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My question is regarding the answers, one and two.

Both answers talk about the impedance of the crystal driver output.

I'm trying to understand how the crystal driver output impedance will be related to the load capacitors of the crystal.

The first answer has this, "The impedance of 6 pF at 32.8 kHz is 810 kΩ. Now the impedance of the crystal driver is certainly not zero, but quite likely significant relative to 810 kΩ."

Can someone tell me how this impedance of 810k is relevant or how to understand this with the remaining part of the answer.

Basically, I want to understand the relation between the crystal drive impedance and the load capacitance of crystal.

  • About the "*impedance of the crystal driver*"...I assume you're referring to the inverter's internals. This is something rarely spec'd by chip manufacturers. It varies a lot. I've measured large-driver output Z to be in the thousand-ohm ballpark. Weaker drivers can have much, much higher output Z. A high-Z driver is sufficient for low-power crystals like the 32kHz tuning fork types. – glen_geek May 02 '23 at 12:37

1 Answers1

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Basically, I want to understand the relation between the crystal drive impedance and the load capacitance of crystal.

The phrase "crystal drive impedance" is a bit ambiguous so, I'll define it as the output resistance of the inverting gate used in the Pierce oscillator. I mention Pierce oscillator because that's pretty much what all MCU/logic chips use. Pierce oscillator circuit diagram: -

enter image description here

Image from here.

So, "crystal drive impedance" is the resistance above shown as \$R_S\$. Resistor \$R_F\$ turns the unbuffered logic gate into a linear amplifier (due to negative feedback). I believe you know what \$C_a\$ and \$C_b\$ are i.e. they are the so-called crystal loading capacitors.

Now, where you might be getting confused is in the terminology. For instance, the term loading capacitance is badly named because in fact, they are part of a phase shifting network. The "true" (as in actual) loading capacitor for the crystal is \$C_a\$. Capacitor \$C_b\$ does not "load" the crystal; it loads the output of the inverting gate via its output resistor \$R_S\$.

So, there are in fact two phase shifting networks: -

  • \$R_S\$ and \$C_b\$
  • The crystal's reactance and \$C_a\$

Together they produce 180° of phase shift at one particular frequency. This produces oscillation and all is good. The crystal's reactance is inductive (at oscillation frequencies) and, along with \$C_a\$ shift the phase by nearly 180° (not enough on its own for oscillation to occur). \$R_S\$ and \$C_b\$ shift the phase by a few degrees more and hence, you get oscillation.

So, to answer your question, the relationship between \$R_S\$ and \$C_b\$ (half the so-called loading capacitance) is to produce a few degrees of phase shift and allow the circuit to oscillate.

Maybe you should read what I put on my basic website about crystal oscillators.

Can someone tell me how this impedance of 810k is relevant or how to understand this with the remaining part of the answer.

The 810 kΩ is the impedance of \$C_b\$ (above) at 32.768 kHz and it will form a phase shift network with \$R_S\$ as explained earlier.

Andy aka
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  • Thank you very much for the answer! Could you throw some light on the values and the related question which I mentioned in my actual question? –  May 02 '23 at 10:06
  • @Newbie you need to be much more explicit than that. You mentioned two other questions from this site and, I'm not going to do a tear-down of their full answers in the hope that I target the particular part you need explaining. Hopefully you see my problem in what you ask. Have you looked at my basic website explanation. There is a considerable amount of BS about pierce oscillators BTW. I have added one part however. – Andy aka May 02 '23 at 10:08
  • Thank you. Yes, I understand your point. Just a quick follow-up. In your edited answer, how is the 810k value relative, like how big or small (relative) is it to the desired or required impedance value? 810k compared to what impedance value? –  May 02 '23 at 10:21
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    I can't ask Olin to explain that; all I can say is that \$C_b\$ forms a phase shift network with \$R_S\$ i.e. it acts like a low pass filter. If \$R_S\$ was 810 kohm then there would be 45 deg phase shift at 32 kHz. But, 45 degrees won't be needed so, if \$R_S\$ is around 200 kohm it would fit the bill. Remember that Olin is talking about a 32 kHz crystal and, it naturally uses big impedances to keep power into the sub microwatt regions. – Andy aka May 02 '23 at 10:37
  • Thank you for the clarifcation –  May 02 '23 at 10:39