I am currently testing a 50MHz 3.3V CMOS oscillator in my circuit, which is connected to two pins/loads – MCU pin and Ethernet PHY pin. The oscillator is rated for 15pF output load, which I could be exceeding by a few pF perhaps. Things are working fine in my circuit and so is the Ethernet Communication.
Now, the oscillator Vcc is supplied by 3.3V. When I am observing the waveform at the output pin of the oscillator, I am getting a 4V high, 0.8V low 50MHz sine waveform. I am wondering what is causing the voltage to go to 4V when my supply vltage is 3.3V? I have a 10R resistor at the oscillator output which is provided for termination adjustment.
Note: I am testing the waveform using an oscilloscope of 100MHz bandwidth, because I don't have any other oscilloscope at hand. But I think that should actually result in lesser amplitude.
Diagram:
simulate this circuit – Schematic created using CircuitLab
Could anyone please help me in identifying the reasons for this higher output voltage?
Let me know if I missed providing any important information required for the analysis.
Thank you.
EDIT: The higher voltage is irrespective of Ethernet cable connection.
Best Answer
A pure square wave of amplitude 3.3 volts p-p when perfectly filtered to remove its harmonics will result in a sinewave whose amplitude is 27% higher at 4.2 volts p-p: -
Notice that the fundamental is higher amplitude than the perfect square wave. If you do the math you'll find that the fundamental is \$\dfrac{4}{\pi}\$ times higher that the square wave.
Somewhere between perfect filtering and the filtering provided by your o-scope you get what you get.