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I am trying to chose the parallel capacitors for the crystal on my micro-controller. The crystal is a 9.6MHz quartz crystal (chosen because 9.6MHz is good for UART communication). Going through the equations I get a value of 32.29pF. This seems higher than what I typically see on boards. This value is also higher than the 15pF recommended value in the Clock App note (image below). Can someone explain to me if I am doing the calculations correct?

enter image description here enter image description here

MCU = ATXMEGA128A1U
MCU Clock App Note = here

Crystal = 407F35E009M6000

From application documents for the micro controller I get these numbers and equation:

Ce = 2CL - Ci - Cs

where:

Ce - is the external capacitance needed

Ci - is the pin capacitance

CL - is the load capacitance specified by the crystal vendor

Cs - is the total stray capacitance

ATxmega128A1 IBIS file (typical values):

C_comp = 4.44 pF

C_pkg = 0.27 pF

Ci = C_comp + C_pkg

Ci = 4.71 pF

Cs = ~3pF (typical of board traces)

CL = 20pF (specified for crystal).

Using that equation and Ce works out to 2*20 - 4.71 - 3 = 32.29pF.

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  • \$\begingroup\$ Where did "Ce = 2CL - Ci - Cs" come from? \$\endgroup\$ – Andy aka Nov 18 '13 at 9:46
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From what I can tell you have some mistake in your final equation. You forgot a bracket and the two parasitic values are subtracted from each other.

Using your values it should be more like

\$ Ce = 2*(Cl - (Ci + Cs)) \$

\$ Ce = 2*(20pF - (4.71pF + 3pF)) = 24.58pF \$

So I would probably select \$ Ce=22pF \$.

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