fpga 怎么读取epcs 里的id

2019-03-25 07:10发布

我的altera epcs6E22 用epcs  M25p16.
我想加密我的系统, M25P16 有一个唯一的ID, 如果可以读取,我就可以加密了。
可我不知道怎么在FPGA 里读取。
datasheet 是这么说的:
Instructions M25P16
6.3 Read Identification (RDID)
The Read Identification (RDID) instruction allows to read the device identification data:
„ Manufacturer identification (1 byte)
„ Device identification (2 bytes)
„ A Unique ID code (UID) (17 bytes, of which 16 available upon customer request).
The manufacturer identification is assigned by JEDEC, and has the value 20h for Numonyx.
The device identification is assigned by the device manufacturer, and indicates the memory
type in the first byte (20h), and the memory capacity of the device in the second byte (15h).
The UID contains the length of the following data in the first byte (set to 10h), and 16 bytes
of the optional Customized Factory Data (CFD) content. The CFD bytes are read-only and
can be programmed with customers data upon their request. If the customers do not make
requests, the devices are shipped with all the CFD bytes programmed to zero (00h).

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17条回答
白丁
2019-03-25 19:45
liwenz 发表于 2016-10-13 21:54
谢谢版主回答。我不明白主动配置方式,我就照开发板给的电路, 以及qsys 设计。
我想这里附图都不可以。 ...

可以看下fpga的主动配置和被动配置,时间太长了那几个控制管脚已经忘了,主动配置就是fpga上电主动从norflash读取配置数据,被动配置就是利用个mcu去通过管脚配置fpga,具体有好多种配置方式,altera有个官方文档的,可以找来看看

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