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I'm a newbie in using this LTC software.

Here is a crystal oscillator circuit. I have been searching for a tutorial and guide to put in the value or what type of analysis should I use, but I can't find any of them.

Can anyone guide me what functions should I choose for the voltage source? Either pulse or sine? I tried many values, but the output didn't shows a repetition of sine wave. This circuit expected to produce an f0 ~ 2 MHz.

2 MHz Crystal Oscillator Circuit

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  • \$\begingroup\$ where is crystal in your image? \$\endgroup\$ Apr 1 '19 at 17:56
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The two voltage sources you have for the opamp rails should be DC sources.

Use the transient analysis type.

Oscillators often cause problems for a simulator. Two possibilities. When they do the initial DC solve step, they fail to converge, as the point about an oscillator is that it's unstable. Or in the real world, the oscillator starts by feeding back noise, and an ideal simulator is noiseless, so there's nothing to start oscillation.

You can fix both these problems by setting the 'initial voltage' conditions for one or more capacitors or nodes. This can break the feedback loop of an unstable circuit, and give a kick to one that needs to be started. Use the .ic directive. C1 and C2 would be ideal candidates for this intervention.

C1 and C2 look awfully big for a 2MHz oscillator, are you sure you have the right values?

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  • \$\begingroup\$ For that C1 and C2 values, actually this circuit is referred from my textbook and the values were given. How do I assigned the initial voltage for the capacitor? \$\endgroup\$ Nov 29 '16 at 7:55
  • \$\begingroup\$ How do I assign the initial voltages for the capacitor? Use the .ic directive. You could also RTFM, and google for instructions and examples for using the .ic directive. \$\endgroup\$
    – Neil_UK
    Nov 29 '16 at 7:59
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I'll try and keep this simple: -

  • With RL equal to 100 kohm and C2 equal to 100 nF you create a low pass filter with cut-off frequency equal to 159 Hz.
  • At 1.59 kHz the attenuation is 20 dB and at 15.9 kHz the attenuation is 40 dB (i.e. rising at 20 dB per decade in frequency).
  • At 159 kHz, attenuation will be 60 dB and at 1.59 MHz attenuation will be 80 dB.

So at 2 MHz, the attenuation will be a bit more than 80 dB - how can you expect this circuit to oscillate when the gain (formed by RF and R1) is only ten? Think about what you are doing here.

But it gets worse because, for series resonance (that's what this design is attempting to emulate), there is a further level of attenuation caused by CS and C1. Both CS and C1 form the capacitive branch of a tuned series circuit but given that CS is only 0.0122pF and C1 is 10 nF, there is a further attenuation of 820,000 or 118 dB at C1.

Hopefully you should realize now that RL, C1 and C2 are totally inappropriate values. If, as you say in a comment, that your book gave these values you should either buy some reading glasses or throw away the book.

Can anyone guide me what functions should I choose for the voltage source ?

The only voltage sources I can see are the DC supplies to the LT1001 and, at 15 volts these seem adequate but, now that we are discussing the op-amp, take a look at the data sheet and you will see that it has a gain-bandwidth product (GBP) of typically 0.8 MHz. This means that the op-amp runs out of steam and is unable to provide voltage amplification at a frequency greater than 800 kHz.

In other words, the op-amp is totally unsuited for operation as a Pierce oscillator at 2 MHz.

So, choose an op-amp with a GBP greater than (say) 50 MHz, make RL more like 100 ohms (not 100 kohms), and make C1 and C2 more like 10 pF (pico farads) and it might just work.

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  • \$\begingroup\$ Perhaps Adam has tried to scale frequency up from a much lower-frequency textbook example. Before trying to scale, learn how the circuit works. Andy aka has outlined how you can get into trouble. \$\endgroup\$
    – glen_geek
    Nov 29 '16 at 14:52
  • \$\begingroup\$ Andy aka , i've modified the circuit as you told, and use an opamp with much greater gain, but, why did the the gain drop to negative value in the graph db vs Hz which i'm trying to observe the oscillation frequency at 3db. I had changed the load capacitance C1 and C2 to many values but still didn't work. \$\endgroup\$ Dec 1 '16 at 9:02
  • \$\begingroup\$ I have no idea what graph you refer to or what new component values you have tried. Without a schematic diagram of precisely what you have got, I'm also in the dark. \$\endgroup\$
    – Andy aka
    Dec 1 '16 at 9:06
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First read Neil_UK's answer and then this.

An additional challenge for simulating a Crystal (or a model thereof) is that a Crystal has a very high Q or quality factor.

This means that it takes many cycles to change the amount of energy in the Crystal resonance circuit. This means that it is very impractical to simulate such an oscillator in a transient (time) simulation as it will take very long to reach a stable state as amplitude changes take extremely long (many cycles of that 2 MHz).

Before you can even think of simulating a crystal oscillator and get meaningful results out of it I strongly recommend you to gain experience using non-crystal oscillators first.

To simulate an oscillator properly you have to know how to determine if it has enough loopgain. With a crystal oscillator you can very easily be fooled by the high Q because when you excite the crystal, it appears to keep oscillating in the simulation while in practice the oscillation will die out.

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