Question 8.2: As a continuation of Example 8.1, predict and simulate the p...
As a continuation of Example 8.1, predict and simulate the phase shift for an injection current of 100 \muA. Also compare this to the phase shift when i_{inj} and i are equal.
With an oscillating voltage of 2.1V, as determined in Example 8.1, total current is found by multiplying by g_{m} the feedback transconductance, found in Example 8.1 to be 3.346 mA/V. Thus, total current is 7.03 mA and phase shift can be predicted to be
\phi _{inj}= \sin ^{-1} \left(\frac{i_{total} }{i_{inj}} \sin \phi _{osc} \right) = \sin ^{-1}\left(\frac{7.03mA}{0.1mA} \sin \left\{-\tan ^{-1}\left[\left(\omega C- \frac{1}{\omega L} \right)R \right] \right\} \right)This is compared to simulations with results in Figure 8.21. As discussed above, R is set equal to R_{p} (314\Omega ) with the result that there is near perfect agreement between simulations and theory. As can be seen, at 1.0003 GHz, theoretical and simulated phase is 0°. Note that 1.0003 GHz is the free-running frequency and is also equal to the resonant frequency of the parallel tank circuit. As seen in the figure, when phase shift reaches approximately 90°, the oscillator is no longer locked. For comparison, the curves for 50 \mu A and 200 \mu A are also shown, verifying that, with more current, the phase shift is lower.
The phase shift can also be explored as a function of the magnitude of the injected signal with the input frequency at a constant 1.002 GHz. The results are shown in Figure 8.22. This figure shows that with larger injected amplitude, phase shift is decreased. It was observed from both theory and simulation that the minimum current for injection locking at this offset frequency is about 240 \mu A.
To verify that the system is not locked for injected currents less than 240 \mu A, 235 \mu A is injected, and the resulting phase plotted as a function of time is shown in Figure 8.23. This shows the obvious cycle slipping, verifying that the oscillator is not locked.
