I am trying to understand the inner workings of crystal oscillators, but I do not understand why the current has 0° phase shift at series resonance. To my knowledge this is the equivalent circuit for an crystal oscillator.


simulate this circuit – Schematic created using CircuitLab

Series resonance occurs when C1 and L1 resonate. Because the reactance of C1 and L1 cancel each other at this frequency they can effectively be removed from the circuit as is shown in the circuit below.


simulate this circuit

Microchip application note AN826 agrees that the above circuit models crystals at series resonance.

Recall that series resonance is that particular frequency which the inductive and capacitive reactances are equal and cancel: XL1 = XC1. When the crystal is operating at its series resonant frequency the impedance will be at a minimum and current flow will be at a maximum. The reactance of the shunt capacitance, XC0, is in parallel with the resistance R1. At resonance, the value of XC0 >> R1, thus the crystal appears resistive in the circuit at a value very near R1.

If I calculate the impedance of the circuit at series resonance I get a value very close to R1, but it is not purely resistive as is predicted by AN826. $$X_{C0}=-j\:\frac{1}{(1.1254*10^{6})(2*\pi *4*10^{-6})}$$ $$X_{C0}=-35355.1944j$$ $$R_{1}=75\Omega$$ $$X_{C_{0}}||R_{1}=74.999-0.1591j$$

However, this graph pulled from The Art of Electronics indicates that the current phase is zero at series resonance. Furthermore the caption states:

Note that the graphs of impedance and phase are unaffected by external capacitance.

This "external capacitance" is equivalent to C0, so essentially this caption is saying that the value of C0 has no effect on the current phase at resonance. However, this completely disagrees with the model presented in AN826 in which the current phase at series resonance is dependent on the value of C0.

If someone would please explain what the current phase of a crystal is at series resonance and why this is the case your help would be greatly appreciated.

enter image description here

Update: It appears many people believe that the current phase of the crystal at series resonance is just slightly capacitive. I too would like to believe this is the case, however if we analyze the circuit just below 1.1254Mhz the L1 and C1 combination will be capacitive. Now if we analyze the circuit just above the resonant point of L1 and C1 the series combination of L1 and C1 has inductive impedance. Because the RLC branch is in parallel with C0 the branch with the lesser impedance dominates the circuit. Near resonance of L1 and C1 the impedance of the RLC branch is much lower than the impedance of C0 (35355.1944 ohms @ 1.1254Mhz). Thus, the RLC branch should dominate the impedance of the crystal around resonance of L1 and C1.

Now, here is where I get confused. At a frequency just below 1.1254Mhz the crystal's current phase is capacitive, if we analyze the crystal's current phase at exactly 1.1254Mhz the current phase is just slightly capacitive, but if we analyze the crystal's current phase just above 1.1254Mhz the current phase becomes inductive. How could the current phase just switch from capacitive to inductive without passing through zero phase shift? The picture below is from https://www.electronics-tutorials.ws. It shows that the crystals impedance is capacitive below series resonance and inductive above series resonate. Can anyone please explain how the phase behavior of a crystal below, at and above series resonance all fit together?

enter image description here

  • \$\begingroup\$ The phase angle is -0.12 deg, i.e. approximately zero. \$\endgroup\$
    – Chu
    Jun 17, 2021 at 0:16
  • \$\begingroup\$ At series resonant frequency Co has least effect. At any other frequency (barring spurious resonances, and harmonics) Co dominates. We can easily measure Co with a capacitance meter, since it is unlikely that the instrument probes the crystal terminals with a frequency near series resonance. \$\endgroup\$
    – glen_geek
    Jun 17, 2021 at 4:51
  • \$\begingroup\$ Now compare the two results you calculated: -j35k with 75-j0.16. This is typical of a bandpass or bandstop transfer function -- the phase is zero at resonance. Here, you have an additional shunt capacitor, so it adds a tiny bit (read: negligible) of phase. \$\endgroup\$ Jun 17, 2021 at 5:54
  • \$\begingroup\$ The current phase does not switch. It swings through zero as the impedance changes from capacitive, through resistive to inductive, then swings back through zero as it goes back through resistive to capacitive during parallel resonance. The Crystal Reactance against Frequency graph in the Electronics-tutorial is very confusing, giving the impression that the current switches phase during parallel resonance. It should have shown an upward peak with a line drawn through the middle where it goes from inductive, through resistive to capacitive like the graph above. \$\endgroup\$ Mar 9 at 3:24

2 Answers 2


It is exactly the other way around: The frequency at which an L-C series combination exhibits a phase angle of zero is defined as series resonance frequency.

  • \$\begingroup\$ From what I understand you are saying that series resonance of a crystal is defined by the series resonance frequency of L1 and C1 (this is equivalent to the frequency at which L1 and C1 have phase angle of zero). Did I understand you correctly? \$\endgroup\$ Jun 17, 2021 at 11:56
  • \$\begingroup\$ Yes - that is correct. At resonance, the input resistance of a frequency-dependent circuit must be real. This is identical to imag. part=0 and zero phase angle \$\endgroup\$
    – LvW
    Jun 17, 2021 at 13:03

For the impedance transfer function, the series resonance always occurs 1st from L1C1 with resonance at 0 deg. The 2nd resonance is parallel with C1 in series with C2 making it slightly smaller at the slightly higher f with 180 deg phase shift, thus requiring an inverter for an oscillator.

Gain margin for stability depends on the attenuation at resonance which is limited by Rs in series and the external Rs for parallel that must be less than the gain (f) of the amplifier.

  • 1
    \$\begingroup\$ Could you please elaborate why the current phase of the entire crystal is perfectly zero at series resonance and not -0.12 degrees as is predicted by the AN826 model described in the original question? \$\endgroup\$ Jun 17, 2021 at 16:42
  • \$\begingroup\$ Which page? I did not see that. But the net loop series resonance must be 0 deg so if the amp is + 0.12 deg then the loop shifts the crystal in the opposite direction. For Parallel, if inverter is 181 deg then Xtal is 179 deg. Since CB & CC amps have wider BW, at unity voltage gain, they have lower phase shift at resonance than inverting transistors. \$\endgroup\$ Jun 17, 2021 at 17:00
  • \$\begingroup\$ Sorry if I was unclear, my question is not asking about a crystal in an oscillator circuit. I am simply trying to understand the current phase behavior of a crystal around series resonance if it were measured on a VNA as is done in youtube.com/watch?v=G9zZRNzhsEE at 7:30. When I referred to AN826 I was referring to my quote from Microchip's application note AN826 describing the behavior of a crystal at series resonance. \$\endgroup\$ Jun 17, 2021 at 19:12
  • \$\begingroup\$ Sorry I don’t follow your references to -0.12 deg at series resonance. It may just be the resolution of a digital VNA \$\endgroup\$ Jun 17, 2021 at 22:35

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