6. The feedback network used in Colpitts oscillator is given below:
The value of \(C_2\) and feedback factor '\(\beta\)' for oscillating frequency of 1 MHz is:
The oscillating frequency \(f_0\) of a Colpitts oscillator is given by:
\(f_0 = \frac{1}{2\pi\sqrt{L_{eq}C_{eq}}}\)
where \(L_{eq}\) is the total inductance and \(C_{eq}\) is the equivalent capacitance.
In this case, \(L = 20 \text{ mH}\) and \(C_1 = 4 \text{ pF}\).
The equivalent capacitance for two capacitors in series is \(C_{eq} = \frac{C_1 C_2}{C_1 + C_2}\).
The given frequency is \(f_0 = 1 \text{ MHz}\).
Substituting the values:
\(1 \times 10^6 = \frac{1}{2\pi\sqrt{20 \times 10^{-3} \times \frac{4 \times 10^{-12} \times C_2}{4 \times 10^{-12} + C_2}}}\)
Solving for \(C_2\):
Squaring both sides and rearranging, we get:
\((2\pi f_0)^2 L = \frac{C_1 C_2}{C_1 + C_2}\)
\(\frac{C_1 C_2}{C_1 + C_2} = \frac{1}{(2\pi \times 10^6)^2 \times 20 \times 10^{-3}}\)
Let's calculate the right-hand side value first:
\(\frac{1}{(2\pi \times 10^6)^2 \times 20 \times 10^{-3}} \approx \frac{1}{(39.48 \times 10^{12}) \times 20 \times 10^{-3}} \approx \frac{1}{789.6 \times 10^9} \approx 1.266 \times 10^{-12} \text{ F} = 1.266 \text{ pF}\)
So, \(\frac{4 \times C_2}{4 + C_2} = 1.266\) (where \(C_2\) is in pF)
\(4 C_2 = 1.266 (4 + C_2)\)
\(4 C_2 = 5.064 + 1.266 C_2\)
\(2.734 C_2 = 5.064\)
\(C_2 = \frac{5.064}{2.734} \approx 1.85 \text{ pF}\)
This is closest to 1.8 pF.
The feedback factor \(\beta\) is given by \(\beta = \frac{C_2}{C_1}\).
\(\beta = \frac{1.8 \text{ pF}}{4 \text{ pF}} = 0.45\)
Therefore, the correct option is 1.8 pF and 0.45.