Understanding ultra-low phase noise oscillators
Keywords:electrical noise? oscillator? jitter? Gaussian? Gaussian?
However, an engineer can easily over-specify the oscillator, and hence the key is to understand exactly how the oscillator phase noise (or jitter) limits the system performance. To help with this understanding, a tutorial on phase noise and jitter is in order.
Phase noise and jitter in oscillators: A tutorial
In an oscillator, phase noise is the rapid random fluctuations in the phase component of the output signal. The equation of this signal is:
Where:
A_{0} = nominal peak voltage
f_{0} = nominal fundamental frequency
t = time
?(t) = random deviation of phase from nominal C "phase noise"
Above, ?(t) is the phase noise, but A_{0} will establish the signal-to-noise ratio. Figure 1 illustrates this.
Figure 1: The signal to noise ratio is a function of A_{0}. |
The noise floor
Noise signals are stochastic and, in a broad sense, noise can be characterized as any undesired signal that interferes with the main signal to be processed or generated. It can disturb any physical parameter such as voltage, current, phase, frequency (or time), etc. Therefore the idea is to maximise the signal and minimise the noise for a high signal-to-noise(S/N) ratio.
Noise power is quantified as
Where
K is the Boltzmann's contant = 1,38 x 10^{-23} (J/K)
T is the absolute temperature inK
And f= B is the bandwidth in which the measurement is made, in hertz
In the absence of any signal, there is thermal noise floor. This floor level can be specified in a variety of units: Watts, V^{2}/Hz, V/ Hz , dBm/Hz to name a few. For oscillators, it is convenient to use dBm/Hz to define noise density.
Before defining dBm/Hz we need to first define dBm. dBm refers to decibels above 1 milliwatt in a 50? system and is given by
Thus from above equation, 1 milliWatt is equal to 0 dBm.
Equation 2 gives us the magnitude of thermal noise and substituting for K and T we get:
Where B is the bandwidth of interest, for which we will use 1Hz to normalise the result. Using the equation of dBm (Equation 3) , and using the result from above we have:
Setting the bandwidth B to 1Hz will give us the final result in dBm/Hz, and since the log(1) is zero, we have:
Related Articles | Editor's Choice |
Visit Asia Webinars to learn about the latest in technology and get practical design tips.