And to make this work, what connections would be required? Just brainstorming here, any expert feedback is welcome…
As inspiration I looked at the FEZ Duino schematics:
Here is the connection table:
USB-C connector <-> SCM20100E
------------------ --- -------------
VBUS <-> STPS140A
<-> LDO (LM1117MP-3.3) <-> 3V3 (pin 32)
GND <-> GND2 (pin 31)
CC1 <-> R1 (5.1K) <-> GND2 (pin 31)
D- <-> USBC_N (pin 45)
D+ <-> USBC_P (pin 46)
CC2 <-> R2 (5.1K) <-> GND2 (pin 31)
And here for each connection the reasoning …
The VBUS provides +5V. In order to safely connect this to +3V3 pin 32, it must be regulated.
On the FEZ Duino this is done like in the 2 images above. First through the STPS140A and then using a LM1117MP-3.3. I assume the STPS140A provides extra safety for spikes / voltage reversals?
Since the SCM20100E does not have a +5V pin somewhere, the FEZ Duino setup is followed. As a result, a 3V3 is supplied to the SCM20100E on pin 32.
CC1 and CC2
On the FEZ Duino the CC1 and CC2 pins of the USB-C connector are each connected to a 5.1K resistor and then to GND.
This is a safe way of signalling to the USB host that nothing is going to be charged here? CC1 and CC2 are charging signals right?
These signals will be routed as in the SCM20100E prototype setup. They will go through 5.1K resistor to GND. They will not go into SCM20100E.
D- and D+
The D lines go directly to pin 45 and pin 46 of SCM20100E. On the schematics of the SCM20100E is see the following:
This means we have a resistor between the pin and the chip on board already.
So it is safe to connect directly from the USB connector to PIN 45 and 46?
Again … feedback welcome!