I'm trying to connect 32 I²C "root of trust" NXP A71CH devices.

  • I'm thinking about a good way that does that, I mean what kind of communication should I use?
  • A digital multiplexer that selects a device? but based on what?
  • switching on the VDD source? but that needs reinitialization of the device and that takes time.
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    \$\begingroup\$ Why are you needing 32 of those? They only seem to have 2 addresses so you would need to have at least 16 addressable bus segments. \$\endgroup\$ – r_ahlskog Jun 7 '19 at 11:12
  • \$\begingroup\$ @r_ahlskog it's a requirment for a project. which bus do you recommend to use ? \$\endgroup\$ – Andre Jun 7 '19 at 11:22

Really short: I²C with a bus multiplexer; this is a common problem and it's solved with a bus multiplexer.

However, seriously, the point of having a root of trust is having one root of trust, not 32; hence, your project requirements make me suspect you're either building a test rig for these devices, or you're testing a device that is supposed to work with one of these devices, or you've misunderstood your project requirements.

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    \$\begingroup\$ I concur with your doubts about the need for 32 such devices in a single system. I don't see any case where it makes sense, but maybe I'm not imaginative enough. \$\endgroup\$ – dim Jun 7 '19 at 12:35
  • \$\begingroup\$ maybe they are doing a large testbed to figure out how much entropy the internal state of the device has after key generation or something like that. No idea. \$\endgroup\$ – Marcus Müller Jun 7 '19 at 12:42
  • \$\begingroup\$ @MarcusMüller I would like to parallel the signing of transactions using those 32 devices, I have no solution at my hand right now with FAST speed to connect those 32devices in parallel, signing a transaction costs 7 seconds.. do you have a solution ? \$\endgroup\$ – Andre Jun 12 '19 at 9:49
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    \$\begingroup\$ @Andre that's not what you use a root of trust for! You use that root to sign a key that you then use to sign the data. That key can be under control of your CPU. \$\endgroup\$ – Marcus Müller Jun 12 '19 at 16:50

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