Cross-talk square-wave?

[Jon Elson]

On 03/30/2017 10:07 PM, David Bridgham via cctalk wrote:
>> It's not clear C-coupling is what's going on here (the wave shape looks pretty sharp for what I understand of the circuit/layout).
>> Notably though, C-coupling would remove any DC bias, but David's screen shot indicates a DC bias on the line.
>>
>> Is this line currently connected to the FPGA, or is it just the wire and R?
>> Perhaps the bias is coming from the FPGA, with C-coupling of the wave via the wire.
>> Or perhaps it's all crosstalk from within the FPGA, 'visible' because of the high load R.
> Yes, the wire is connected to the FPGA at one end.  That FPGA I/O pin is
> *supposed* to be configured for high-Z but that's the only place I can
> see the DC bias coming from.
Don't trust ANYTHING!  Recent Xilinx FPGAs have permanent 
"weak keepers" on all pins, they can not be turned off.
What this is is a non-inverting receiver on the pad, that is 
driving back to the pad with about a 50K Ohm resistor.
Plays hob with analog stuff like crystal oscillators.  The 
weak keeper would PERFECTLY explain your square wave!
When it gets a narrow pulse to high, it holds the line 
high.  When it gets a narrow pulse to low, it will switch to 
holding the line low.  So, if you are using a Xilinx FPGA of 
recent vintage, or some of their CPLDs, they will do exactly 
this.

Jon