Let me investigate on this and I will reply as soon as I have an answer.
Thanks, that would be great.
My guess is my bet is pin J20-1 (V_SYS).
With the 12v input I have measured some of the IO pins using meter and see:
J17-4 (VIN) = 12v as expected
J18-5(V_BAT...) = 2.497
J19.2(V1p80) = 1.8
J20-2(V_SYS) = 4.39
J20-2(V3p30) = 3.4
This makes some sense. There is a 5V_SYS Bus, but I don't see any connectors that connect to it.
I do see on page 6, that it feeds into a chip with an enable, that than feeds into USB (VBUS). But again don't see any IO pins for this (there is a test point). On system could probably plug USB hub into Edison, and then use usb cable to grab the voltage.
Hopefully there might be other things I am missing here.
When you are feeding your board with an external power jack (J22), power goes to a DC-DC converter that transforms that voltage to 5V, as shown on Figure 4 of https://communities.intel.com/docs/DOC-23252 . This goes to a battery recharger which limits the output voltage to 4.4V. I think this is the maximum voltage that you can get from any pin in the breakout board. If you take a look at table 2 you’ll see that the labeled outputs are of 3.3V (J20-2, V_V3P30) and 1.8V (J19-2, V_V1P80). V_SYS and VIN are both inputs and that power is transferred to a lower voltage than 5V by IC in the board. According to your readings, the closest value to 5V would be 4.39 from V_VSYS, and I think that is the closest to 5V as you can get. The method of using an USB hub might work, but I haven’t tried it yet.