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Debugging JTAG daisy chain

LBeau.1
Associate III

Hello, I have a 4 identical microcontrollers daisy chain wich seems to work...exept I detect 8 devices?... My setup is the folowing...I have a segger J-link edu mini and 4 wire JTAG. basically I dont got JTRST but NRST connected to the J-link... In tis event should my reset strategy be Type 0:Normal or None?? Boot 0 is shorted to GND.

Something seems to work, as I can enter the debugging session and get this log below. But my true goal is to precisely select ny microcontroller on this chain and debug this and get to work. I seem to be only able to program tap #auto(position 0?) and position 2... Why IRPre9?? No idea...Saw this in a post.

J-Link found 8 JTAG devices, Total IRLen = 36

JTAG ID: 0x4BA00477 (Cortex-M3)

Connected to target

Waiting for GDB connection...Connected to 127.0.0.1

Reading all registers

Read 4 bytes @ address 0x080004D8 (Data = 0xB580E7FE)

Read 2 bytes @ address 0x080004D8 (Data = 0xE7FE)

Connected to 127.0.0.1

Reading all registers

Anyways, this is not my main problem, I am just confused as to why I get this effect, no numerical combination of position and IRPre seem to work.

Any help would be appreciated, Please shed light on this issue?0693W000006F127QAC.jpg

6 REPLIES 6
LBeau.1
Associate III

helllloooooooooooooooooooooooooooo

Uwe Bonnes
Principal II

A CortexM Device Jtag chain is already two devices. One for debug access and one for Boundary Scan/IO-access.JTAG ID:0x4BA00477 and similar is Cortex-M3debug. 0x064xx041 are the boundary scan devices.

LBeau.1
Associate III

so what are my 4 numerial possibilities to make this work and go on?

I have 2-9 and auto...Missing some.

Uwe Bonnes
Principal II

List the chain and tell your debugger to connect to the right device.

LBeau.1
Associate III

What do you mean? How?

I am using these 2 tickboxes, that is all.

Uwe Bonnes
Principal II

Learn about Jtag chains. Some tools offer you more insight about the chain. I do not know about the segger tools. There are probably 5 bit used for the Debug TAP and 4 bits for the Boundary Scan TAP. So you have 9 bits before then second chip.