Question

The output frequency of an LC tank oscillator circuit employing a capacitive sensor acting as a capacitor of the tank is 100 kHz. If the sensor capacitance increases by 10% the output frequency becomes ________ kHz.

Answer 1

The frequency of oscillation is equal to the resonance frequency of the tank circuit is:

\({f_1} = \frac{1}{{2\pi \sqrt {L{C_1}} }}\)   ---(1)

The capacitance is increased by 10%, i.e.

⇒ C₂ = C₁ + 0.1 C₁

C₂ = 1.1 C₁

New resonant frequency will be:

\({f_2} = \frac{1}{{2\pi \sqrt {L{C_2}} }}\)

\(f_2= \frac{1}{{2\pi \sqrt {1.1L{C_1}} }}\)  ---(2)

Dividing Equation (2) with (1), we get:

\(\frac{f_2}{f_1}=\frac{1}{\sqrt{1.1}} \)

With f₁ = 100 kHz, f₂ is:

f₂ = 95.346 KHz