Why do crystals have 0° phase shift at series resonance?

Question

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. 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.

Figure 7.36. Impedance and phase in the neighborhood of the series resonance, for the 10.0 MHz crystal of Figure 7.35. Note that the graphs of impedance and phase are unaffected by external capacitance. The high-𝑄 resonance of our sample, with its 𝑅₁=4.7 Ω, is considerably degraded for a crystal with worst-case specified 𝑅₁=50 Ω.

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, but if we analyze the circuit just below 1.1254 Mhz 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 @ 1.1254 MHz.) 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 1.1254 MHz the crystal's current phase is capacitive, if we analyze the crystal's current phase at exactly 1.1254 MHz the current phase is just slightly capacitive, but if we analyze the crystal's current phase just above 1.1254 MHz 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 crystal's impedance is capacitive below series resonance and inductive above series resonance. Can anyone please explain how the phase behavior of a crystal below, at and above series resonance all fit together?

Answer 1

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.

Answer 2

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.