THS8135: THS8135 on C6748 LC DEV KIT VER A7E

Part Number: THS8135

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Hello, 

As shown in the picture below, could you please help me to understand why BCB0-4 and BCB5-9 are both configured for LCD_1V8_D0-4, and GY0-3 and GY6-9 are both configured for LCD_1V8_D_7-10?

Thanks,

Michael

2 Replies

  • Michael,
    I am not familiar with the C6748 LC DEV KIT but this bit duplication is a common way of expanding dynamic range when the source has a lower number of bits per channel than the target.

    If you have a source which has, for example, 6 bits per color channel and you want to connect to a target that has 10 bits per color channel there are a few ways to expand the 6 bits to 10 bits.

    The first is to attach the source MSB to the target MSB, then then next bit etc... i.e. T9=S5, T8=S4 etc..., then tie off the lowest target bits to '0'. This results in a black source gives black on the target, and a linear ramp, but when the source is at max (111111) the target will not be full range (1111110000).

    The next approach is to do as above, but connect the LSBs to 1 instead of 0. This results in full white source = full white target but blacks are not black. e.g. source = 111111 gives target 1111111111 but source = 000000 gives target 000000111111. This also has a linear ramp.

    Next, the source LSB can simply be replicated in the target LSBs so source = 000000 gives 0000000000, source 000001 gives 0000011111. This results in source black gives target black, and source white gives target white, but the ramp is not linear (there is a DC bias which can throw off colors)

    A common way is to duplicate source MSB bits in the lower target bits so that the full dynamic range is achieved (black = black, white = white) and the ramp error is minimized at the expense of a slight non-linearity.
    e.g.
    Source S5-S4-S3-S2-S1-S0 is transposed to T5-T4-T3-T2-T1-T0-T5-T4-T3-T2

    A more elaborate bit expansion is to dither, but this cannot be achieved by simple wiring connectivity.

    Looking at the above schematic I am guessing that the source if a 16 bit color interface (5 red, 6 green, 5 blue)

    BR,
    Steve

    Please mark this post as answered via the Verify Answer button below if you think it answers your question.  Thanks!

  • In reply to Steve Clynes:

    Thank you so much for your help Steve.