I've included a diagram of typical Pierce oscillator circuit taken from ST Application Note AN2867:

pierce oscillator circuitry

The note suggests that C_L1 and C_L2 should be chosen such that the following equation holds:

enter image description here

Since C_L1 and C_L2 are typically equal, this equation can be rewritten to solve for their value: C_Lx = 2(C_L-C_S).

As a rule of thumb, the stray capacitance is usually taken to be 2pF - 5pF, although I've seen a datasheet suggest 10pF as a rough estimate.

The question: By choosing an oscillator with a low load capacitance C_L, e.g. equal to the expected stray capacitance, C_Lx goes to zero. Is there any problem with choosing an oscillator with a small load capacitance and eliminating the need for the two capacitors C_L1 and C_L2 from a design by depending solely on the stray capacitance?

  • 1
    \$\begingroup\$ Yeah. Watch your oscillator stop if somebody looks at it funny. No sneezing, farting, or laughing allowed near your MCU. \$\endgroup\$
    – JRE
    Commented Apr 30, 2019 at 20:21
  • \$\begingroup\$ Please provide evidence for the comment that C_L1 and C_L2 are typically equal. That is sometimes true but very well not true for some carefully tuned and characterized crystal oscillators. \$\endgroup\$ Commented Apr 30, 2019 at 20:23
  • 1
    \$\begingroup\$ @MichaelKaras The example in AN2867 solves for C_L1 and C_L2 assuming that they are equal, and all of the designs I've seen choose equal C_L1 and C_L2. I'd be interested in seeing documentation for a selection process that produces different values for C_L1 and C_L2. \$\endgroup\$
    – mablem8
    Commented Apr 30, 2019 at 20:47
  • \$\begingroup\$ @JRE Can you expand on your comment? Using the guidance in the datasheets, I don't see any reason why depending only on stray capacitance wouldn't work. In fact, the equations leave no other option for oscillators with very small load capacitances. \$\endgroup\$
    – mablem8
    Commented Apr 30, 2019 at 20:48
  • 2
    \$\begingroup\$ @mablem8 If you have a ground plane under Xtal and traces, this adds much more capacitance than without. In that case , you can compute C from track width and dielectric thickness in pF/mm then you are ok to consider eliminating caps but without doing that and just no caps then you have 0.5nH/mm inductance and more proximity effects like touching it. \$\endgroup\$ Commented Apr 30, 2019 at 20:53

1 Answer 1


The ESD diodes are part of the stray capacitance. These, being installed on the silicon SUBSTRATE, inject all the MCU trash into the XTAL's energy storage loop.

You'll have lower jitter (phasenoise) if the ESD diodes are swamped out by external capacitors. The oscillator will build up the stored energy until cycle-by-cycle losses equal the cycle-by-cycle newly-injected energy.

Note you have the same resonant current circulating in both capacitors of the PI network. This can be a reason to make the Cinput (OSCin, CL1) be substantially smaller, so the voltage across Cinput is quite LARGE. Of course, smaller caps are more vulnerable to MCU trash injected thru the ESD diodes/structures. Experiment with this tradeoff.


simulate this circuit – Schematic created using CircuitLab

  • \$\begingroup\$ This answer is precisely addressing the question at the end of OP's text. \$\endgroup\$ Commented May 1, 2019 at 6:11
  • \$\begingroup\$ Why the downvote? \$\endgroup\$
    – Unknown123
    Commented May 1, 2019 at 6:14
  • \$\begingroup\$ I expect that this answer was downvoted (not by me!) because (1) it is kind of hard to understand (what's phase noise? what's 'MCU trash' supposed to mean? What's a "PI network"? I know the answers to some of these questions, but they're not going to be obvious to the questioner necessarily without being explained.) And (2) it uses boldface and capital letters in quirky ways, which comes across as kind of aggressive. \$\endgroup\$ Commented May 1, 2019 at 6:18
  • \$\begingroup\$ @analogsystemsrf Thanks for addressing the MCU side of the circuit! The downvote is not mine. The question remains unanswered - if an oscillator requiring a 2pF - 5pF load capacitance is used in a design, is there any problem depending on the stray capacitance only? It's not clear why that would result in higher phase noise. It seems like clever (or maybe even not-so-clever) use of traces could eliminate the need for two components from the design. \$\endgroup\$
    – mablem8
    Commented May 1, 2019 at 13:19
  • \$\begingroup\$ One might indeed proceed to design 2pF or 5pF or 18pF capacitors into the copper foils. The areas involved make the "parasitic capacitors" very vulnerable to Efield aggressors which, again, increases the injected non-synchronous energy and thus boosts the (deterministic) jitter (phasenoise). \$\endgroup\$ Commented May 1, 2019 at 15:29

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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