Question 3.8: Design the GB oscillator in Figure 3.29(a) to oscillate at 1......

With I_{C}=1\;{\mathrm{mA}}, it follows that

b_{i b}={\frac{V_{T}}{I_{C}}}={\frac{25  \times  10^{-3}}{1  \times  10^{-3}}}=25\Omega

and

g_{m}={\frac{I_C}{V_{T}}}={\frac{1}{b_{i b}}}={\frac{1}{25}}=40\ \operatorname{mS}

From (3.65)

C_{2}\geq{\frac{10}{\omega b_{i b}}}                                  (3.65)

C_{2}\geq{\frac{10}{\omega b_{i b}}}={\frac{10}{2\pi  \times  10^{7}\ (25)}}=6.36\ \mathrm{nF}

and from (3.64) letting the loop gain be 3, we obtain

\beta(j\omega_{o})A_{v}(j\omega_{o})={\frac{C_{1}  +  C_{2}}{C_{1}}}                                          (3.64)

C_{1}={\cfrac{C_{2}}{2}}={\cfrac{6.36\times10^{-9}}{2}}=3.18~{\mathrm{nF}}

Then,

C_{T}=C_{1}\parallel C_{2}=2.12\;\mathrm{nF}

The value of L is

L={\frac{1}{\omega_{o}^{2}\,C_{T}}}={\frac{1}{(2\,\pi  \times  10^{7})^{2}(2.12  \times  10^{-9})}}=0.12~\mu\mathrm{H}

and

R_{P}^{\prime}\approx b_{i b}{\bigg(}{\frac{C_{1}  +  C_{2}}{C_{1}}}{\bigg)}^{2}=25(3)^{2}=225\Omega

Since R_{P}^{\prime}=225\Omega, any R_{L{}} such that R_{L}\gg225\Omega can be used.

The simulation results are shown in Figure 3.29(b) where the fundamental frequency of oscillation is 9.977 MHz. The actual Q point values are: V_{C E}=7.5\mathrm{V} and I_{C}=1.07~{\mathrm{mA}}.