Topic: Analog Circuits
An LC tank circuit resonates at a frequency 'fr'. If the temperature coefficient of permittivity of the dielectric material used in the Capacitance is $6ppm/^{\circ}C$, then what is the temperature coefficient of frequency
- A. $-3~ppm/^{\circ}C$
- B. $-6~ppm/^{\circ}C$
- C. $3~ppm/^{\circ}C$
- D. $6ppm/^{\circ}C$
Answer: A. $-3~ppm/^{\circ}C$
Explanation: The resonant frequency of an LC tank is $f_r = \frac{1}{2\pi\sqrt{LC}}$. Since capacitance $C$ is directly proportional to permittivity $\epsilon$, a $6~ppm/^{\circ}C$ increase in permittivity causes a $6~ppm/^{\circ}C$ increase in $C$. The relationship between frequency and capacitance is $f_r \propto C^{-1/2}$. Taking the derivative, the fractional change is $\frac{\Delta f_r}{f_r} = -\frac{1}{2}\frac(\Delta C}{C} = -\frac{1}{2} (6~ppm/^{\circ}C) = -3~ppm/^{\circ}C$.