I am wondering about a signal integrity issue with a 100MHz clk generated by an FPGA.
This is the set up:
- FPGA generates a 100MHZ square signal with 50% of duty cycle.
- The IO constraint is LVCMOS 3.3V.
- The PCB track has a length of 20cm and it finished in a Sub-D25 connector. The track has a series termination resistor (Rt).
In the next pic you can see the problem:
It does not matter the value of the Rt, the CLK signal always looks like that. So it has a 1.5V offset and it has an amplitude around 800mV.
What do you think it is happening? I should set up a special constraint in the FPGA side to get this clock like a square signal with an amplitude of 3.3V?
That is why one creates impedance controlled traces.
Without knowing the parameters for your trace impedance it's hard to help. (Trace width, GND plane, PCB material, thickness...)
You say there is a series termination resistor. But no value. Does it match the trace impedance?
The DSUB most probably is not the "end" of the signal. Thus it makes not much sense to measure the signal there.
Install the circuit at the other side of the DSUB, with proper termination. Then do the measurement again.
Be careful ... that your scope probe does not harm the signal.
There may be other issues. Like not suitable scope probe (impedance, bandwidth...) or wrong scope setup (bandwidth limitation)...
Beside all other possible problems, you need an oscilloscope with 500 MHz, better 1 GHz bandwidth and a suitable probe (active or resistive) to reproduce a 100 MHz square wave.
Hi! Thanks for your answer. There is a 0hm series termination resistor, and the Trace is routed with Zo=50ohm. I tested a 47ohm series termination and the waveform almost doesn't change.
On the other hand, I have also connected the device to the SubD-25 and the aspect of the waveform doesn't change either...
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The oscilloscope should not be the problem because its characteristics are 100MHz and 1Gs/s
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I have also tried to measure a 120MHz square signal in other track and I was able to see a perfect square signal with the same oscilloscope...
I am a bit lost right and I don't know what else can I try...
Ok. Maybe it was with other model of oscilloscope. But I checked a 80MHz signal with this one I have observed the same attenuation and the same Offset as the 100MHz signal... what do you think about it?
80MHz square wave is built from a lot of sinewaves (overtones, read about fourier series):
240MHz, 400MHz, 560MHz, 720MHz, ...
100MHz square wave is built from a lot of sinewaves (overtones):
300MHz, 500MHz, 700MHz, 900MHz, ...
You see, even the first overtone is at 240MHz or 300MHz respectively.....
For a square wave to "almost look like a squarewave" you need at least the 10-fold bandwidth than it's fundamental frequency.
totally understood. I need a faster OC to see a perfect square signal. But what about the offser? Why is it having a 1.8V offset? It is also because the OC is very slow to sample that signal?
OK, then do another measurement:
* use the high speed scope
* use the high speed probe: select 1:10
* connect probe_GND next to the FPGA
* connect probe to the FPGA-Pin with the 100MHz output ... (as close as possible to the FPGA, best directly at the FPGA pin)
Take a photo oft the scope picture including scope setup.
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May I ask why we don´t see exactly 100MHz? it´s just about 96MHz.