# If I parallel two crystal filters together, will that double the bandwidth? [closed]

Will the bandwidth be doubled ? And will the impedance be halved ?

• no. But that single word might not be the answer you're looking for. Can you explain why you think that would be the case? Why do you care? – Marcus Müller Aug 28 '19 at 14:00

As pointed out, the short answer is no. The Q value for the series resonance of a crystal can be calculated from $$Q = \frac{2\pi f_s L_m}{r_m}$$. Q value is of course related to the bandwidth: $$Q = \frac{f_s}{BW}$$. If you take the equivalent circuit for a crystal, add another similar next to it, and then simplify the circuit as far as you can, you can see that both $$L_m$$ and $$r_s$$ are halved and the $$Q$$ is not affected.

The answer about bandwidth is "not doubled, but not the same".

As demonstrated in a previous answer, Q of two identical crystals in parallel remains the same.

But BW of the filter is determined not only by crystals' Q (unloaded Q) but also by generator and load impedances.

If those are << than Rm (xtal series R at series resonance) then BW with 2 xtals is the same.

But if >> then it tends to 2x because they are lowering the "loaded Q" of the filter.

What are your design needs?