There's a lot of seemingly conflicting information regarding how the load capacitance for a parallel resonant crystal is calculated. Faced with a lot of problems in recent years with non-oscillation and frequency inaccuracy, I'm asking the community's help to get to the bottom of the problem.

How exactly should the values of external load capacitors be calculated?

What's the reason if the oscillator completely fails to start with crystals from some vendor and only completely removing the external load capacitors help? Other vendors' crystals work fine and they advertise seemingly similar parameters (load capacitance, fundamental mode, parallel resonant).

Are all integrated XTAL oscillators in, say, microcontrollers, always Pierce oscillators? Does it have any relevance on the issue?

For reference, here's some information that I've found from the Web regarding the calculation of load capacitance. One IC vendor defines it like this:


One crystal manufacturer defines it like this: enter image description here

Another website has this to show for an answer:

    The equation is C=2(CL)-(CP+CI)

C = crystal capacitor value
CL = load capacitance
CP = parasitic capacitance (wires, socket, traces)
CI = input capacitance (mcu itself)

Quite many seem to think that the load capacitance is the value the crystal manufacturer recommends for the external capacitors. This, to me, seems completely incorrect. (But, as it turns out, might still work perfectly ok).

One web page puts an emphasis on knowing the oscillator inverter's input and output capacitances and gives this answer: enter image description here

Is there a one true answer to the question? It all seems very frustrating to me. Why doesn't an oscillator start? Why does removing the external load capacitors make it start? How should the external capacitors' value be calculated?

PS. Sorry I can't tell you exactly which ICs I'm working with. But I've seen this happen with quite a few during the years.

  • \$\begingroup\$ You left a comment in electronics.stackexchange.com/questions/159820/… asking about how to calculate the external load capacitance. The last diagram you show is the more accurate representation. Since all the external capacitors are connected in the middle (to ground), CL = (Cin+C1) in series (Cout+C2). For example, in the answer to the link, I gave the estimate CL = (C1 in series to C2) + (Cin in series to Cout) was not as accurate because it disregarded the middle connection, but it was close enough. \$\endgroup\$
    – rioraxe
    Commented Aug 4, 2016 at 21:46
  • \$\begingroup\$ You have gotten two good answers already. Additional possible explanation for oscillator not starting: the loop gain at low signal level must be greater than 1 to get the oscillation started. The external capacitors does have a tendency to reduce the loop gain, therefore taking out a capacitor may allow the oscillator to start up if the inverter/amplifier is starved for gain (perhaps because it needs to operate at high frequency relative to its gain bandwidth). \$\endgroup\$
    – rioraxe
    Commented Aug 4, 2016 at 22:04
  • 2
    \$\begingroup\$ The only formula that takes the internal gate capacitance into account is the last one you quoted. I wonder why this is ignored in almost every other example I found. The values seem significant. Atmel specifies Cin/Cout as 8pf/18pf for some controllers, for example. \$\endgroup\$
    – Rev
    Commented Nov 22, 2017 at 14:50

3 Answers 3


Is there a one true answer to the question? It all seems very frustrating to me. Why doesn't an oscillator start?

A crystal oscillator will fail to start when the crystal and the capacitors attached either side do not fully produce a 180 degrees phase shift back to the input of the inverter inside the chip.

The inverter produces effectively 180 degrees phase shift so, for oscillation to begin, the two capacitors and the crystal together must form an extra 180 degrees phase shift AND there must be an overall voltage gain greater than 1.

Look at this response - it mimics a crystal and one capacitor but it doesn't quite reach 180 degrees: -

enter image description here

V1 is the driving voltage source and R2 (100 ohms) represents the output impedance of the gate involved in the oscillator. Look carefully, the phase angle doesn't quite reach 180 degrees and this will mean NO OSCILLATION.

The extra few degrees of phase shift come from the output capacitor on the invertor - the 100 ohms (or whatever the output impedance of the inverter has) AND this extra capacitance push the phase shift past 180 degrees and the oscillator will then oscillate.

Here's a picture showing the effect of increasing input and output capacitance from 1 pF up to 20 pF: -

enter image description here

The X axis is at 9.9 MHz FYI. As you can possibly see, only when capacitance is 10 pF or 20 pF does the circuit produce 180 degrees of phase shift. This means the oscillator will oscillate at the left hand point on each phase curve that the response crosses 180 degrees (parallel resonant point tuned by the external capacitance).

So, you need capacitors to make this type of oscillator work and the manufacturer tells you what to use but, in my humble opinion, there are a lot of subtleties around that some manufacturers maybe either don't fully know or won't tell you. I'll also add that there appear to be very few web articles about what really is going on and the true importance of each capacitor.

Why does removing the external load capacitors make it start?

Maybe the self capacitance of the tracks and gate input capacitance are sufficient. It depends also on the Q of the crystal and is hard to speculate on. Maybe the inverter's slew rate is too slow?

  • \$\begingroup\$ Thank you for your answer and great analysis, I really appreciate your effort. Let's say I have determined that the small signal DC gain at the inverter bias midpoint is about 20 to 22. How does this sound? Any thoughts about the effect of Miller capacitance and transconductance? \$\endgroup\$
    – PkP
    Commented Aug 8, 2016 at 11:26
  • \$\begingroup\$ Miller capacitance is all part of the small signal gain so it's irrelevant. If you take the 20 pF scenario in my lower graphs, at -180 degrees where the relevant red curve crosses, project upwards to the blue curve and the net loss (there is in fact gain in my sim) should be cancelled to a net gain by the small signal gain of 21 to achieve oscillation. \$\endgroup\$
    – Andy aka
    Commented Aug 8, 2016 at 11:39
  • \$\begingroup\$ you're right. Sorry, I was assigned another project with some haste and I forgot to come back to check this. \$\endgroup\$
    – PkP
    Commented Sep 3, 2017 at 12:02

AN2867 from ST is also a great app note that you should read. Besides the oscillator transconductance, almost ALL IC makers do not specify the oscillator input and output capacitances. Those should really be included the load capacitor calculation and can make a quite a difference. Naturally, PCB capacitances should also be accounted for, and those are easier to estimate using various PCB design tools (Saturn PCB toolkit is a good free app).

The only real way to guarantee full oscillator functionality (after doing the design calculations) is to test it over the full temperature range and operating voltages. Some crystal manufacturers also offer services to do exactly this. Abracon is one that I know of, and the current cost is around 800USD for full characterization of one crystal on one application circuit.


The condition of oscillation, as Andy mentions, is that the Barkhausen criteria is met (phase shift around the loop an integral multiple of 2\$\pi\$ and gain of \$\ge\$1).

A good crystal manufacturer will give you the information you need to guarantee oscillation, however there is one missing piece- the transconductance \$g_m\$ of the CMOS amplifier - how much output current change you get for a given amount of input voltage change. IC manufacturers don't want to specify this (yes, I've asked major MCU makers). It changes with temperature, of course, and varies from unit-to-unit. The end result is that nobody (except you) gets stuck guaranteeing your product will actually function with an external crystal.

I suggest you read this technical note and referenced materials for starters. There is at least one entire book devoted to crystal oscillators.

As far as load capacitors go- the equations you give are equivalent (the second one assumes the usual situation with the capacitors the same value). If you want the crystal to oscillate as close as possible to the marked frequency, on average, you match up the total load capacitance to what the xtal maker has specified by specifying the two capacitors (usually they are both the same value, so you have only one degree of freedom). If the capacitors cause the oscillator to be unreliable, choose a different crystal or a different chip, or live with a slight error in frequency. Or specify a crystal oscillator module and shift the risk to an outside supplier.

Whether the crystal will actually oscillate reliably under all conditions with all instantiations of a given chip part number is really an orthogonal question (and probably the more important one- may applications don't care about +/-0.01% error in crystal frequency, but the crystal oscillator not starting is a serious issue).

Not part of your question, but maximum drive power is also a factor that needs to be checked, especially with smaller crystals and tuning fork crystals. The older crystals could often take 1mW of power, but smaller new ones have maximums that are much lower. Often a series resistor (from the output) is useful to reduce the drive power, but of course it will reduce the gain, so at some value the oscillator will fail to start.

  • \$\begingroup\$ Thank you for your answer, I really appreciate your effort. Let's say I have determined that the small signal DC gain at the inverter bias midpoint is about 20 to 22. How does this sound? Any thoughts about the effect of Miller capacitance and transconductance? \$\endgroup\$
    – PkP
    Commented Aug 8, 2016 at 11:25
  • \$\begingroup\$ You can't infer the transconductance from the voltage gain. \$\endgroup\$ Commented Aug 8, 2016 at 12:09
  • 3
    \$\begingroup\$ For anyone else reading this, the linked doc seems to have moved here: statek.com/wp-content/uploads/2018/03/tn31.pdf \$\endgroup\$
    – Matt Gaunt
    Commented Feb 3, 2019 at 0:02

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