adcintel
Looking at the schematics of the Intel Edison mini breakout board, there are no pins for analog inputs, e.g. for reading state of a potentiometer.
For a hardware workshop based on the mini breakout board, I am thinking to discuss analog inputs. What circuit or breakout board would provide analog-to-digital conversion in a "cheap" way? What about power rails for the ADC? I2C on the Edison has 1.8V while system voltage can be higher but is not stable/accurate enough for ADC maybe?
There are various options, one reasonably simple circuit is this (bias/filter caps and bias voltage circuit not shown for simplicity):
Simulation:
For both graphs the x axis shows the input voltage - top graph is the input voltage itself (not necessary really), and the bottom graph is the output voltage versus the input voltage.
Note that the circuit inverts the signal, so 1.8V = 7V in and 0V = 16V in, but this is easily dealt with in a micro.
The circuit divides by 5 and shifts so the 1.8V span is between 0 and 1.8V. To learn how to calculate and design such circuits, OpAmps for Everyone is a good free book to start with.
1.8V is enough to drive a transistor which can be used to power all kinds of loads. Example:
simulate this circuit – Schematic created using CircuitLab
Best Answer
Use an external ADC connected to one of the digital interfaces of the Edison, for example I2C. Sparkfun sells PCBs that are designed to attach to the Edison in a stack, and here is one for a 4 channel (single ended, 2 channel differential) Delta Sigma 12 bits ADC. https://www.sparkfun.com/products/13327 I haven't looked around, there might be cheaper alternatives.
Nothing prevents you from making you own "edison block" with other ADC chips / analog multiplexers. It's probably cheaper, but very often your time is more valuable than a 10 dollars savings.