Question 8.1: Injection Locking Demonstrate injection locking with a simpl...
Injection Locking
Demonstrate injection locking with a simple model of a transconductor.
The injection-locking circuit is as shown in Figure 8.16. We start by choosing an inductance and capacitance of 5 nH and 5.06 pF to result in an oscillating frequency close to 1 GHz. The calculated frequency using (8.20) is predicted to be 1.0006 GHz. Note that there is a 600-kHz offset from 1 GHz. With a Q of 10, the inductor has a parallel resistance of about 314Ω. Noise injection due to the resistor is
\omega _{\omicron } =\frac{1} {\sqrt{LC} } (8.20)
i_{n} =\sqrt{\frac{4kT}{R} } = \sqrt{\frac{4\times 1.38\times 10^{-23} J/K \times 300K} {314\Omega } } =7.26 PA/\sqrt{Hz}
To represent noise inputs, 40 current sources are placed in parallel, separated by 100 kHz from 0.98 GHz to 1.02 GHz, and each has a current amplitude of 2 nA, representing the noise from a 314Ω resistor in a 100-kHz bandwidth. The transconductor is modeled with a voltage-controlled current source with transfer function:
i_{\omicron } =0.005 v_{i} -0.0005 v^{3}_{i}The third-order term is sufficient to produce g_{m} compression and amplitude limitation. The simulated free-running frequency is 1.0003 GHz, as shown in Figure 8.18. Note that the time step is 5 ps, and total simulation time is 40 ms for a resolution of 25 kHz. From (8.28), the amplitude is predicted to be
V=\sqrt{\frac{4}{3k_{3} } \left(\frac{1}{R_{p} }-k_{1} \right) } (8.28)
V =\sqrt{\frac{4} {3k_{3} } \left(\frac{1}{R_{p} } -k_{1} \right) } =\sqrt{\frac{4} {3 \cdot \left(-0.0005\right) }\left(\frac{1}{314} -0.005 \right) } =2.20 VTime domain plots show an amplitude of 2.1V, in agreement with the prediction and with the spectrum shown in Figure 8.18. Note that the frequency is not all in one frequency bin; thus, the total power must be taken to derive the amplitude from the spectrum. In this case, two bins are dominant, with about 1.4V each, for a total of just over 2V, in agreement with the time domain simulation. Thus, the effective g_{m} can be determined from (8.27):
g_{m} =k_{1} +\frac{3k_{3} V^{2} } {4}=0.005+\frac{3\cdot \left(-0.0005\right) \cdot 2.1^{2} } {4}
=0.005-0.001654
=0.003346
The effective value of g_{m} is 3.346 mA/V, which is down from the small-signal value of 5 mA/V. With these values, for an input signal injected at 1 GHz, assuming this is outside of the oscillator bandwidth, (8.36) predicts that the required injection current i_{inj}for locking is
v_{out} =\frac{i_{inj} }{j2C\Delta \omega } (8.36)
i_{inj} \gt v_{out} \cdot 2C\Delta \omega =2.2\times 2\times 5.06p \times 2\pi \times 300k =41.95 \mu AThis can also be determined from (8.31), but to use this equation, R and B must be known. Equivalent closed-loop parallel resistance R can be determined from (8.24) as
i_{inj} \gt \frac{\overline{i_{n} \ } \left(f-f_{0 } \right) \sqrt{B} }{ 2 \left(f_{c} -f_{0 }\right) } =\frac{\overline{i_{n} } \left(f-f_{0} \right) \sqrt{B} } {2\frac{B}{2} \frac{1}{2\pi } } =\frac{\overline{i_{n} } \cdot 2\pi \left(f-f_{0} \right) }{\sqrt{B} } =\frac{v_{\omicron } }{R} \frac{\left(f-f_{\omicron } \right) }{\left( f_{c} -f_{\omicron } \right) } (8.31)
\sqrt{v^{2}_{\omicron } } =\sqrt{\int\limits_{\infty }^{\infty } {\left|v_{out} \right| ^{2} } d\omega } =\sqrt{\int\limits_{\infty }^{\infty }{\frac{i^{2}_{n} R^{2} }{1+\left(\frac{\omega }{\omega _{c} } \right) ^{2} } } d\omega } =\omega _{c} i_{n} R \sqrt{\left[\tan ^{-1} \left(\frac{\omega }{\omega _{c} } \right) \right] ^{\infty }_{-\infty } } =\overline{i_{n} } R \sqrt{\frac{B}{2\pi } \cdot \frac{\pi }{2} } =\frac{\overline{i_{n} } R \sqrt{B} }{2} =\frac{\overline{i_{n} } }{2} \sqrt{\frac{R}{C} } (8.24)
R= \frac{4v^{2}_{\omicron } C }{\overline{i^{2}_{n}} } =\frac{4\times 2.1^{2} \times 5.06p }{\left(\frac{2n}{\sqrt{100k} } \right) ^{2} } =2.2315\times 10^{12} \Omega
Bandwidth is determined to be
B=\frac{1} {RC} =\frac{1} { 2.2315\times 10^{12} \Omega \times 5.06 p } =0.88565 rad/s or 0.014096 HzThen, the required current for injection can be found from (8.31):
i_{inj} \gt \frac{\overline{i_{n} \ } \left(f-f_{0 } \right) \sqrt{B} }{ 2 \left(f_{c} -f_{0 }\right) } =\frac{\overline{i_{n} } \left(f-f_{0} \right) \sqrt{B} } {2\frac{B}{2} \frac{1}{2\pi } } =\frac{\overline{i_{n} } \cdot 2\pi \left(f-f_{0} \right) }{\sqrt{B} } =\frac{v_{\omicron } }{R} \frac{\left(f-f_{\omicron } \right) }{\left( f_{c} -f_{\omicron } \right) } (8.31)
i_{inj} \gt \frac{v_{out} }{R} \frac{\left(f-f_{\omicron } \right) } {\left(f_{c} -f_{\omicron } \right) } =\frac{2.2\times 300k} {2.2315 \times 10^{12} \times 0.014096/2 } =41.96 \mu A
This is in agreement with the previous calculation. These calculations also demonstrate the extremely high resistance and narrow bandwidth of oscillator circuits. A series of injected tones is applied, with the results shown in Figure 8.19. The simulated current required for locking is 44 μA, just slightly higher than the
predicted current.
