Question 3.11: (a) A design for the GB oscillator in Figure 3.27(b) at fo =......

(a) For this oscillator the feedback factor is

\beta=\frac{{ C}_{1}}{{ C}_{1}+{ C}_{2}}=\frac{1.142  \times  10^{-9}}{1.142  \times  10^{-9}  +  80  \times  10^{-9}}=0.0141

and

n={\frac{C_{1}  +  C_{2}}{C_{1}}}={\frac{1}{0.0141}}=71

The total capacitance {{C}}_{T} is

C_{T}={\frac{C_{1}C_{2}}{C_{1}  +  C_{2}}}=1.126\ \mathrm{nF}

and the resonant frequency is

f_{o}=\frac{1}{2\pi}\,\sqrt{\frac{1}{L C_{T}}}=\frac{1}{2\,\pi}\,\sqrt{\frac{1}{10  \times  10^{-6}(1  \times  10^{-9})}}=1.5\,\,\mathrm{MHz}

The Q point value of I_{C} is

I_{C Q}\approx I_{E Q}=\frac{10-0.7}{22  \times  10^{3}}=0.423\ \mathrm{mA}

and the small-signal {{g}}_{m} is

g_{m}={\frac{I_{C Q}}{V_{T}}}={\frac{0.423  \times  10^{-3}}{26  \times  10^{-3}}}=16.27\ \mathrm{mS}

From (3.82) the amplitude V_{1} stabilizes when

G_{m}(x)={\frac{n}{R_{T}}}                                    (3.82)

{\frac{G_{m}(x)}{g_{m}}}={\frac{n}{g_{m}R_{T}}}={\frac{71}{(16.27  \times  10^{-3})\,(8  \times  10^{3})}}=0.545

and it follows from Figure 3.43 that x = 3. Therefore,

V_{1}=x\,V_{T}=3(26\times10^{-3})=78\ \mathrm{mV}

V_{o}=n V_{1}=71(78\times10^{-3})=5.54V

and the output voltage is

v_{o}=5.54\ \mathrm{cos}\ \omega_{o}t                          (3.84)

It is of interest to observe that for this example the value of V_{1} (i.e., the dc value across {{R}}_{E}) is

V_{\lambda}=10-0.7=9.3\lor

and therefore, (3.79) is satisfied.

V_{\lambda}\gg\frac{\ln\,\,I_{o}\left(x\right)}{V_{T}}                              (3.79)

The simulation of this oscillator is shown in Figure 3.47(a). The Q point is at V_{C E} = 10.6V and I_{C} = 0.426 mA. The output waveform, shown in Figure 3.47(b), oscillates between ±5.4V at a fundamental frequency of 1.498 MHz in fairly good agreement with (3.84). The base-to-emitter voltage and collector current waveforms are shown in Figure 3.47(c). The peak value of v_{b e} shows that V_{\mathrm{1}} ≈ 78 mV. From the discussion in Chapter 2, the pulse width and amplitude of the i_{C} waveform when the BJT conducts affects significantly the phase noise of the oscillator. The SSB phase noise for this oscillator is shown in Figure 3.47(d).

(b) If {{R}}_{L} = 20 kV, then

{\frac{G_{m}(x)}{g_{m}}}={\frac{n}{g_{m}R_{T}}}={\frac{71}{(16.27  \times  10^{-3})(20  \times  10^{3})}}=0.218

In this case, from Figure 3.43, the value of x is 8.8, and it follows that

V_{1}=x\,V_{T}=8.8(26\times10^{-3})=228~\mathrm{mV}

and

V_{o}=n V_{1}=71(0.228)=16.1{\mathrm{V}}

But  V_{o} cannot be 16.1V. In this case the BJT saturates and the amplitude {{V}}_{o} is limited to approximately V_{C C} = 10V. With {{V}}_{o} = 10V, it follows that{{V}}_{1} = 141 mV and x = 5.42. This value of x implies that the large-signal value of G_{m}(x) is

{\frac{G_{m}(x)}{g_{m}}}=0.33

or

G_{m}(x)=16.27\times10^{-3}(0.33)=5.37~\mathrm{mS}

The reason that G_{m} settles at 5.37 mS is due to the saturation of the BJT. The corresponding {{v}}_{o} is

v_{o}(t)=10\ \cos\ \omega_{o}t                                (3.85)

The simulation of the oscillator with R_{T}=20\mathrm{~k\Omega} is shown in Figure 3.48(a). The output waveform is limited by saturation and in agreement with (3.85). The

base-to-emitter voltage and the collector current waveforms are shown in Figure 3.48(b), and the SSB phase noise is shown in Figure 3.48(c).