Can LPC43S70 due to errata OTP.2 brick itself?

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Can LPC43S70 due to errata OTP.2 brick itself?

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mateuszspychala
Contributor I

Hello,

In LPC43S70 AES key is stored in OTP region of memory. 

According to errata OTP.2 problem OTP region can change its value.

When programmed AES key change, the chip will be bricked.

In this region of memory also boot source can be programmed. 

When boot source change, also chip will be bricked.

Are above scenarios possible?

Best regards,

Mateusz Spychała

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victorjimenez
NXP TechSupport
NXP TechSupport

Hello Mateusz Spychała,

The scenarios you mentioned are possible, since the information of the OTP may change at some point. But, if you follow the work-around  you shouldn't have any problems. Also, please notice that this problem was fixed in the revision D devices. 

Hope it helps!

Victor.

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victorjimenez
NXP TechSupport
NXP TechSupport

Hello Mateusz Spychała,

The scenarios you mentioned are possible, since the information of the OTP may change at some point. But, if you follow the work-around  you shouldn't have any problems. Also, please notice that this problem was fixed in the revision D devices. 

Hope it helps!

Victor.

-----------------------------------------------------------------------------------------------------------------------

Note: If this post answers your question, please click the Correct Answer button. Thank you!

----------------------------------------------------------------------------------------------------------------------- 

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mateuszspychala
Contributor I

Hello Victoria,

Thanks for fast answer. Do you have any data about frequency of occurring this OTP problem.

For example 10 units per minion or something like that?

Best regards,

Mateusz Spychała

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victorjimenez
NXP TechSupport
NXP TechSupport

Hello Mateusz Spychała,

Unfortunately we don't have that information. If your MCU is not revision D I highly recommend you to implement the work-around to avoid having problems in your final application. 

Regards, 

Victor.

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mateuszspychala
Contributor I

Ok. Thank you for your help.

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