Single-tuned oscillator: what is the max peak value of Vout?

Single-tuned oscillator

Hi, I was hoping someone could help me understand this problem.

Q: What is the max peak value of Vout in the following picture? and why?

** I believe it's supposed to be 2Vdd, but I don't know why.

Re: Single-tuned oscillator

Yes, that is it.
If you look at the graph, imagine the waveform amplified until the bottom reaches the ground potential. Obviously you cannot drive it lower than that, since the MOSFET is connected to ground.
So the bottom half of the curve can have a maximum amplitude equal to Vdd. Since the waveform si symmetical, the top half also has an amplitude of Vdd, so the peak si 2*Vdd.

As to why this is happening: the LC tank, if not heavily damped, will try to oscillate, that is what an LC tank does, until it dissipates its energy, if no energy is supplied (in this case the transistor supplies more energy, so the oscilation is maintained). When an LC tank produces oscillations, the voltage at one end can become positive or negative with respect to the otherr end. In your case, one end is connected to Vdd, so the other end can swing about Vdd, both in the positive and in the negative direction. When it swings in the positive direction the two voltages add up, resulting in a peak, while the negative swing produces a trough.
As we saw above, the maximum amplitude of either swing can only be Vdd, therefore the peak will be Vdd+Vdd=2*Vdd.