500

Does TI offer an evaporated gold mirror coating option on any of its DMD chips?  I have an application where very high near IR performance would be a necessity. 

  • 0
    TI__Intellectual 2175 points

    Hi Chris,

    Gold coating is not an option at this time.  What is the application?  Is there a certain efficiency you are wanting to achieve?

    Thanks,

    Eric

  • 0 in reply to ELD
    Prodigy 40 points

    My thought was that the ~%98 reflection efficiency at 1.06 microns would allow me to increase my laser power without overheating the mirror structure.

    I found an article about a company that developed a process to remove and replace the optical window on the DMD without causing any mirror failures.  Assuming this was possible in my application, could I have a ~50nm gold layer evaporated onto the mirror array? 

    Would this adversely effect the dynamics of the mirror/actuator? 

    Would there be potential adhesion problems between the gold and the existing mirror coating?

    Thanks for your quick reply to my post!

    Regards,

    Chris

  • 0 in reply to Chris Burak
    Prodigy 40 points

    I understand that TI does not offer a gold coating, and have discovered that "aftermarket" gold overcoating is not an option for a variety of reasons. 

    In order to understand the power handling of existing DMD's, I have been trying to use the example on "sheet 9" of TI DN 2509928 Rev A.  Perhaps someone can help me.

    The example uses a screen illumination of 2000 lumens at 300 lumens/watt efficiency.  That indicates 6.7 watts on the screen.  Using the stated 83.7% illumination hitting the active array, and 87% projection efficiency, I back out 9.2 watts incident on the DMD (Pinc).

    Assuming (worst case) the 87% projection efficiency means that 13% is absorbed in the DMD, and that the remaining 16.3% of the incident light (11.9% on the array border and 4.4% on the aperture) is completely absorbed in the DMD as heat, I come up with a figure of 2.5 Watts absorbed, not the ~5.4 Watts calculated in the app note.

    Pinc * (0.837  x  0.13  + 0.163 x 1.0) = 2.5 Watts.

    I think my assumption about the array border absorption is fairly severe - the array border is made up of mirrors and should reflect about as much incident light as the active array mirrors.  Isn't this true?

  • 0 in reply to Chris Burak
    TI__Guru 54175 points

    Hello Chris,

    Sorry for the delayed response.  There are two things to take into consideration.  The DMD must handle the worst case situation - all mirrors in the "off" position relative to the input illumination.  The example you refer to  Also the DMD circuit also contributes some heat, 4.4 Watts for the 1080p. 

    You can directly download the current datasheet at DLP9500 (DLPS025A).  Please use the numbers in the current revision for comparison.

    Which app not are you referring to?  I would like to review the numbers to see if the app note does not need correcting.

    Fizix.

  • 0 in reply to Fizix
    Prodigy 40 points

    Hi, Fizix.

    The app note I used is referred to as "Product Preview TI DN 2509928 Rev A June 2008".  The numbers used in the new data sheet appear to be the same.

    The 4.4 watts generated by the internal circuitry was excluded - the 5.4 W was the remaining power (absorbed from the incident light) computed in the app note. 

    Using the same numbers as before (Pinc=9.2W, etc), you would have to have over 50% absorption by the active array when the mirrors are all in the off state.  Does that sound about right?

    Thanks.

  • 0 in reply to Chris Burak
    TI__Guru 54175 points

    Chris,

    The actual number is closer to 365 Lumens/Watt, but this is the absorption rate, not the Lumens/Screen Watt.  Notice in the calculation that QILL = CL2W x SL where QILL is the absorbed energy in Watts.  So this constant is the array absorption rate, not the Illumination efficiency.  So read this as Lumens per absorbed Watt.

    A Metal Halide lamp (comonly used is around 87 Lumens/Watt)  So that the input power (if the system were 100% efficient - no geometric or optical losses except for the 5.4 Watts aborbed) would have to be at least 23 Watts + the 5.4 absorbed.  That would be 28.4 Watts illumination which would be about 19% absorption. 

    I hope this helps.

    Fizix