The boot process for VLYNQ is as follows:
1. If FASTBOOT is enabled bootloader configures PLL according using appropriate PLL multiplier.
2. Bootloader makes sure VLYNQ is enabled.
3. Bootloader executes empty loop polling for BOOTCMPLT.CMPLT flag.
4. VLYNQ Host downloads application to DSP.
5. VLYNQ Host writes application start address to DSPBOOTADDR register.
6. VLYNQ Host writes 1 to BOOTMCPLT.CMPLT register flag.
7. Bootloader detects BOOTCMPLT.CMPLT and branches to start address in DSPBOOTADDR.
DM6446 can downloads application to DM6437 through VLYNQ interface.
DM6437's VLYNQ Remote Memory Map is 4C00 0000h: 4FFF FFFFh
DM6446‘s VLYNQ Remote Memory Map is 0C00 0000h: 0CFF FFFFh.
DM6446 acts as host and DM6437 acts as slave.
Does it mean DM6446 can only downloads application to 4C00 0000h: 4FFF FFFFh of DM6437's Memory?
But the address of DSPBOOTADDR in DM6437 is 0x01C4 0008, the address of BOOTCMPLTin DM6437 is 0x01C4 000C. The address was not between 4C00 0000h and 4FFF FFFFh. How to visit these two registers?
Or DM6446 can visit any memory of DM6437?