Question 19.11: Consider the common source amplifier in Fig. 19.14(a). Trans......

The small-signal equivalent circuit is shown in Fig. 19.14(b). The voltage noise source v_{n1} models the flicker noise of the transistor Q_{1} and has a noise spectral density,

V^{2}_{n1}(f) = \frac{K}{WLC_{ox}f}

The current noise source i_{n} includes the thermal noise of both the transistor and resistor,

i^{2}_{n}(f) = 4kT\left\lgroup\frac{2}{3} \right\rgroup g_{m}+\frac{4kT}{R_{d}}

The resulting voltage at v_{0} takes on the form of (19.55)—a sinusoidal voltage in additive noise.

v _{t} = A\sin (\omega _{0}t) + n(t)      (19.55)

The sinusoidal component has amplitude \left|V_{s}\right| g_{m1}(R_{d}\left|\right|r_{o1} ) . The voltage noise spectral density at v_{0} is obtained by a routine noise circuit analysis of Fig. 19.14(b).

Assuming that the zero-crossings of the sinusoidal waveform at v_{0} are disturbed only a small amount by the noise, the approximation in (19.56)

τ _{k} \cong \frac{n_{k}}{A\omega _{0}}        (19.56)

may be applied. In terms of phase,

\phi _{k} \cong \frac{n_{k}}{\left|V_{s}\right| }

Where n_{k} is the sampled noise voltage v_{o,n} . Therefore, the phase noise is,

S_{\phi }(f) = \frac{(R_{d}\left|\right|r_{o~1})^{2}}{\left|V_{s}\right|^{2} } \left[\frac{K}{WLC_{ox}f} g^{2}_{m~1} +4kT \left\lgroup\frac{2}{3} \right\rgroup g_{m} + \frac{4kT}{R_{d}} \right]

Note that it has a 1/f component and a white component. ^{14}

14. Strictly speaking, \phi _{k} is a discrete-time random process; hence, there may be some aliasing of the white noise in v _{o,n} causing the phase noise to be larger than this. Naturally, the white noise is assumed to be eventually bandlimited by some capacitance a v _{o} ensuring it has finite power.