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islamh
Newbie
Newbie
6,401 Views
Registered: ‎09-07-2011

Memory Bits calculation

 

Hi All,

 

i need to calculate the memory bits for any of the FPGAs of Xilinx.

 

I have opened "FPGAs Overview" document, I have found out that there are 3 columns (Fields) under the "RAM BLOCK".

 

I am not sure which of these 3 columns ( 18Kb, 36Kb and Max(kb) ) will represent the Memory Bits.

 

should I take the summation of both 18Kb and 36Kb ? or just take the "Max" only ?

 

Would you please help me in that  (write an equation for me please) ?

 

 

Thanks in advance.

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4 Replies
austin
Scholar
Scholar
6,394 Views
Registered: ‎02-27-2008

i,

There is more than one kind of memory, which do you want to count?

The BRAM is listed in the datasheets, as a maximum for all the BRAM in the device (by part number). Remember that total number of bits = 1024 * "Mbytes" specified (K=024 bits for memory specification).

How much of the BRAM you use depends on how many BRAM blocks are used by your design. That information is in the device usage reports provided by the synthesis, place and route tools.

Configuration memory, or CRAM, is the memory which is used to configure (or reconfigure) the device.

To estimate the CRAM size, you may take the total bitstream size (from the datsasheets), subtract the BRAM size, and take 90% of that number (~10% of the bitstream are commands, and not contents).
Austin Lesea
Principal Engineer
Xilinx San Jose
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gszakacs
Instructor
Instructor
6,390 Views
Registered: ‎08-14-2007

What may not be so clear in the product brief is that the 18Kb and 36Kb blocks are really

counting the same resources.  So for example for V6 you can see in the table:

 

 

showing for the smallest part 312 18Kb and 156 36Kb block RAMs.  In reality there are only 156 memory

blocks, each of which can be configured as either 2 x 18 Kb or 1 x 36 Kb, and the Max (Kb) column will

show the total block RAM bits, which is the same as 156 x 36 Kb in this case.  You do not add the 18Kb and

36Kb columns.  The Max Kb column shows the actual total number of block RAM bits in the device.

 

The total distributed memory, for devices that show it, is not part of the block RAM bits total.

 

-- Gabor

-- Gabor
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austin
Scholar
Scholar
6,386 Views
Registered: ‎02-27-2008

Gabor,

Good point about the 18Kb, and 36 Kb BRAMs (two modes of the same block).

Distributed RAM are the look up tables, used as RAM, and are part of the bitstream configuration memory (CRAM).
Austin Lesea
Principal Engineer
Xilinx San Jose
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islamh
Newbie
Newbie
6,375 Views
Registered: ‎09-07-2011

Thanks a lot Gabor and Austin. Both Explanations and answers are great.

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