This message was posted on behalf of Intel Corporation
Thanks for reaching out!
I would like to quote a section taken from section 4 of the Arduino Expansion Board Hardware Guide (http://download.intel.com/support/edison/sb/edisonarduino_hg_331191007.pdf):
"...You can power the Intel® Edison kit for Arduino* using any of the following:
- An external power supply on J1;
- DCIN via shield header pin VIN;
- A USB cable via micro USB connector J16; or
- A lithium-ion battery connected to J2.
When power is applied to J1 or VIN, the external power must be in the range of 7 to 17 V. The power is converted to 5 V via a switching power supply, which powers the rest of the system. This supply was designed for a 1 A continuous supply..."
So, you indeed can power your Edison with an 11.1v, I highly suggest you to read this section of the hardware guide to avoid issues when choosing the correct connector. That's because not all batteries can be used on J2. But, for example, your battery can be connected directly to J1 because of it power.
If you correctly power the Arduino expansion board, it will indeed power both the board and Edison module.
I hope this information helps you,
Still, take care when using Li-Po (normally pouches as used in cell phones) or Li-ion (metal can cylindrical cells) batteries. I hope you have a factory built battery with built-in protector circuit? Or you really know what you are doing..
A protector circuit will prevent overcharging (which will cause Li-Po to bulge and eventually burst into flames and Li-ion to explode), or underdischarge (causing permanent damage that may result in fire later in the batteries life). Limits for overcharge are in the 4.1 - 4.2V range and underdischarge 2.7 - 3.0V.
So, your 3x3.7V battery may actually reach 3x4.2 = 12.6V.
And if you are charging in-circuit, when full the protector circuit disconnects the cells and the full charger voltage (constant V - constant I curve) will be on the Edisons input. A correctly built charger will of course limit this voltage to below the protector limit, in your case probably 12.0V
And then there is the matter of charge balancing...