Question 3.3: Design the GG oscillator in Figure 3.12(a) to oscillate at 2......

Let the Q point be at I_{D} = 4.5 mA, and V_{D S} 5V. For this transistor, typical values are {}V_{P} = −3.5V and I_{D{SS}} S = 12 mA. Therefore, at I_{D} = 4.5 mA the value of V_{G S} is approximately −1.36V and \mathbf{g}_{m} = 4.2 mS. The Q point design requires that

R_{S}={\frac{-V_{G S}}{I_{D S}}}={\frac{1.36}{4.5  \times  10^{-3}}}=302\Omega

and

R_{D}={\frac{V_{D D}-V_{D S}-V_{S}}{I_{D S}}}={\frac{12-S-1.36}{4.5  \times  10^{-3}}}=1.25~\mathrm{kΩ}

For the inductor, select L = 1 μH having {{Q}}_U = 80 (or R_{{p}} = 10 kΩ). From (3.35)

\omega_{o}={\frac{1}{\sqrt{L C_{T}}}}                                  (3.35)

C_{T}={\frac{1}{\omega_{o}^{2}L}}=63.4{\mathrm{~pF}}

Letting the loop gain be 3, it follows from (3.36) and (3.38) that

C_{T}={\frac{C_{1}C_{2}}{C_{1}  +  C_{2}}}                                (3.36)

\beta(\omega_{o})A_{v}(\omega_{o})=\frac{{ C}_{1}  +  { C}_{2}}{{ C}_{1}}                                (3.38)

C_{2}=3C_{T}=3(63.4\times10^{-12})=190\ \mathrm{pF}

and

C_{1}={\cfrac{C_{2}}{2}}={\cfrac{190  \times  10^{-12}}{2}}=95~{\mathrm{pF}}

Since

n={\frac{C_{1}  +  C_{2}}{C_{1}}}=3

and with R_{S}\|(1/g_{m})\approx1/g_{m}, it follows that

R_{P}^{\prime}={\frac{n^{2}}{g_{m}}}=238(9)=2.14\ {\mathrm{k\Omega}}

Hence, the equivalent resistance of the tuned circuit is

R_{P}\parallel R_{L}\parallel R_{P}^{\prime}=10\times10^{3}\parallel10\times10^{3}\parallel2.14\times10^{3}=1.5\mathrm{~k\Omega}

hich shows that the approximation (3.37) is not exactly valid. However, making n = 3 should be sufficient to start the oscillation.

A_{\upsilon}(\omega_{o})\approx g_{m}n^{2}\biggl(R_{S}\left\|\frac{1}{g_{m}}\right)\ ,                              (3.37)

The simulation of this oscillator is shown in Figure 3.13. The Q point is at V_{G S} = −1.41V and I_{D} = 4.7 mA, and the fundamental frequency of oscillation is 19.95 MHz.