Version of the C API for LabVIEW FPGA 2011

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  How can I correctly address the 128 - bit DRAM memory?  I have the Bank DRAM 0 set as a memory of 128 bits, set up in my design as a CLIP.  I realize it's a wide RAM on 32-bit.  I had a National Instruments AE do the original design I've been adding to.  He said that the addresses needed to incrementing by four with each entry.  Example: if I had to write in consecutive addresses, I would write to the address: 0, 3, 7, 11, 15 etc, and I would like to send 128 bits to each address.  My address is calculated as: (number of pixels in a line of video + line * (number of pixels per line) for a picture of the video).  So I take my calculated address and add 4.

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  Sets the address in external memory for reading or writing. The physical data bus for external memory is 32 bits wide (4 bytes). Each unique address value represents 4 bytes of data. Therefore, the total number of unique addresses in external memory is equal to (Memory Size in bytes)/4.  
  
  Note  The memory interface exposed to LabVIEW FPGA is 128 bits wide. As a result, each memory write or read operation accesses four different address locations in memory. The memory controller latches this signal value only when you issue a new memory write command by asserting the Command_Write_Enable signal.
  

  I'm confused by the 2nd paragraph "every Scripture memory or read operation four access address locations of memory."  Does that mean I increment the address by 1 to get 128 consecutive bits 'locations' (Yes, I know, that's 4 words of 32 bits in memory), or do I increment the address by 4, in the order of words of 32 bits 4 by 128-bit single transfer?

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  I want to clarify my previous post.  There are two ways to access memory DRAM, CLIP (that you have described is what you do) and using the memory node.  As noted before, the DRAM is 128 bits wide.  When you write to the CLIP you basically write pieces that is the width of the databus (in this case 32-bit).  Therefore, when you write a total of 128-bit DRAM, you place 32 bits in each address.  The address being the width of the databus, then you write with a writing & the address 0, 1, 2, 3.  Then the next write will be 4, 5, 6, & 7 and then address 8, 9, 10, & 11 and so on.  In this case, you must increment your address by 4 whenever you write.  Note that you start at 0, then 4, then 8, etc 12.  In your previous post, you were out of a figure.

  There is also another way to write in the DRAM memory, and it is through the node of memory, which is what is used in the example that you are pointing out.  Here, LabVIEW takes on some of the thought, and instead of being the width of the databus address, they are the width of the entire segment of 128 bits.  So when you write to DRAM here, you only increment 1 whenever address because they refer to any segment of memory.  This contrast with the CLIP, address 0 of the memory node interface match the addresses 0, 1, 2, 3 in CLIP mode & and address 1 of the memory node would correspond to 4, 5, 6, 7 in CLIP mode addresses &.  If you do not write an integer of 128 bits for the memory node, then the remaining addresses in the data block are filled with "junk" so that the address remains constant.

  As I mentioned previously, it is the most effective writing in chucks of 128 bits so that you don't waste all of the DRAM.  I hope you find this explanation clearer.

  Brandon Treece

  Technical sales engineer

  National Instruments

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