Electronic – How to calculate the value of the inductor and capacitor in a parallel LC circuit

We've actually thought about how to do this from a microcontroller to make a cheap L/C/R meter. It needs to be cheap and small because we plan to make a business card that has a useful circuit on it.

Anyway, the answer is probably different from doing this manually with a signal generator and scope or automatically with a microcontroller.

Manually, you can set up a L-R low pass filter. That means input signal to one side of L, R to ground and output other side of L. By feeding in a square wave and looking at the result on a scope you can measure the exponential time constant. The time constant is T = L/R. When L is in Henrys, R in Ohms, then T is in seconds. This will be the time a step gets to 1 - 1/e of its final value, or about 63%. It may be easier to measure the 1/2 decay time, which occurs in .693 time constants. From that you can find the time constant, and from the equation above the inductance by knowing the resistance.

The automated way we will probably use is to measure the magnitude of a known AC signal fed thru the same filter. Higher frequencies will be attenuated more. If you feed in a square wave instead of a single pure frequency (sine wave), then you'll have to do a little more math. But the inductance can be calculated if you know exactly what you stuck in, the value of the resistance, and the magnitude of the outcoming AC signal.

Inductors store energy in the magnetic field generated when current passes through them. When the supply is removed, the collapsing magnetic field induces a current flow in the same direction that it was traveling when it generated the magnetic field in the first place. This is why it is used as one of the storage devices in switching power supplies; the capacitor maintains the same voltage, and the inductor maintains the same current.

^{simulate this circuit – Schematic created using CircuitLab}

(But don't try to actually build this circuit.)