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DLP4500NIR: 4 diffraction patterns are generated from ON state

Part Number: DLP4500NIR
Other Parts Discussed in Thread: DLP4500

Hello Community/Support

 

I have a laser at 1.37um illuminating a DLP4500NIR from 10cm away with 24 degree AOI. The input beam is collimated and expanded to cover the entire active area of the DMD. I am using the DMD as an external monitor and I have the DMD displaying a white circle with a black background for the image. I measured the reflections using a Thorlabs infrared detector card at 10cm from the DMD.

As expected, the OFF state displays 1 image and there is a reflection off the DMD surface. But I have a problem with the ON state as it is reflecting 4 diffracted images instead of 1. The ON state should only produce 1 diffracted image. And as I move the infrared card away, the same 4 images continue to diverge.

Has anyone seen something similar, perhaps with the visible DLP4500 using a monochromatic source? Could this be from the geometry of the diamond pixel array?

The attached image is the result. From left to right: 4 circles in ON state, 0th order diffracted reflection off DMD surface, Conjugate image in OFF state.  

 

Cheers,

Vincent 

  • Hello Vincent,

    I will discuss it with my colleagues and get back to you.

    regards,

    Vivek

  • Hello Vincent,

    Thank you for your patience. DLP4500 uses 7.56 um pixel and 1.37 um wavelength happens to be between blaze conditions. That why you are seeing energy distribution on various diffraction order.

    You may experiment with different AOI angle and may be able to close to blaze condition. You have to move by a large angle. 

    If you could choose a different wavelength then please consider using 1.1 um.

    Hope this helps.

    regards,

    Vivek 

  • Hi Vivek

    I'm trying to calculate the blaze condition. How did you get the recommended wavelength of 1.1um? I keep getting 1.58 um as the blaze condition. I'm probably setting the angles incorrectly.

    a*( sin(theta_m) + sin(theta_i) ) = m*lambda

    1st order (m=1) 

    7.56um * (sin(0) + sin(12)) = 1.58um

    I set theta_m=1 to 0 degrees which would make the 1st order reflections at the blaze angle (same as 0th reflection).  

    Please advise. 

    Cheers,

    Vincent 

  • Hello Vincent,

    Thank you for your patience. Our response has slowed down due to recent conditions.

    I will get back to your question about the calculation after consulting our expert. 

    Typically shorter wavelength will higher diffraction efficiency. For example 740nm will be blaze condition and efficiency is ~87% for F/2.4. The next blaze condition at 1100 nm and it efficiency is ~86%. 1580 nm could have close to next peak but efficiency will be significantly low.

    regards,

    Vivek

  • Hello Vincent,

    The equation is correct. However few values are not correct -

    Pitch - It will half of   pixel diagonal i.e. 7.56 x sqrt(2)/2 ; it distance between two vertical  rows of pixels or distance between two closest tip of pixel.

     m - please use value of 2 for m.

    i - should be 24 

    7.56um * (sin(0) + sin(24)) ((sqrt(2)/4)= 1.09um

    Hope this explains.

    regards,

    Vivek

  • Thanks for the clarification Vivek! Appreciate it! 

  • You are welcome.

    regards,

    Vivek