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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liwenz 你好呀 我现在的情况也是可以读写flash 使用epcs read id 指令就会进入一个死循环,你这个问题找到解决办法了吗?
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