Yes, most likely the body diode in the regulator’s MOSFET is conducting the current back through the regulator. So, if one Mini is powered down, the other will be powering both of them. The MOSFET is probably robust enough to handle this current with causing any harm.
It would be best if both units were powered up when connected, though.
Another option, as I’ve said previously, is to use a QWIIC cable without the 3.3V wire and only SDA, SCL and GND connected.
No, you need pullups somewhere. In I²C mode the data and clock output drivers are “open collector”. They only pull the signal low and are high impedance for a high signal. Therefore, you need a pullup to pull the line high, otherwise it would be floating.
I suggest you include the pullups on the Mini and preferably include bridged solder jumpers that can be cut if you only want pullups elsewhere.
Well I dunno there is input pullup and this guy says it works I’m gonna test it. I think it can’t work in situation where the other chip is not strong enough to pull down the signal but I think one arduino can pull down another.
@bateske,
Have you done any testing of the diode (D1) when powered via USB, and MOSFET (Q3) when battery powered, to see if they can handle the specified current (226mA) to the Qwiic interface?
You should also check the 3.3V regulator for the Qwiic interface to see how it holds up.
You could put a 15Ω 1W resistor across the 3.3V and GND pins of the Qwiic connector to draw 220mA.
the light usually comes on even when there is no battery (even though I would prefer no light)
battery only
dim LED
no LED
battery charging
bright LED
bright LED
3v3(?) power
dimmer LED
no LED?
I’ll get the multimeter out and do some tests, but I what I’m suspecting is happening is the mosfet is not blocking the voltage, and so it’s trying to charge itself?