Maybe I am just too slow, but I can't understand exactly how do you connect an ATMEL-ICE (6 pin connector, not the 10 pin) to a chip. I am using a SAMD21G (http://www.atmel.com/Images/Atmel-42181-SAM-D21_Datasheet.pdf)

I made a board, where I take the pins needed for the connection to the ATMEL-ICE. I am following this doc: https://github.com/femtoduino/femto-usb

As connection, I did connect all the ground (pin 5,18, 35 and 42), and connected the power pins (pin 6, 17, 36, 43 and 44) to the power rail, which give 3.3V to each of these pin.

Then I have the pin 45 and 46, which are SWCLK and SWDIO

From the doc linked, the pin should be:

ICE SAM port         Microcontroller
Pin 1 3.3v           3,3v (main power rail)
Pim 2 SWDIO          SWDIO (pin 46)
Pim 3 GND            GND (main GND line)
Pin 4 SWDCLK         SWDCLK (pin 45)
Pin 5 GND            GND (main GND line)

But I do not see the green led on the ICE go on; also when I read the voltage in Atmel Studio, I see -3.2 in the target voltage; which tell me that the connection is not matching the right pins.

I am using the cable included in the ICE, using the SAM port as described in the document. The user manual of the ICE has so many different connections, but I can't figure out exactly the correct pins to start to program the microcontroller.

Looking for some advice about how to connect the microcontroller to the Atmel-ICE at this point; since the pinout on that document seems to be wrong.

  • 1
    \$\begingroup\$ Are you providing power to the device externally or trying to do so through ATMEL ICE ? \$\endgroup\$
    – efox29
    Commented Jun 23, 2016 at 8:26
  • 1
    \$\begingroup\$ FYI, the 10-pin half-pitch connector on the Atmel ICE intended for ARM Cortex parts is almost standard - essentially it is the standard pinout, but for some reason they did it with the notch on the wrong side. So it's actually a 180 degree rotation of the standard and requires a cable with as similarly rotated end. \$\endgroup\$ Commented Jan 7, 2017 at 8:13
  • \$\begingroup\$ I see; That would factor too in the confusion. BTW I am not making a board like the reference board; mine is minimal, so that's why I am programming the chips first, before they get on the PCB. Not going to make industrial production, so taking space for a connector on the board was not that important for me \$\endgroup\$
    – rataplan
    Commented Jan 9, 2017 at 18:19

1 Answer 1


So, for some reason, the pinout are totally messed up because they are 2 different connections. The standard ICE cable holds 2 different connector, where one has less pin and a larger pin pitch (standard 100 mil but only 6 pin), while the other has all the pins but the pin pitch is very small (50 mil and 10 pin), but the connector on the ICE is 10 pin 100 mil, so you need an adapter.

I ended up solving the issue, making a self-made adapter for a 10 pin connector on standard 100 mil pitch, buying the flat cable and connectors, and using a breakout board that has the 10 pin connector ready for usage. The next step was to map the pins, and that came easy since the ICE guide has the pinout of the SAM port using SWD mapping:

SWDIO     2
Reset     10
VTG       1
GND       3

These are all the pin needed; once connected the flat cable to the breakout board, I did connect these wires and I was able to write on the chip without problems. The voltage was correct, and the chip was correctly recognized in Atmel Studio.

I just wish someone would tell me what pin set to use, what port and suggest me to ditch the original cable that came with the ICE from the start; but hope that this will help others like me, that just started with microcontrollers

  • \$\begingroup\$ SAM JTAG Pinout (Cortex-M debug connector) When designing an application PCB which includes an Atmel SAM with the JTAG interface, it is recommended to use the pinout as shown in the figure below. Both 100-mil and 50-mil variants of this pinout are supported, depending on the cabling and adapters included with the particular kit. Figure 4-2. SAM JTAG Header Pinout. \$\endgroup\$
    – efox29
    Commented Jan 7, 2017 at 7:38
  • \$\begingroup\$ That was right from the manual UserGuide (atmel.com/Images/Atmel-42330-Atmel-ICE_UserGuide.pdf). We understand that you are new, but there wasn't much activity on your question for a debugging to begin. At some point, someone might have found out sooner than 6 months. If you post a question, follow up with it, and if someone asks you a question, answer it, it keeps the question alive and on the front page. You'll get it next time. \$\endgroup\$
    – efox29
    Commented Jan 7, 2017 at 7:40
  • \$\begingroup\$ Yes, I am aware that the question may not be followed; I was not complaining that nobody replied (it didn't take me 6 months to figure it out, but it took me weeks; I just remembered that I had an account here and was asking questions, so I did post my answer only now). I did read that manual many times; and the only thing that was pointed out often was to get the 99 dollars adapter. Now, I did get the flat cable and the breakout for 2 dollars...not sure what kind of gold is used for the official Atmel connector \$\endgroup\$
    – rataplan
    Commented Jan 7, 2017 at 7:50
  • \$\begingroup\$ Also if you start with a bare chip, how do you end up with a prototype with a JTAG connector on it? That make sense when you have a reference board (which Atmel sell), but if you buy the chip and need to start to experiment, is not like you prepare a prototype on a circuit board, and for the sake of it put also the JTAG connector. I am new into this world, so probably that's why I do not understand the workflow. Thanks for the comments BTW! \$\endgroup\$
    – rataplan
    Commented Jan 7, 2017 at 7:52
  • 1
    \$\begingroup\$ @newbiez actually, you do put a jtag or SWD connector on any board you design. If you put the MCU on some generic package breakout you put in a breadboard, then you probably split the jtag cable to individual wires or flying leads. \$\endgroup\$ Commented Jan 7, 2017 at 8:10

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