1
$\begingroup$

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

$\endgroup$
  • 2
    $\begingroup$ 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? $\endgroup$ – Marcus Müller Aug 28 '19 at 14:00
3
$\begingroup$

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.

enter image description here

| improve this answer | |
$\endgroup$
1
$\begingroup$

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?

| improve this answer | |
$\endgroup$

Not the answer you're looking for? Browse other questions tagged or ask your own question.