Here is what we are dealing with:
Looking inside my unit, I see even pins 4-20 indeed connected to ground.
Pin 2 goes to a jumper (which is in place). With the jumper in, this gets a 3.3 volt supply, with it out, it is isolated. I am pulling this jumper and leaving it out so this is a N/C pin as per the Segger document.
Checking with a meter, all the odd pins measure 0 volts.
Pins 3-17 all connect to the 245 chips via 22 ohm resistors.
These are signals.
Pin 1 is called "VTref" and should be connected to the power rail on the target system. The Segger uses this to sense the levels on the target system and to configure itself.
pin 19 goes to what looks like a transistor.
There is 5 volts on another terminal of the transistor, so as the document says,
this "may" be used to provide 5 volts to the target.
Segger pin 7 (TMS) -- SWDIO Segger pin 9 (TCK) -- SWCLK Segger pin 15 (RESET) -- RESETMy devices with SWD don't include reset, but it would be handy to have it. If you have access to your reset pin, you can just connect it to pin 15.
What this guy does is to get rid of the transistor controlling the 5 volts (since as he says, these days over 99 percent of projects are using 3.3 volts) and adds a jumper wire to carry 3.3 volts to pin 19 all the time instead. He also makes a permanent connection from pin 19 to pin 2. Then he can install a jumper to tell pin 1 (VTref) that he is running on 3.3 volts.
What I could do, inspired by him, is to remove the transistor and then just connect pin 19 to pin 2 (since I can put 3.3 volts there on my board already via the jumper I mention above. Then I install his jumper from 1-2 to tell VTref what I am doing and let pin 19 supply my target.
I'll note that on my Segger, an AMS1117 device is the 3.3 volt regulator. He suggests this can supply up to 800 mA, which may be true for the device, but he is forgetting the 500 mA USB power limit. Also the Segger itself does use some current, so we will be able to get maybe 400 mA as my guess.
Tom's Computer Info / [email protected]