请问Spartan6的BRAM硬核位宽是多少，其他FPGA系列呢？

在xilinx官网上下载对应手册，效率要高很多。

好像SPARTAN6只有9Kb和18Kb的bram， 位宽的话看设置为什么模式，simple dual port 还是true dual port, 看 spartan-6 memory resources user guide 会更清楚！

    官方数据手册”Spartan-6 FPGA Block RAM Resources User Guide“，里面说”The block RAM in Spartan-6 FPGAs stores up to 18K bits of data and can be configured as either two independent 9 Kb RAMs, or one 18 Kb RAM.“;    文档”Spartan-6 Family Overview“里面也说:
         Programmable Data Width
• Each port can be configured as 16K × 1, 8K × 2, 4K × 4, 2K × 9 (or 8), 1K × 18 (or 16), or 512 x 36 (or 32).
• The x9, x18, and x36 configurations include parity bits. The two ports can have different aspect ratios.
• Each block RAM can be divided into two completely independent 9 Kb block RAMs that can each be configured to any
aspect ratio from 8K x 1 to 512 x 18, with 256 x 36 supported in simple dual-port mode.
    这里面说的位宽都是经由Core Generator配置生成，也就是说在BlockRAM的基础上添加外围电路（?），但是我想知道的是BlockRAM本身的位宽是多少，18K×1?或者说BlockRAM的结构是什么样的，为什么一配置位宽就可以改变呢！

有两种，一种RAMB16BWER是32位，一种RAMB8BWER是16位。
The Spartan-6 FPGA block RAM library primitives, RAMB16BWER and RAMB8BWER,
are the basic building blocks for all block RAM configurations. Other block RAM
primitives and macros are based on these primitives. Some block RAM attributes can only
be configured using one of these primitives. See the Block RAM Attributes section.
Figure 8 illustrates all the I/O ports of the 18 Kb true dual-port block RAM primitive
(RAMB16BWER). Figure 9 illustrates the 9 Kb dual-port block RAM primitive
(RAMB8BWER). Table 7 lists the 9 Kb and 18 Kb primitives.

DS160 (v1.6)中说道，BlockRAM的基本大小为18Kb，可配置为两个独立的9Kb BlockRAM！