ADIS16448 Writing to Registers (example DIN = B504 for self-test enable)

Question

Hello! I hope you can help me out with my confusion. I am wondering as to how is the command 0xB504 formed for the activation of the self-test function? From the timing diagram, DIN has the format: 
 
 [15] - !Read/Write [14:8] - Address [7:0] - DC 
 I'm trying to create a function for writing to the registers of ADIS16448 and if I was to follow the format from here and use the MSC_CTRL as an example my data would be: 
 [15] - !Read/Write - 1 [14:8] - Address - 011 0100 or 0x34 [7:0] - DC - ? my first 8 bits would be 10110100 or 0xB4, I'm having trouble as to what my DC or bits 7 to 0 would be and how the final command for the self-test is equivalent to 0xB504? 
 Thank you for your help!

NevadaMark · Accepted Answer

Thank you for your post. I think that I see where there might be some confusion. Please see the red coloring, from your message, along with my comments, which are in blue . 
 
 [15] - !Read/Write - 1 [14:8] - Address - 011 0100 or 0x34 
 NevadaMark >> Bit 10 is actually located in the upper byte, which is at address 0x35. In this register, bits 7:0 are located at address 0x34 and bits 15:8 are located at address 0x35, which is why you would want to use 0x35 here, not 0x34. 
 [7:0] - DC - ? my first 8 bits would be 10110100 or 0xB4, I'm having trouble as to what my DC or bits 7 to 0 would be and how the final command for the self-test is equivalent to 0xB504?

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ADIS16448 Writing to Registers (example DIN = B504 for self-test enable)

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