Question 3.7: Design the BJT Hartley oscillator in Figure 3.25(b) to oscil......
Design the BJT Hartley oscillator in Figure 3.25(b) to oscillate at 1 MHz using the 2N3904, a general purpose BJT. The Q point is to be at V_{C E}=6V and I_{C} = 3 mA. The 2N3904 lists f_{T}\gt 300 MHz and a typical beta of 100.
Let V_{C C}=12\mathrm{V}, then
R_{E}={\frac{12-6}{3 \times 10^{-3}}}=2{\mathrm{~k\Omega}}
Let
R_{T H}=\beta\frac{R_{E}}{10}=100\frac{2,000}{10}=20\mathrm{~k\Omega}
and it follows that
V_{T H}\approx6+0.7=6.7\mathrm{V}
R_{1}=R\,_{T H}\frac{V_{C C}}{V_{T H}}=\left(20\times10^{3}\right)\frac{12}{6.7}=20\mathrm{~k}\Omega
and
R_{2}=\frac{R_{T H}}{1-\displaystyle\frac{V_{T H}}{V_{C C}}}=\displaystyle\frac{(20 \times 10^{3})}{1-\displaystyle\frac{6.7}{20}}=45.3\ \mathrm{k\Omega}
Let C = 5 nF, then from (3.57)
\omega_{o}={\frac{1}{\sqrt{(L_{1} + L_{2})\,C}}} (3.57)
L_{1}+L_{2}={\frac{1}{\omega_{o}^{2}C}}={\frac{1}{\left(2\pi10^{6}\right)^{2}5 \times 10^{-9}}}=5.1\ \mu\mathrm{H}
Letting the value of the loop gain be 3, it follows from (3.58) that
\beta(\omega_{o})A_{v}(\omega_{o})=\frac{L_{2}}{L_{1} + L_{2}}g_{m}n^{2}r_{e}\approx\frac{L_{1} + L_{2}}{L_{2}} (3.58)
L_{2}={\cfrac{L_{1}+L_{2}}{3}}={\cfrac{5.1\times10^{-6}}{3}}=1.7~\mu\mathrm{H}
and
L_{1}=2L_{2}=3.4~\mu\mathrm{H}
The simulation of the oscillator is shown in Figure 3.26. The fundamental frequency of oscillation is 999.5 kHz.
