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Simulators >> RF Simulators >> Thermal noise floor in oscillators
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Message started by Lieutenant Columbo on Oct 1^st, 2002, 9:49pm

Title: Thermal noise floor in oscillators
Post by Lieutenant Columbo on Oct 1^st, 2002, 9:49pm

I am using SpectreRF to find the phase noise of an oscillator. I see the 1/f³ and 1/f² regions, but I don't see a white noise floor as I would expect. In fact, I don't know that I have ever seen a white noise floor when simulating oscillators with SpectreRF.
Why is that?

Title: Re: Thermal noise floor in oscillators
Post by Ken Kundert on Oct 1^st, 2002, 10:39pm

In my experience one usually does not see the thermal noise floor because the noise analysis is setup to compute the noise at the resonator. In other words the output of the oscillator is taken at the resonator, and there is no buffer amplifier present. In this case the there is no thermal noise floor because it is shorted out by the resonator. If you add a buffer stage, you will see the thermal noise floor. Designers are use to always seeing the thermal noise floor because high-frequency spectrum analyzers have 50 Ohm inputs, and so there must be a buffer between the resonator and the analyzer.

-Ken

Title: Re: Thermal noise floor in oscillators
Post by Lieutenant Columbo on Oct 1^st, 2002, 10:43pm

But won't the oscillator have a resistive component in its output impedance? Wouldn't the thermal noise from the resistive component create a noise floor?

Title: Re: Thermal noise floor in oscillators
Post by Ken Kundert on Oct 1^st, 2002, 10:48pm

Consider a differential LC oscillator. In this case, the output is shunted by a parallel LC resonator. Often the inductor is connected to an ideal voltage source and the capacitor connected to ground. Thus, low frequencies are shorted by the inductor and high frequencies are shorted by the capacitor. If you added a resistance in series with the capacitance to model its finite Q, then you might see a white noise floor, but it will likely be at a very low level, perhaps too low to see. Notice that if you add the resistance in series with the inductance, you would see a noise floor at low frequencies, not at high frequencies.

-Ken