[SOLVED] Crystal Oscillator problem

Can anyone point out what's wrong with this circuit ? I'm only getting a DC signal at the output.

This looks something like a Pierce oscillator circuit, but I think you need a small phase-shifting capacitor (about 100 pF) in parallel with R2.

DasPreetam said:

Can anyone point out what's wrong with this circuit ? I'm only getting a DC signal at the output.

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Simulation models of oscillators often do not self-start, even though the circuit is correct.

This is because oscillators rely on some small imbalance in the circuit to get it going, and in real life circuits this is provided naturally by noise or PS ripple etc.

To get your model 'going', try giving it a very small kick by capactive injection of a single pulse somewhere into your circuit.

@Kripacharya : Yes I know that simulation of crystal oscillators are difficult... that's why I used a sinewave function generator of 1.5 MHz, but the oscillations damped to zero as soon as I turned it off... I assembled the circuit on a breadboard and checked the output with a CRO, it is giving a DC voltage, nothing else.

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@Tunelabguy : I tried with a 100pf capacitor across R2, it still gives the same output.

Who designed the circuit?
How do you saying this circuit will work for 1.5 MHz?
Do you have any design reference?
How did you choose 10uF on feedback, Did you starting trial and error method?

Do you know the working principle of the oscillator - and did you design it according to the oscillation criterion (Barkhausen)?
Did you consider the dampuing properties of R1||R2=5kohms?

If you design your circuit with the correct values, it will simulate. The problem with crystal oscillator circuits is that you need a small-time step and a lengthy oscillation time due to the slow built-up of oscillations caused by the very high Q of the crystal.
The 22 pF capacitor can be a trimmer to make slight frequency corrections.

So be prepared to wait quite a while if you don't use some injection to start it up.

@Venkadesh_M : The circuits I looked up on the internet does not specify the values of the component, so yes, i'm literally in the dark... Any reference texts/blogs would be greatly appreciated !

@LvW : No, I can only find the gain of the amplifier... from Rc/Ie formula. I have no idea how to find the gain of the feedback path.

@E-design : Can you please elaborate how your circuit works ? I'm a newbie in this field so please bear with me..

DasPreetam said:

@Venkadesh_M : The circuits I looked up on the internet does not specify the values of the component, so yes, i'm literally in the dark... Any reference texts/blogs would be greatly appreciated !

@LvW : No, I can only find the gain of the amplifier... from Rc/Ie formula. I have no idea how to find the gain of the feedback path.

@E-design : Can you please elaborate how your circuit works ? I'm a newbie in this field so please bear with me..

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Your circuit as originally posted certainly has some unusual component selections, and I noticed that though it does oscillate, it dies down very quickly. E-design's circuit is different from yours, and is a Colpitts variation.

For your own circuit, here's what seems to work :

1) Connect a new capacitor of value 385pF from collector to ground.
2) Change your inductor to 30uH.

These two then form an LC tank which helps store energy & sustain oscillations. You can use any combination of L & C which satisfy the resonance formula, but try keep the reactance somewhere in the range 100 - 300 ohms.

In addition to kripacharya´s comments I only can emphasize that (quote E-design) "the problem with crystal oscillator circuits is that you need a small-time step and a lengthy oscillation time due to the slow built-up of oscillations caused by the very high Q of the crystal."

@Kripacharya : What I built is a pierce crystal oscillator, and going through the schematics, I found there using a FET, but also states that the FET can be replaced by a BJT. So why doesn't it work ? I want to know how to find out the gain of the feedback circuit.

If you use a pierce configuration you must ensure that you have enough phase shift around the loop to satisfy oscillation criteria. While you're working with a design, it is often better to use a crystal equivalent circuit with a lower Q that will allow oscillations to start more quickly during simulations.

The attachment shows the crystal equivalent block and circuit for this pierce design. Components R2,3,5 and C2,3 together with the crystal help to ensure that there is enough phase shift around the loop to allow oscillator action.

This illustrates the problem with simulating oscillators involving high Q resonators like crystals.

Even with the crystal equivalent circuit having a much lower Q than an actual crystal, transient analysis shows a simulation time of more than 5 mS to reach final operation conditions. Now this combined with a small time-step takes a lot of CPU time to complete. It is not unusual for high Q crystal oscillators to take in excess of 50 mS to start up.

If you were only simulating for 1 mS you will be convinced that your circuit wasn't working. ;-)

That 5Kohm base node impedance isn't helping any. The
crystal wants as high a Q as possible and for the input this
means high DC impedance.

You want to play some with simulator settings, too loose
and you may lose the oscillation just to numerical noise.
I do not know what settings make it better, but guys who
have done working XOs tell me this. I'd bet you could find
some finer guidance with a well keyworded search.

With the pierce circuit the crystal operates in series mode and having equivalent series load resistance lower than the internal Rs of the crystal actually aids in short term frequency stability. Many high stability pierce designs aim to get the series load resistance below 15% of the crystal internal Rs value. By doing so you obtain in-circuit Q values that approach the Q of the crystal and much improved short term frequency stability.

With crystals operating in parallel mode you don't want loading and hence low frequency crystal oscillators with a high internal crystal Rs need a jfet or something similar to prevent loading that will impede start-up and frequency stability.

That 5 Kohm base node impedance isn't helping any. The crystal wants as high a Q as possible and for the input this means high DC impedance.

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The base node impedance is even lower because the common emitter circuit input is connected in parallel.

But that's not the reason why the original circuit isn't working. The reason has been shown by E-design and others. It's the lack of external capacitors that are necessary to shift the inverted phase of the CE amplifier towards 0° to fulfill the oscillation condition. If you place e.g. 100 pF capacitors to ground at collector and base node, the original circuit is oscillating.

For clarification we should know the LCR parameterss of the Multisim crystal model, but I guess it's not too far away from the said "lossy" model by E-design.

@LvW : No, I can only find the gain of the amplifier... from Rc/Ie formula. I have no idea how to find the gain of the feedback path.

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Obviously, the question hasn't been answered yet, and it's in fact far from being simple, although the circuit is rather basic. A complete analysis of the problem can be found in the classical paper Analysis of loop gain in feedback systems by Middlebrook.

In a short, you need to break the loop without changing the circuit impedances, or you need to consider the circuit impedance in your simulation. A straightforward way to achieve this in a simulation (not discussed by Middlebrook) can be to replicate the circuit at least two times to allow an open loop measurement with almost identical impedances. (You still need to assume that the reverse loop gain is considerable lower than the forward loop gain.)

Okay let's just get down to the basics. We know that for oscillations, a positive feedback is required. So this can the achived by using a feedback circuit that shifts the phase of the signal by 180°. So the Amplifier gives an 180° shift, and the feedback circuit gives 180° shift... So, total phase shift 360°, the circuit oscillate, right ?

Here, the transistor is giving 180° shift, and the crystal is giving 180°, so why the oscillations not sustaining ? Why we need the two capacitors to shift the phase to 0° when we need 360° ?

DasPreetam said:

Okay let's just get down to the basics. We know that for oscillations, a positive feedback is required. So this can the achived by using a feedback circuit that shifts the phase of the signal by 180°. So the Amplifier gives an 180° shift, and the feedback circuit gives 180° shift... So, total phase shift 360°, the circuit oscillate, right ?

Here, the transistor is giving 180° shift, and the crystal is giving 180°,..

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It is not true that the crystal gives 180 deg. phase shift. The phase shift depends on how close the frequency is to the series resonant frequency and what other circuit components there are. The capacitor that was omitted from the original circuit is necessary to make that phase shift come out to 180 degrees. Without it you never get to 180.

At series resonance a xtal looks like a few hundred ohms. Above resonance it looks like an inductor whose reactance is Q X few hundred ohms, err, 100 X 50,000, or or very large value. So just connecting a xtal from base to collector is like connecting 200 ohms, would it oscillate?, I think not. So you need some capacity between base and earth to give the 180 degrees of phase shift to get it to oscillate. If you also connect a smaller capacitor from collector to earth, you have also got a PI network to better power match the voltage at the collector to the base, so you get lower losses.
Frank
P.S. the 10 MF in series with the xtal is tooooo big, try 1000PF

P.S. the 10 MF in series with the xtal is tooooo big, try 1000PF

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I thought the 10 uF was to block DC, does it have any other applications also ?