5
\$\begingroup\$

I've been looking at the Pierce oscillator and I'm unsure of the role played by the series caps to ground (the so-called pi network).

I've put together an image which attempts to explain my intuition (the 1's and 0's are a simplification; I understand that the crystal produces a sine wave).

Pierce Oscillator Action

I assume the action of the crystal - oscillating back and forth - causes the caps to charge and discharge and, crucially, produce the 180 degree phase shift at the inverter input pin.

The feedback "kick" is produced every half-cycle in response to C1 charging and is in-phase with the crystals current movement during this period.

Incidentally, as I understand it, the large resistor R1 is designed to put the inverter in a highly sensitive state (hovering around Vcc/2) where the slightest tip either way produces a large output (relatively speaking) in the opposite direction. This is possible because an inverter output of Vcc could feed the input, in tiny increments though the feedback resistor, from 0 to just shy of Vcc/2 before it would "flip". Then, of course, it would draw current from the input and so on until finally it settled at a balancing point around Vcc/2.

Is that more-or-less how it is?

\$\endgroup\$

2 Answers 2

1
\$\begingroup\$

Without capacitors how can you expect the crystal to produce a phase shift of 180 degrees at its resonance. There needs to be 180 degrees of phase shift because you need a total of 360 degrees and the inverter provides only 180 degrees hence it's called an inverter.

If you want to read more try this - it's quite a good document on the subject by Microchip entitled AN826 Crystal oscillator basics.

Here is also a very good article about figuring out the series resonant point and the parallel resonant point (all xtals have them and this basically determines the capacitor values chosen.

\$\endgroup\$
4
  • \$\begingroup\$ So I take it that's an endorsement of my intuition, as in - yes? I should have pointed out that I'm very much a beginner and what seems obvious to others takes quite a bit of quiet contemplation for me. I think I understand (with reference to my model) that if the cap values are off, the total phase shift will be off (with respect to how the crystal would expect - that is, it won't oscillate). \$\endgroup\$
    – Buck8pe
    Commented Sep 3, 2014 at 20:58
  • \$\begingroup\$ There's more to C2 - it also works with the inverter's output resistance to make oscillation at overtones less likely because (a) it provides too much phase shift at overtones and (b) reduces the output amplitude on overtones making it less likely to oscillate. Yes, your basic explanation (like mine really) grabs the gist of the caps being needed to add 180 degrees more phase shift. The cap values can be off and it will still basically oscillate very close to the orignal frequency because the xtal can swing huge amounts of phase shift for minimal frequency change (thus compensating). \$\endgroup\$
    – Andy aka
    Commented Sep 3, 2014 at 21:04
  • \$\begingroup\$ Cheers James, I think I understand the subtlety of c2 at some intuitive level and that the "key" here is that the resonant frequency is "selected" because it's in that narrow band of frequencies that evoke the "kick" at the right moment (all other frequencies being attenuated by phase shift "misalignment"). In other words, it finds its own balance. \$\endgroup\$
    – Buck8pe
    Commented Sep 3, 2014 at 21:12
  • \$\begingroup\$ How dare you call me James hehe. Yeah, it self aligns to 180 degrees by shifting it's resonant point by a few hertz - it's got massive Q so it only needs to move a few hertz to invoke a phase change of close to +/- 90 degrees; barely an error at all unless you have really tight constraints on f. \$\endgroup\$
    – Andy aka
    Commented Sep 3, 2014 at 22:01
1
\$\begingroup\$

"I understand that the crystal produces a sine wave"

No - a crystal is a pure passive part which cannot produce anything. The working principle is as follows:

  • The inverter - together with the feedback resistor R1 - forms a linear amplifier (with a finite output resistance Rout - that`s important!

  • No we have a frequency-dependent feedback loop which is a 3rd-order lowpass: First-oder lowpass block: R,out-C2; second-order lowpass block: Lc-C1. Hence, we have a 3rd-order lowpass that can produce a phase shift of -180deg at a certain finite frequency.

  • Note that the crystal works here as an (high-quality) inductor Lc. (A crystal can be used as a capacitor, as an inducctor or as a resonant circuit).

\$\endgroup\$

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.