Is the FPGA Smatfusion2 claimed to be highly secure by its manufacturer (Microsemi, now Mircochip) because it is a non volatile memory FPGA? Meanwhile the SRAM FPGA like the zynq 7000 are considered vulnerable because the content need to be loaded each time power is applied to the system? Is that enough to ensure secure booting?
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3\$\begingroup\$ It is enough for those for which it is enough. That is how marketing works. If you want to know if it is secure, then give a definition of what is secure. \$\endgroup\$– PlasmaHHCommented Sep 17, 2018 at 13:41
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2\$\begingroup\$ Perhaps you should invite Microchip to your office and ask them about the security of their products and how you wish to utilize them for your projects. They're the best source of information, not us. \$\endgroup\$– user103380Commented Sep 17, 2018 at 14:19
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\$\begingroup\$ after your comments I updated it to : is it enough to ensure secure booting \$\endgroup\$– LavenderCommented Sep 17, 2018 at 14:40
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4\$\begingroup\$ The SmartFusion 2 documentation cover all the different secure boot settings and security functions in detail. It's probably best to read through that so you can ask more specific questions. \$\endgroup\$– user185972Commented Sep 17, 2018 at 15:24
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Nothing is absolutely secure. There are degrees of security. While higher security is often advantageous, there are usually tradeoffs to get it. These can be higher cost, more difficulty in working with and doing development using the product, etc.
Storing the configuration inside the FPGA does provide higher security than having the configuration external and having to load it into the FPGA at powerup. Whether that's secure enough for your purposes is something only you can say. Meaningless words like "highly" are just that, meaningless.
answered Sep 17, 2018 at 13:46