Electronic – What happens to RLC circuits at resonance frequency, conceptually


What occurs when $$\omega = \omega_r$$
The circuit frequency equals its natural frequency? So what. And why is it important?

Best Answer

A series RLC circuit has an impedance of: -

\$R + j\omega L +\dfrac{1}{j\omega C}\$

Knowing that "j" in the denominator is the same as "-j" in the numerator means there is a frequency when the inductive and capacitive reactances totally cancel and this means that the only component left is resistance, R.

This occurs when \$j\omega L = -\dfrac{1}{j\omega C}\$

Or \$\omega^2 = \dfrac{1}{LC}\$ or \$\omega = \dfrac{1}{\sqrt{LC}}\$

So what?

It's a really big deal when wanting to filter just one small band of frequencies whilst disregarding the rest; in simple terms, at resonance, a series RLC circuit lets through a small band of frequencies and progressively attenuates frequencies that are not in the "pass-band".

And why is it important?

It's a big deal for radio receivers and transmitters so it's really very, very important.

If you analyse a parallel RLC circuit it has the same formulas and, depending on the circuit configuration, it can do pretty much the same as a series RLC circuit. The "R" in a parallel circuit is a different way of expressing losses and it functions inversely to the "R" in a series RLC circuit.

Resonance can do bad things too - if not controlled it can produce voltages that cause semiconductor devices to breakdown - voltage regulators is a device that springs to mind.

Have you seen the video of that suspension bridge rocking backwards and forwards and then collapsing - that was uncontrolled resonance and mechanically it has virtually the same formula - it is a second order filter.