i.MX Processors中的主题 Re: The question about the JTAG_MOD

The question about the JTAG_MOD

王剑翰 — Mon, 15 Sep 2014 11:07:21 GMT

Dear,

1,As the above table said, if I don't want use JTAG function, should the JTAG_MOD pin be connected with pull-up resistance? what is the resistance ?

2,I also find there are three different JTAG security modes:

Mode #1: No Debug-Maximum Security.

Mode #2: Secure JTAG-High security.

Mode #3: JTAG Enabled-Low security.

I am puzzled about these. I think the JTAG security modes only can be select by burning the related e-fuses if the JTAG_MOD pin is connected to GND.

If we connect the JTAG_MOD pin with pull-up resistance, we can not select the mode. Is my understanding right?

If I don't want to use JTAG function, which method will be recommended?

Thank you~!

igorpadykov — Mon, 15 Sep 2014 11:32:43 GMT

Hi 王剑翰

you can connect JTAG_MOD pin to ground if JTAG function

is not used. For bsdl testing JTAG_MOD should be high.

JTAG security modes implies:

No Debug-Maximum Securit - jtag just disabled

Secure JTAG-High security.- access with secret patterns

JTAG Enabled-Low security - usual jtag usage

Best regards

igor

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王剑翰 — Tue, 16 Sep 2014 02:08:17 GMT

Hi igor,

Thank you for your quickly answer.

As the table says, I think connecting JTAG_MOD pin to pull-up resistance is also used for no JTAG debug function. Am I right?

I think the difference between JTAG_MOD=0 and JTAG_MOD=1 is chain length.

No matter we set JTAG_MOD=0 or JTAG_MOD=1, the boundary scan test can be used.

So I think there are two method for no JTAG function:

1) Connect JTAG_MOD pin to ground and then burn the related e-fuses to enter No Debug-Maximum Securit mode;

2) Connect JTAG_MOD pin to pull-up resistance, so that we don't need to care the related e-fuses;

Above is my understanding, could you kindly correct me if I am wrong?

igorpadykov — Tue, 16 Sep 2014 06:48:16 GMT

if JTAG function is not used at all you can just leave it unconnected

~igor

王剑翰 — Tue, 16 Sep 2014 07:50:18 GMT

Thank you for your answer. I think I understand it.