A circuit I am simulating shows off MOS switches as the biggest noise contributor in my ciruit.  One side of the cap is connected to the signal where I am measuring phase noise.  The other side of the cap is connected to a MOS switch that provides either a low impedance or a high impedance between the cap and a supply (switch cap in or out).

My phase noise simulations point to the MOS devices as the largest contributor of phase noise when the MOS switch is turned off.  I don't understand how an off device can create noise unless the junction diodes are the problem.

Should I debug my circuit or my simulation (SpectreRF)?

I have seen this problem twice before. In both cases, the circuit in question was a differential oscillator with band switching, and the "off" transistor was used to connect capacitors to the resonator in order to change the resonant frequency and switch the band.

In both cases I found that problem was due to a poor flicker noise model interacting with a largely unrecognized aspect of the circuit's behavior. The poor flicker noise model resulted from using flickmod=0. Using flickmod=1 gives results that are more in line with what we expect.

However, from my perspective, it is the circuit behavior that is the most interesting. Here is what is going on in a nutshell.

The switch in question is not off all the time. It can't be. If it were always off, then the voltage on the capacitor would be undefined (would float off to infinity). Rather the switch turns on briefly in each cycle in reverse subthreshold conduction mode. During this brief period of time, it produces a very small amount of noise.

So now the question becomes, how can this small amount of noise come to dominate the output noise of the oscillator. The answer is, there is a largely unrecognized effect that is called "parametric amplification" occurring that causes the noise to be amplified to a huge degree.

Parametric amplification is normally a very prized form of amplification because it involves no resistive components and so is noise free. Ironically, this noise free amplification is a parasitic effect that in this case is amplifying noise.

I have a write-up that describes this problem. I'll try to post it in the next few days.

-Ken