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DS90UB914 Parallel Interface

KK Kim
Prodigy 110 points


Dear members

I am reviewing the datasheet and application related to DS90UB914Q.

But, I have some questions.

Please advise for me.

Q1) I understood that DS90UB914Q can support 12bit or 10 bit YUV interface.

Can DS90UB914 support YUV 8bit interface if the processor can only support 8-bit mode YUV?

If R[0:7] is connected to CPU , is that possible?
Is there resister to change to set 8 bit mode

Q2) I heard that Serialize and Deserializer have to use the pair products/

If serialiser is use DS90UB913 , deserializer have to use DS90UB914 of TI.

I want to know why the pair product have to use.

Can I use another deserializer of TI?
Ex) I have to use deseries of TI , but it isn't the serialize of another product.
How to choice TI product to desing with another products?

Thanks

  • 0
    TI__Intellectual 2010 points

    Hi Kim,

    1) You can send 8bit on either 10bit or 12bit mode.  There is not a unique mode for 8 bit. You are allowed to use less than the full number of bits. Ex) R[0:7]

    2) For the DS90UB913 you have to use the DS90UB914. At this time no other serializers or deserializers are compatible with the pair. 

    I do want to point out that there are two versions of these parts, the DS90UB913Q + DS90UB914Q and the DS90UB913A + DS90UB914A.  The Q parts support STP and have a minimum pclk of 10MHz.  With the A parts you can use either STP or Coax and the minimum PCLK is 25MHz.  You are allowed use a Q part with an A part but you are limited to STP with a minimum PCLK of 25MHz.

    Let me know if you have any other questions.

    Regards,

    Mike

  • 0 in reply to Michael Mellitt
    Mastermind 9165 points

    Dear Kim, Mike,

    Let me join the discussion, because of my customer also has similar ideas of usage of DS90913A/914A. From your discussion above, I understand that any of bits out of 12bit can be used for 8bit transmission, also no specific register settings for 8bit mode (either setting can be used).

    My customer intends to allocate the bit as drawing below, the bit is transmitted and output as it is, means straightforward. Is my understanding correct?

    Thanks for your reply in advance.
    Regards,
    Ken
  • 0 in reply to Kenichiro
    TI__Intellectual 2010 points

    Hi Ken,

    Yes this is correct.  Each DIN bit on the 913A will be outputted on the corresponding 914A ROUT bit.  With your current allocation you can transmit the 8 bits in either 10 bit or 12 bit mode.  In 10 bit mode DIN[10] and DIN[11] are not used.

    Regards,

    Mike

  • 0 in reply to Michael Mellitt
    Mastermind 9165 points

    Thanks, Mike!

    Since it's not my thread, I cannot do "verify Answer", sorry : (

    Regards,
    Ken