In some application notes, there is a recommendation to use an external resistor in series with the OSC_OUT pin to reduce the power dissipated by the crystal, if necessary to do so. A good example is from AN2867 STM32 oscillator design guide called "R_Ext".

enter image description here

Note there is never a resistor on the OSC_IN (EXTAL) leg in any application note and thus I never considered it for any transconductance calculation (Note 1).

So I've been a little bit thrown out by the S32K datasheet which explicitly shows an extra (internal) 280R resistor on the OSC_IN (the S32K calls the pin EXTAL) pin (p27, Figure 8):

enter image description here

So, my questions are:

  1. Does this "resistor" have any effect on the transconductance (gain margin) of the circuit design? EDIT - Supplementary question: If it does affect the transconductance, how is this calculated?
  2. Is it likely to introduce any other effects that needs to get taken into account in the design?
  3. Would there ever be a reason to deliberately design in a resistor connected the inverter input?

Unfortunately, the 280R is 2-3 times larger compared to the ESR of crystals in the 16MHz range (in the order of 100R, e.g. this or this), so if it does indeed affect the transconductance calculation, the effect is non-trivial to the point where the circuit may refuse to oscillate.

Note 1:

The transconductance calculation I am referring to is the modified form that takes R_Ext into account as explained in the STM32 document referenced earlier:

enter image description here

Part of the question is whether the resistor on OSC_IN needs to be included anywhere in this calculation.

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    \$\begingroup\$ Are you talking about the 280 ohm resistor? \$\endgroup\$ – Andy aka Jun 11 '18 at 11:30
  • \$\begingroup\$ Since answers currently have to guess which resistor you are talking about (there's no OSC_IN on the second figure, and a couple of resistors where at least three are connected to what looks like the input) I'm voting to close until fixed, so we don't get plenty of unrelated answers. Feel free to ping back if it's updated. \$\endgroup\$ – pipe Jun 11 '18 at 15:27
  • \$\begingroup\$ @pipe, the question clearly states in the title, "on the inverter INPUT". So there is no need to re-formulate anything. \$\endgroup\$ – Ale..chenski Jun 11 '18 at 20:18
  • \$\begingroup\$ @AliChen One of three answers got it "right" and you think the question is crystal clear. \$\endgroup\$ – pipe Jun 11 '18 at 20:24
  • \$\begingroup\$ @pipe, I think we all must pay a better attention to formulations. At first I also missed the essence, but after a second read and some comments I realized that I was mistaken. Formally the question is correct, but there always is a room for improvements, always. \$\endgroup\$ – Ale..chenski Jun 11 '18 at 20:27

The questions are about the 280-Ohm resistor on input. The picture clearly states that it is "ESD pad", and the value "280" likely represents the lump model for this pad cell.

  1. Does this "resistor" have any effect on the transconductance (gain margin) of the circuit design?

Yes, the resistor generally reduces overall gain of the amplifier (due to small parasitic cap on the internal inverter input), but the loss of gain is likely compensated in the inverter design.

  1. Is it likely to introduce any other effects that needs to get taken into account in the design?

Not really, see above.

  1. Would there ever be a reason to deliberately design in a resistor connected the inverter input?

This is just a model for ESD protection on the pin. Since the actual internal input impedance is likely in mega-Ohm range, the 280-Ohm shouldn't have much of impact. Even if the inverter input has 1 pF parasitic capacitance, the RC cut-off will be in ~3 Ghz area, or well outside the typical working frequency of 20-40 MHz

Therefore, the input resistor should be of no concern for the overall design.

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  • \$\begingroup\$ Thanks for the answer. I am trying to reconcile your answer with @Annie's answer. In your answer, you're suggesting the (280R) "resistor" will mostly affect the circuit due to the parasitic capacitance which is small in the grand scheme of things, rather than the effect of the resistance. Annie suggests that the "negative resistance" of the circuit and thus reduce the gain. Would the 280R be "seen" with respect to the amplifier being (more or less) high impedance? \$\endgroup\$ – Damien Jun 12 '18 at 0:11
  • \$\begingroup\$ @Damien, the 280R is in series with input impedance of ~1 MOhm, adding 300R to 1M doesn't affect anything, other than small cut-off due to input gate parasitics. The R(Ext) is different, it is a part of overall feedback, part of C1-XTAL-C2 transformer (or whatever is is called when caps ratio defines voltage instead of winding ratio, and the R(Ext) is the resistor to worry about. \$\endgroup\$ – Ale..chenski Jun 12 '18 at 1:59
  • \$\begingroup\$ Ali, this answer makes intuitive sense to me and I am inclined to accept it. However, since it is contradictory to @Annie's answer, I am waiting to see her answer is modified - there may be some additional insight on offer. \$\endgroup\$ – Damien Jun 12 '18 at 2:46
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    \$\begingroup\$ @Damien, for more amusing things about the Pierce oscillator, see this answer, electronics.stackexchange.com/a/371101/117785. \$\endgroup\$ – Ale..chenski Jun 12 '18 at 3:37
  • \$\begingroup\$ However, for my opinion, non of the answers in the given link correctly explain the role of the crystal as well as the feedback resistor. The crystal is simply used as an inductance (thus forming a 3rd-order lowpass together with the associated components) and the feedback R is necessary for a correct DC bias point (in the middle of the quasi-linear regin of the inverer). \$\endgroup\$ – LvW Jun 12 '18 at 9:21

The crystal does NOT vibrate at some harmonics. It is the only purpose of the crystal to provide its inductance to form a 3rd-order lowpass (Rext+rout)-CL2-L-CL1. Only such a lowpass is able - as mentioned in Andy aka`s comment - to provide the required 180deg phase shift at one single frequency only.

(Such a crystal can always be used as a high-Q inductance at a frequency somewhere between the serial and parallel resonant frequency.)

Together with the inverting properties of the active element this allows to realize the required positive feedback for oscillation.

Remark: As the questioner has observed, in some application notes, there is a recommendation to use such an external resistor - and it some applications it is not included. The necessity of an external resistor depends on the applied active element element: In case of an opamp with a very small output resistance, an external resistor is always needed - in contrast to the collector (drain) of a transistor where we have a finite output resistance which can be exploited for forming the 3rd-order lowpass.

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In your top diagram RExt does two things:

  1. It reduces the drive level seen by the crystal. Some small crystals are rated for low drive levels, and can be damaged by a 5 V square wave, for example.

    Note the description for RExt. It even says "RExt: external resistor to limit the inverter output current".

  2. It provides some guaranteed minimum impedance for CL2 to work against. Unlike what most references say, CL2 is not really a crystal "load" capacitor. Its job is to filter out some of the harmonic content coming from the inverter output.

    In the ideal case, the inverter output has 0 impedance. In that case, a capacitor on its output won't change the voltage at all. CL2 becomes irrelevant, and the crystal is always driven by the square wave out of the inverter.

    RExt and CL2 together form a low pass filter that attenuates the harmonics in the output of the inverter. After all, only the fundamental does anything useful. Crystals can vibrate at some harmonics. If the loop gain is over unity at any of these harmonics, then the whole oscillator might run at that frequency. In some cases, this effect is harnessed deliberately to run the crystal in "overtone" mode.

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  • \$\begingroup\$ Two things - the op I believe is talking about the OSC_IN resistor that I take to be the 280 ohms inside the chip on his last diagram. The 2nd thing is that an output resistor is always needed to give 180 degrees to be able to oscillate. It and CL2 do tip the phase shift so that 180 degrees is met at close to the XTAL's labelled frequency. \$\endgroup\$ – Andy aka Jun 11 '18 at 11:48
  • \$\begingroup\$ @Olin, thanks for the answer. The question should have been clearer that it was referring to the 280R resistor that is on the oscillator input. The question has been updated accordingly. \$\endgroup\$ – Damien Jun 12 '18 at 0:05

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