Q2Analog Electronics

Question

(B) A Colpitt's oscillator is designed with C1= 110 Pf and C2=7500 Pf. The inductance is variable. Determine the range of inductance values, if the frequency of oscillation is to vary between 750 kHz and 2050 kHz.

Answer 1

For a Colpitts oscillator, the equivalent capacitance of C1 and C2 in series is:

C_{e q} = \frac{C _{1} C _{2}}{C _{1} + C _{2}} = \frac{110 \times 7500}{110 + 7500} = \frac{825000}{7610} \approx 108.4 pF

The oscillation frequency formula: $f_{0} = \frac{1}{2 π L C _{e q}}$ . Rearranging for L:

L = \frac{1}{( 2 π f _{0} ) ^{2} C _{e q}}

L_{min} = \frac{1}{( 2 π \times 2050 \times 1 0 ^{3} ) ^{2} \times 108.4 \times 1 0 ^{- 12}} \approx 55.5 μ H

L_{ma x} = \frac{1}{( 2 π \times 750 \times 1 0 ^{3} ) ^{2} \times 108.4 \times 1 0 ^{- 12}} \approx 414 μ H

Therefore, the inductance must be variable over the range 55.5 Î¼H to 414 Î¼H to cover the frequency range of 750 kHz to 2050 kHz.

Q2Analog Electronics

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