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Is there any reason for choosing a quartz crystal with or without metal lid? Some datasheets (this one for instance: link) claim better EMI performance and other do not write at all about it. So is there a reason I should choose quartz crystals with metal lid over crystals in a wholly plastic package? Here is what I have found on that topic: link

I am aware, that the second link is about a quartz crystal oszillator but I have not found anything else.

Considering the answers so far, especially the second one, I understand that shielding of the quartz itself is not needed. But then I am wondering why the quartz's manufacturer states the following (see first link and there note 2):

Terminations #2, #4 and the metal lid are connected internally. End user may connect these pins to circuit ground for EMI suppression.

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  • \$\begingroup\$ The first link goes to a quartz crystal. The second link goes to a quartz oscillator. There is a big difference how they work. \$\endgroup\$ – Justme Jun 15 at 6:11
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A crystal like this works at quite low voltage/current/power levels, and they usually have sine wave signal at the terminals, so the protection exists more to protect the sensitive crystal circuitry from external interference.

So in a harsh environment with lots of EMI, it might make a difference, but in that case a crystal with metal shield is not enough by itself to solve the issue, as the PCB design is just as important.

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  • \$\begingroup\$ So the question is what harsh really is. Is 30 V/m already harsh? Do you have any experience? PCB design is done carefully anyway in order to guarantee EMI robustness, therefore a quartz crystal with a shielded metal lid should be used in order to optimize for good EMI performance of the whole product, I think. What do you think? I mean designing for best EMI/EMC performance is always an optimization since you really never know how much a certain measure will contribute to that performance.. Or am I totally wrong? \$\endgroup\$ – stowoda Jun 15 at 6:37
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The last link you provided is a link to an oscillator, whereas your question aims at quartz crystals.

An oscillator has very sharp output signal transitions and therefore it can be a good thing to shield it with a metal housing.

A crystal is often used as part of a colpitts oscillator with two capacitors and a driver circuit. Therefore the crystal itself does not form a circuit (It can be considered as a discrete element). Thus there is no need to shield crystals.

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  • \$\begingroup\$ Yes I am aware that they are talking about oscillators, but thats the only source I have found on that topic. So basically you say that the quartz itself will not emit significant EM because it is passive.. Thx. \$\endgroup\$ – stowoda Jun 15 at 6:30
  • \$\begingroup\$ Yes, a quartz itself will not emit significant EM and is (as a single component) not susceptible to EMI. But the circuit including the quartz is susceptible to EMI. \$\endgroup\$ – Stefan Wyss Jun 15 at 7:33
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If you want a low_jitter (low phase noise) system, you may want to use a metal_can crystal.

Assume your quarts is 1cm by 1cm plate (about right) and there is an interferer at distance 1cm, with 5volts in 5 nanoseconds edge rate.

How much timing upset will occur? Let use compute those zero_crossing shifts. We need to know the SlewRate of the signal we will perturb. Assume the signal is 10MHz and +-1 volt from the crystal.

We'll assume 22pF Cload on output of the crystal.

We compute the induced current into the crystal, and assume all that current flows out thru the 22pF capacitor.

The coupling is thru opposite sides of a 1cm cube, with air as dielectric.

C = Eo * Er * Area/Distance

C = 9e-12 farad/meter * (1cm * 1cm)/1cm

C = 9e-12 farad/meter * 1cm = 9e-14 ~~ 1e-13 farad.

The induced current, using I = C * dV/dT is

I = 1e-13 farad * 5 volts/5nanoseconds = 1e-13 * 1v/nanosecond

I = 1e-14 amps during that 5nanosecond edge.

What is the voltage variation across the 22 pF Cload?

Q = C * V = I * T

V = I * T / C = 1e-4amps * 5nanoSec / 22pF

V = 1e-1 milliAmps * 5 nanoSec / 22pF

and the milli_nano is canceled by the pF

V = 1e-1 * 5 / 22 = 0.5 / 22 = 1/44 ~~ 0.022 volts

Now for the timing jitter.

We know Tjitter = Vnoise / SlewRate

Slewrate of 10MHz sin with 1 volt amplitude is 2 * PI * 10Million volt/set

SlewRate = 63,00,000 volt second

Tjitter = 0.022 / 63 Million = 1 / (44 * 63,000,000) = 1/30Billion

Tjitter ~~ 33 picoSeconds.

Should you care? Its not random.

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