Why do you think it is "unsolved". The datasheet shows what you can expect from the chip. Remember, the relay draws 16.8mA *when it has 3.3V across the coil* ( it is a 196 ohm resistance). If you put less than 3.3V across the coil, the draw will be less. If you measure 1.8V across the coil, then the PIC is sourcing 9 mA. Since the coil has 1.8V across it, the PIC has 3.3V - 1.8V = 1.5V.
The datasheet says that at 9mA, you aren't guaranteed to be able to source any current at all.

This is not "black magic".


And if you look at the datasheet for the PIC 18F1320, you will find that it is *NOT* guaranteed to work either. You may find some that do work, but the datasheet shows they wont. You need to SINK, not source the current from the relay to guarantee your circuit will work.


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