Xilinx Virtex6 block ram width

Question

I'm confused about Virtex-6 BlockRAM.

I want to implement a BRAM with 15 bit address (32,768 words) and 12 bit write and read data per word. When I explore the implemented design, I founded that 12 BlockRAM are used for this. Does this mean each BRAM in a Virtex-6 has 1 bit data?!

What is the data width and capacity of each BlockRAM in a Virtex-6?

May be easier to think that you have a RAM that stores 32,768x12 = 393,216 bits = 384 kB and each RAMB can store up to 36 kB (Paebbels reply details the size and ratios). This means you need 384/36 = 10.667 RAMBs, for which the tool has adjusted to 12 RAMBs. As a rule of thumb, if this calculation fails (which is not your case), you may need to explicitly write your code to adapt to the RAMB format. — suoto, Feb 29 '16 at 12:41

score 4 · Accepted Answer · edited Feb 29 '16 at 12:33

Each embedded BlockRAM (RAMB36E1) has 36 KiBit. It can also be split into two 18 KiBit primitives (RAMB18E1).

The internal organisation is 1024 × (32+4). 1024 rows with 4 bytes + 4 parity bits.

The BlockRAM macro has a variable layout from 32.768 × 1 up to 512 × 72, which is also depending on the mode: single port, simple dual port or true dual port.

Here is the full list of address to data width ratios:

64 Ki x 1, (cascaded adjacent BlockRAMs)
32 Ki x 1,
16 Ki x 2,
8 Ki x 4,
4 Ki x 8,
4 Ki x 9,
2 Ki x 16,
2 Ki x 18,
1 Ki x 32
1 Ki x 36 or
512 x 64 (simple dual-port mode) and
512 x 72 (simple dual-port mode)

See:

for more details.

When I explore the implemented design, I founded that 12 BlockRAM are used for this. Does this mean each BRAM in a Virtex-6 has 1 bit data?!

The used area (memory size) is equal to other organizations, but using 12 × (32.768 × 1) doesn't require multiplexers to combine the 12 BlockRAMs. This saves resources and is faster.

Xilinx Virtex6 block ram width

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