What can cause an oscillator output voltage to go higher than expected?

Question

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.

Answer 1

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.

Answer 2

Your scope has insufficient bandwidth for such measurements. The square or almost square wave (that your CMOS oscillator makes) has amplitude of the first harmonic higher than the amplitude of square signal. Your scope filters out the next (third) harmonic and all higher ones. So you see more volts than you expect.

Guess your circuit works fine, no need to change it :)

Answer 3

If the connecting wires are long, when pulse risetime are fast, more than few inches, they behave as transmission lines and could reflect back some energy increasing, due to resonance effect, the voltage at the transmitting end. This could also damage the line driver device, so it is important to terminate with the right impedance the wires, generally around 50 to 75 ohm are good. Obviously the reflected waveform is generally sinusoidal near the resonant frequency of the line.