## 1. A frequency determining circuit in an L-C sinewave oscillator consist of a high Q inductance L = 200μH in parallel with a capacitor of C = 450pF. Determine the frequency range of the oscillator if this capacitor, can be varied between 20pF-150pF.

Solution

Inductance, L = 200μH

= 200 × 10–⁶H

Capacitance, C = 450pF

= 450 × 10–¹²F.

LC = Inductance × Capacitance

200 × 10–⁶H × 450 × 10–¹²F = 9 × 10–¹⁴

Find the root of LC, √(LC) = 3 × 10–⁷

Therefore fo = 1/[2π√(LC)] = 1/[2π × 3 × 10–⁷]= 1/[2 × 3.142 × 3 × 10–⁷]= 1/0.0000018852

= 530447.7Hz

Approximately, = 530.4KHz

**Continuation When the variable capacitor C1 is connected in parallel with the circuit, the total capacitance in parallel with the inductor is increased to C+ C1, thus the corresponding oscillator frequency changes to**

_{o}= 1/[2π√(L(C + C1)] Hz

**Using C2 = 150pF, applying the same method as shown**above

f

_{o}= 1/[2π√(L(C + C2)] Hz

### 2. The resonant circuit of a tuned-collector oscillator has a resonant frequency of 5MHz if the value of the capacitance is increased by 50%. Calculate the new resonant frequency.

SOLUTION

f_{o}= 1/[2π√(LC)] Hz ………………………….. (1)

5 × 10⁶ = fo = 1/2π√LC ………………………. (2)

fo = 1/2π√L×1.5C …………………………….. (3)

Divide equation (3) by equation (2)

f_{o}/5×10⁶, = 1/√1.5 or f_{o}= 4.08MHz

3. A certain X-cut Quartz crystal resonate at 450KHz. It has an equivalent inductance of 4.2H and an equivalent capacitance of 0.0297pF. If its equivalent resistance to 600ohms, calculate Q factor.

Solution

Q = wL/R =( wavelength × Length, L)/resistance

Wavelength = 2πF

Q = 2πFL/R = (2×3.142×450,000×4.2)/600

11,876,760/600 = 19,794.6

4. A tuned-collector oscillator has fixed inductance of 100μH and has to be tuneable over the frequency band of 500KHz to 1500KHz. Find the range of the variable capacitor to be used.

Solution

Parameters given, inductance,L = 100μF ;= 100×10–⁶F

Frequency band = 500KHz – 1500KHz

Resonant frequency is given by;

f_{o}=(oscillator frequency) = 1÷2π√(LC) or 1/2π√(LC)

C = 1 ÷ 4π2f02L or 1/4π2f02L

Where L and C refer to the tuned circuit, therefore when f_{o}= 500KHz

C = 1/4π2 × (500 × 10³)² × 100 × 10–⁶

= 1015pF

f_{o}= 500KHz

C = 1/4π2 × (1500 × 10³)² × 100 × 10–⁶

= 113pF

The range of the variable capacitor to be used is = 113 – 1015pF

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