Size of inductors in passive filters

I was reading about why we need active filters even when we have passive filters with us. I got to know that one of the main reasons for doing so, is that at lower frequencies the size of the inductor is large and it becomes bulky so it can not be fabricated on IC.

So can anyone please tell me how the size of inductor is related to the frequency of operation.

• Just as an extra note - for large 'inductance' at low (audio) frequencies you could take at look at 'gyrator' circuits which simulate the inductance using opamps and a few passive components. eg. sound.westhost.com/articles/gyrator-filters.htm – JIm Dearden Mar 29 '15 at 12:32

For a typical passive 2nd order filter the resonant frequency is given by this formula: -

f = $\dfrac{1}{2\pi\sqrt{LC}}$

So immediately you can see that for f to be small the product of L and C has to be large. Next is the Q factor. For an inductor Q factor is: -

Q = $\dfrac{2\pi f L}{R}$ where R is the resistance of the coil.

This tells us that high Q circuits require a decent ratio of L to R. Now also consider the effect on Q when the "lossy" inductor is combined with a capacitor to make a series resonant circuit: -

Q = $\dfrac{1}{R}\sqrt{\dfrac{L}{C}}$

For a high Q circuit, we want L to be significantly larger than C so, going back to the first formula for resonant frequency, we actually want C to be small and L to be big to get a decent resonant peak whilst maintaining the product of L and C at a value to get the desired resonant frequency.

Therefore lower frequencies means a lot of emphasis on the inductor if we want a filter with a sharp resonance. Note - If the desired resonant frequency reduced by 10 the product of L and C has to increase by 100.