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DLP9000X: Minimum display time of 8-bit gray pattern

Zhiguang Liu
Prodigy 60 points
Part Number: DLP9000X
Other Parts Discussed in Thread: DLPC910

I am writing to ask the 8-bit gray scale illumination of DLP9000X. I am wondering the meaning of a maximum frequency 1873Hz in 8-bit gray patterns. Does it mean if I set an input picture with 8-bit gray scale, the minimum diaplay time should be 1/1873 second? So the minimum modulation step will be 1/1873/256=0.023 ms? Thanks!

  • 0
    TI__Mastermind 43125 points

    Hello Zhiguang,

    Welcome to the DLP section of the TI-E2E forums.

    You are correct.  However it does assume that you have a modulated light source.  

    Also, there are no defined sequences in the DLPC910.  You have to build an 8-bit grayscale sequence on your own which includes modulating a light source to create the shortest bits.  This would be done on your front end (APPS_FPGA) on our reference design.

    Fizix

  • 0 in reply to Fizix
    Prodigy 60 points

    Hi Fizix,

    Thanks a lot for your reply.

    I know the shortest modulation step now. However, I am confused about the defined sequences of light source to create the shortest bits.

    To my understanding, the modulation is set to the input iamge, while the light source is invariable. If a pixel in a 8-bit grey input image is pure white, the corresponding mirror will experience a 100/0 duty cycle in the display time period (for example, the shortest display time is 1/1873=0.53ms). If the pixel is 1/256 in gray scale, the landed-on time of corresponding mirror will be 0.53/256=0.023ms, the landed-off time will be 0.53*255/256=0.528ms. Is it correct?

    Could you please address with more details about what should I do to obtain the shortest bits in a 8-bit gray display? Thank you very much!

    Zhiguang

  • 0 in reply to Zhiguang Liu
    TI__Mastermind 43125 points

    Hello again Zhiguang,

    It is possible to create bits that have less "light time" than the pattern switching time.  For example the time you listed is 530 µs, but an 8-bit sequence will have an LSB (least significant bit) of 530 µs / 256 = 2.07 µs. 

    The DMD can switch very fast, but not that fast.  Therefore if you modulate the illumination (LED or laser) to illuminate for only 2 µs, then you will have dark time while the LSB+1 pattern loads.  Very quickly the display time of the higher bits will be as large as the pattern load time and modulation will not be needed for the higher bits.  Usually by the 3rd or 4th bit.  

    I hope this helps.

    Fizix

  • 0 in reply to Fizix
    Prodigy 60 points

    Hi Fizix,

    Thanks for the very helpful reply! 

    If the light source (LED or laser) is unchanged, in other words, without any modulation, what's the shortest display time step of DLP9000X in a 8-bit grey mode? 

    I am also wondering the pattern load time of DLP9000X.

    Thanks!

    Zhiguang

  • 0 in reply to Zhiguang Liu
    TI__Mastermind 43125 points

    Hello again, Zhiguang,

    The minimum display time without modulation if for a simple linear 8-bit image is [ ( 2^n - 1 ) * binary pattern time].  Where n=8 is the bit depth.

    You do not need to load 255 times, only 8.  Each bit starting with the LSB will be shown for twice as long as the bit before it until you reach the MSB.

    This should answer both of your questions.

    For further information search for "bit planes",  There is a great deal of information out there on this.

    Fizix