MAX10 Dual Boot.

Hi,

Refer the links and demo projects.

AN 730: Nios II Processor Booting Methods in MAX 10 FPGA Devices

https://fpgacloud.intel.com/devstore/platform/16.0.0/Standard/dualboot-design-example/

https://fpgawiki.intel.com/uploads/0/00/Using_Dual_Boot_MAX_10_Evaluation_Kit.pdf

https://www.intel.com/content/dam/www/programmable/us/en/pdfs/literature/an/an741.pdf

Let me know if this has helped resolve the issue you are facing or if you need any further assistance.

Best Regards,

Anand Raj Shankar

(This message was posted on behalf of Intel Corporation)

I did as described in the document. However it dosen't work after programming prd file into CFM1. The master project should be configured as a single or a double image?

prd file is a binary file. should I reverse bytes before programming? and what is the order of bytes?

this

word_to_write = (UInt32)((reversed_byte[raw_data[i]] << 24) | (reversed_byte[raw_data[i + 1]] << 16) |   (reversed_byte[raw_data[i + 2]] << 8) | reversed_byte[raw_data[i + 3]]);

or this

word_to_write = (UInt32)((reversed_byte[raw_data[i + 3]] << 24) | (reversed_byte[raw_data[i + 2]] << 16) |   (reversed_byte[raw_data[i + 1]] << 8) | reversed_byte[raw_data[i]]);

May be I burn in the wrong sector? in my module flash_update generics are

SECTOR3_START_ADDR         => 16384,
SECTOR3_END_ADDR             => 114687,
SECTOR4_START_ADDR         => 114688,
SECTOR4_END_ADDR             => 188415,
SECTOR5_START_ADDR          => 188416,
SECTOR5_END_ADDR              => 360447,

As I understood SECTOR5 is CFM0 and SECTOR3-SECTOR4 is CFM1.

Hi,

1. Project related to dual configuration should be configured as double image.
2. Should I reverse bytes before programming? No.

For more information please check the below links.

https://www.youtube.com/watch?v=0k4AZmdW9Sk

https://www.youtube.com/watch?v=Z0l0GrjXn8w

Let me know if this has helped resolve the issue you are facing or if you need any further assistance.

Best Regards,

Anand Raj Shankar

(This message was posted on behalf of Intel Corporation)

I write the rpd file. Let's take the first bytes from the file 50 FF 47 FF. But when I read it back from the flash I get FF 50 FF 47.

What's going on! Who can me explain for for Christ's sake!

Hi,

1. Bit reversal is required if you have used little-Indian option while generating the .pof file. Which is default.
2. If Big-Indian option is used bit reversal is not required
3. Example design is attached

Best Regards,

Anand Raj Shankar

(This message was posted on behalf of Intel Corporation)

Hi Anand

Thank you for the example.

I have something to clarify

for(address=CFM1StartAddress;address<=CFM1EndAddress;address=address+4)
{
scanf("%2x%2x%2x%2x",&receivedHex[0],&receivedHex[1],&receivedHex[2],&receivedHex[3]); /*Get 4 bytes from UART Terminal*/
for(byte=0; byte<4; byte++)
{
receivedHex[byte] = (((receivedHex[byte] & 0xaa)>>1)|((receivedHex[byte] & 0x55)<<1));		/*Swap LSB with MSB before write into CFM*/
receivedHex[byte] = (((receivedHex[byte] & 0xcc)>>2)|((receivedHex[byte] & 0x33)<<2));
receivedHex[byte] = (((receivedHex[byte] & 0xf0)>>4)|((receivedHex[byte] & 0x0f)<<4));
 	}
 
word = (receivedHex[0]<<24)|(receivedHex[1]<<16)|(receivedHex[2]<<8)|(receivedHex[3]);	/*Combine 4 bytes to become 1 word before write operation*/
	IOWR_32DIRECT(ONCHIP_FLASH_0_DATA_BASE, address, word);					/*Command to write into On-Chip Flash IP*/
	byte = 0;
	word = 0x0;
	while((IORD(ONCHIP_FLASH_0_CSR_BASE, 0) & 0x0b) == 0x02)
		{
			  	printf("Writing CFM1(%d)\n",address);
		}
		if((IORD(ONCHIP_FLASH_0_CSR_BASE, 0) & 0x0b) == 0x00)
			{
			  		printf("Write to %d failed\n", address);
			  		state=20;
			  		address = 700000;
			}
			if((IORD(ONCHIP_FLASH_0_CSR_BASE, 0) & 0x0b) == 0x08)
			{
 
			}
 
	}

This part

for(byte=0; byte<4; byte++)
{
/*Swap LSB with MSB before write into CFM*/
receivedHex[byte] = (((receivedHex[byte] & 0xaa)>>1)|((receivedHex[byte] & 0x55)<<1));		
receivedHex[byte] = (((receivedHex[byte] & 0xcc)>>2)|((receivedHex[byte] & 0x33)<<2));
receivedHex[byte] = (((receivedHex[byte] & 0xf0)>>4)|((receivedHex[byte] & 0x0f)<<4));
}

I shouldn't do if I have Big-Indian option or I should swap MSB and LSB anyway?

Hi,

Yes, Shouldn't do if we have Big-Indian option​.