0
\$\begingroup\$

I am wondering, if the quartz crystal, to be used as part of an oscillator, needs additional components to oscillate at its resonant frequency? And if it does, which are those extra components that help the quartz resonate? I know it doesn't produce any sine wave only by itself.

Also, should the output of the whole oscillator (which is feeding back a part of output signal to the input of quartz) lag for 180° vs it's input, so that quartz crystal actually oscillates?

Example shown below

enter image description here

\$\endgroup\$
  • \$\begingroup\$ There are easier ways to make oscillators oscillate in the MHz region. Such as an ring oscillator. It can also be made with 3 resistors and 3 transistors working in an inverting configuration like the ring oscillator. \$\endgroup\$ – Harry Svensson Oct 10 '17 at 13:24
  • \$\begingroup\$ If you don't need Sine wave output z80.info/uexosc.htm \$\endgroup\$ – Optionparty Oct 10 '17 at 21:34
3
\$\begingroup\$

Your circuit is a Colpitt type oscillator in common collector configuration. Please note that a quartz can be (and is in practice) used in four different modes:

(1) series oscillation, (2) parallel oscillation, (3) high-Q inductor and (4) high-Q capacitor.

In the shown circuit, the quartz is used at a frequency where it works as a high-Q inductor. Thus, together with both capacitots it forms a parallel resonant circuit which has zero phase shift at the resonant frequency wo=1/SQRT(LC) with C=C1C2/(C1+C2).

This allows oscillation at the frequency wo - together with a common-collector stage - because the loop gain has zero phase shift.

Comment: Sometimes, you can find - for transistor-based oscillators - another explanation for the working principle: Negative-resistor oscillator. But explanation of the oscillation principle is - in this case - more complicated. Fortunately, every negative-resistor oscillator can be explained also based on feedback when we identify the corresponding feedback loop.

Note: The input resistance at the base is negative due to the positive feedback effect of the capacitor between base and emitter.

\$\endgroup\$
  • \$\begingroup\$ What is the difference between series and parallel oscillation (besides series and parallel resonant frequency of the crystal)? \$\endgroup\$ – Keno Oct 10 '17 at 13:57
  • \$\begingroup\$ When the frequency-selective part of an oscillator circuit is a parallel resonant (tank) circuit, we can say that we have a "parallel resonant oscillator". And the same applies to series resonant effects. Answers this your question? \$\endgroup\$ – LvW Oct 10 '17 at 14:09
  • \$\begingroup\$ Yup. One more thing. When you mentioned those quartz modes: ...(3) high-Q inductor and (4) high-Q capacitor, were you referring to behavior of crystal described in this graph: electronics-tutorials.ws/oscillator/osc46.gif ? \$\endgroup\$ – Keno Oct 10 '17 at 15:36
  • 1
    \$\begingroup\$ In principle - yes. It becomes more clear when you evaluate the phase response (between voltage and current). \$\endgroup\$ – LvW Oct 10 '17 at 16:24
  • \$\begingroup\$ Nice. And phase shift of the signal feeding back from the output should be 0°, or else output signal would be extra distorted? \$\endgroup\$ – Keno Oct 10 '17 at 18:05
1
\$\begingroup\$

A Quartz crystal is a Passive device.

You should see it as an almost ideal (very little loss) LC-network.

If you put some energy in that network (the Crystal) it will resonate for a very short time but the signal will die out.

The crystal does need energy to maintain a signal across its terminals as it is not completely lossless. Adding this energy is the function of the circuit, to add small amounts of energy so that the crystal can start to resonate and maintain that resonance.

Also, you want to use the signal, this also costs some energy (to drive the next circuit). Providing this signal is another function of the circuit.

In the example circuit you included of a crystal oscillator it cannot be easily seen how it actually works. To make a crystal oscillator it is often enough to make a circuit which will add a bit of negative resistance in parallel (or series) with the crystal. That will overcompensate for the losses of the crystal (which are small anyway) and thus resulting in an oscillator circuit.

There is a more thorough explanation to be found here

\$\endgroup\$
  • \$\begingroup\$ So, it would never oscillate a sine wave if there would be no feedback to its input? \$\endgroup\$ – Keno Oct 10 '17 at 13:21
  • 1
    \$\begingroup\$ Yes - any oscillator needs a closed loop which allows a unity loop gain (zero phase) at one single frequency only. Because each passive circuitry contains damping elements, the loop must contain an active part with gain for compensating damping effects. \$\endgroup\$ – LvW Oct 10 '17 at 13:40
  • 1
    \$\begingroup\$ The crystal itself has no input or output, it is just a resonator circuit with a very high Q factor. In order to make an any oscillator circuit feedback is needed. I challenge you to find an oscillator circuit (or even a mechanical system!) without feedback. You will not find any as these do not exist. What the crystal does is force the oscillator to oscillate at a specific frequency. \$\endgroup\$ – Bimpelrekkie Oct 10 '17 at 13:41
  • \$\begingroup\$ Bimpelrekki - what about a negative-resistance oscillator using a tunnel diode? However, this is the only exception - as far as I know - from the principle you have mentioned. All other negative-resistance oscillators based on active circuits (BJT, opamp) can be explained using both principles: Negative resistance or feedback loop. \$\endgroup\$ – LvW Oct 10 '17 at 14:31
  • \$\begingroup\$ @LvW Good suggestion that tunnel diode. Maybe we can say that it has feedback (or maybe feed forward) as a voltage increase causes a current decrease. In a sense it does respond actively to external parameters. \$\endgroup\$ – Bimpelrekkie Oct 10 '17 at 14:52

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.