Emitter coupled oscillator - Page 1
Forum section: electronics / projects

author: tree
author_rank: Regular Contributor
timestamp: January 26, 2014, 06:15:12 pm
content: Hello folks,I've been trying to understand how crystals work in crystal oscillator circuits. I understand the piezoelectric effect to the following extent: If we apply an electric field to the crystal it will deform and when the field is removed, the crystal will generate an electric field in the opposite direction. I get stuck at the point where the electric field is removed. I can't seem to understand how the electric field is removed in crystal oscillator circuits. I've attached a diagram of a crystal oscillator circuit. I'm looking for a qualitative analysis of the circuit without using the RLC model of a crystal. I'm trying to understand what exactly happens to the crystal during oscillation. What I'm looking for in terms of an analysis, is an explanation similar to the following one: base voltage of Q1 increases, which increases voltage at the emitter which increases current through R1, etc, etc, etc, around the loop.Thanks.

author: T3sl4co1l
author_rank: Super Contributor
timestamp: January 26, 2014, 06:54:56 pm
content: It's not just fields turning on and off. That's stupid: fields are continuously variable, and suggesting that an otherwise ordinary physical material somehow knows the difference between "changing gradually" and "turning on and off" is peculiar at best.The piezo effect provides coupling between mechanical and electric potentials.Typical crystals are flat plates with electrodes on either side. Which direction the quartz flexes when stimulated depends on how it was cut, because the motion is aligned with the crystal structure. I think the usual mode is a shear or torsion force, across the plane of the crystal. Not like the bulk flexing and bending of a piezo disc (which is a different, polycrystalline, material).It takes time for this force to propagate through the crystal, bounce off the other side and return, so there will be some sort of frequency dependent effect, probably a strong resonance.A sine wave will experience peaking (transmission or attenuation, depending on how it's used in circuit), when the reflected wave interferes constructively or destructively with the applied wave. In principle, this occurs at harmonic frequencies (a 1MHz crystal will also peak at 3, 5, etc. MHz), but material properties and geometry result in dispersion, i.e., the speed of sound varies with frequency; the actual harmonic might be 3.02MHz or something like that.A pulse will experience transmission line and reflection effects (at least, if you can observe them well; I don't think crystals are usually well suited to time-domain analysis, because of impedance matching and high Q factor). A step pulse will result in two half-steps (if impedance matched properly) with about 1 microsecond (the round-trip delay) between them. But because the harmonics are not exactly harmonic, there will be odd ringing artifacts, and the pulse won't settle quickly.These are both the exact same effect, viewed through different means: one with a signal generator and spectrum analyzer, the other with a pulse generator and oscilloscope. There is nothing fundamentally different about these two observations.What you need to know regarding the oscillator, its operation and design, is that the crystal looks like a high Q resonant circuit. A good analogy is an open-circuit stub transmission line: waves sent down will propagate, reflect and return; what the circuit sees at the terminals is a whacky impedance and phase shift, which given enough positive feedback, guarantees one characteristic frequency gets selected and amplified.Whereas an RC oscillator is simple to say "V1 increases, V2 decreases, yadda yadda", a circuit with long time constants does not submit so well to such a simplified explanation. The ups and downs are still implicit in the surrounding circuit -- the gain, positive feedback, negative resistance or however you like to view it are all there. But the time dependency is something completely glossed over in such an explanation. In reality, what happens is, a little bit of noise is naturally present in the system (either as residual offsets from applying power relatively suddenly, or from the various noise sources in the system -- resistors and transistors, and potentially, ambient noise as well). The frequency response of the crystal filters out the noises from other frequencies, and because of the phase shift of the crystal and the circuit around it, these noises are amplified. Each time it "goes around" (of course, it doesn't go through iterations, the process is continuous over time), the noise is filtered to an ever narrower bandwidth, and amplified to an ever higher amplitude. Eventually, the circuit's nonlinearity begins to dominate, and the gain drops (usually, the waveform gets clipped, but various methods of controlling gain without adding harmonic distortion are used in high quality oscillators). The drop in loop gain necessarily increases noise at other frequencies, roughly proportional to the crystal's Q factor. Fortunately, this is huge (~100k is common I think), so the noise floor remains low (say ~ -100dB).Why does the oscillator continue to oscillate at a given frequency? Because that's what frequency it's been oscillating at. If, during startup, you added additional noise at a harmonic frequency, you'd stand a good chance of getting the oscillator to operate at a higher mode (harmonic) of the crystal. 3.02MHz or whatever, for example. This can even be a problem: an oscillator with strong harmonic distortion might skip modes, where it starts at 1MHz, distorts heavily (producing components at 3.00MHz, 5, 7, etc.), and with this stimulus, achieve enough gain at 3.something MHz to actually continue at that frequency. When desirable (or required), crystal modes can also be selected, by wrapping the crystal with a bandpass filter near the desired frequency. As far as I know, there is no such thing as a 200MHz crystal, for example; crystals cut for these frequencies are designed for such an oscillator circuit, so the operating frequency is a higher mode. Because modes are dispersive (not exactly harmonic frequencies), such a crystal might be not 40.0, but actually 38.6 or so MHz at its fundamental (or 42.something, depending on whether the dispersion pulls the harmonics higher or lower... I forget). You only get the labeled frequency when operating the crystal in its intended mode and method (there is also the series versus parallel coupling method, which differ by a few kHz).Tim

author: tree
author_rank: Regular Contributor
timestamp: January 26, 2014, 07:25:48 pm
content: An RLC filter, say bandpass, which is what a crystal is, ideally, has the same properties that you mentioned specifically that noise of a particular frequency gets amplified while all other frequencies get attenuated, which is why it can be modeled as an RLC circuit. How would I begin analyzing the circuit that I attached? If you saw the circuit for the first time, how would you begin to analyze its behavior?In order for the circuit to oscillate, the gain must be greater than 1 and the phase around the loop must be a multiple of 360 degrees. Q1 adds 180 degrees of phase as does Q2, but how does the crystal affect those parameters?

author: G0HZU
author_rank: Super Contributor
timestamp: January 26, 2014, 11:54:16 pm
content: I've not got any experience of that type of oscillator but I think it is pretty easy to understand.The idea is that (with the crystal removed) the gain around the loop must be less than unity. There's a strong clue in that the collector resistor is 0.8 x the emitter resistor. This causes less than unity gain in this amp without the crystal. The phase around the loop without the crystal will be somewhere around 360degrees.When you add the crystal it will reduce the emitter impedance a fair bit at the crystal series resonant frequency and the gain around the loop will wink up above unity at frequencies close to the series resonance of the crystal. The frequency it oscillates at will be the one where the phase around the loop is unity with gain >1 as we all know.If I had to analyse that circuit I'd do it as a 1 port analysis looking up into the emitter crystal. I'd expect to see a sharp dip into negative resistance where it will oscillate. The frequency it would oscillate at would be where the negative resistance has no reactive term. eg at resonance looking into the bottom of the crystal.Alternatively, I'd break the loop open and do a 2 port gain and phase analysis of the circuit.Hope this is OK as I've never looked at that type of oscillator before

author: dannyf
author_rank: Super Contributor
timestamp: January 27, 2014, 12:14:55 am
content: Q1's emitter goes up, causing Q1's collector to go down, which causes Q2's base to go down, which causes Q2's collector to go up, which causes Q1's base / emitter to go up -> positive feedback.

author: tree
author_rank: Regular Contributor
timestamp: January 27, 2014, 01:10:21 am
content: Quote from: dannyf on January 27, 2014, 12:14:55 amI got the same result initially, but then what happens when the voltage of Q2's collector reaches the rail, does it then reverse direction and head down towards ground?

author: G0HZU
author_rank: Super Contributor
timestamp: January 27, 2014, 01:17:28 am
content: To add to my last post, the voltage gain of Q1 (with no crystal fitted) will be approx. = -Rcollector/Remitter = -0.8The voltage gain of Q2 will be approx. = -Rcollector/Remitter = -R2/R2 = -1This is because Q2 has equal resistors in its collector and emitter. It is just there to provide about unity gain around the loop and a phase inversion (180degrees)So without the crystal the overall gain around the loop will be less than 1 but the phase will be in the ballpark of 360degrees. So without the crystal it shouldn't oscillate.At resonance the crystal will have a series resistance of maybe 30 ohms. So it can be modelled as a simple 30R resistor at this one frequency..With the crystal fitted the loop gain will peak up at resonance because the emitter of R1 will have the ESR of the crystal across it ( = 30R typical)So with the crystal fitted and at resonance the gain of Q1 will go UP and suddenly you have more than unity gain around the loop. Only one frequency will have zero phase shift around the loop and >unity gain and this is the frequency it will oscillate at. It will be quite close to the series resonant frequency of the crystal but there will be some deviation from this because the phase through the two inverting amplifiers won't be exactly 360degrees at RF frequencies because each (inverting) amplifier will not give exactly 180degree phase shift due to the real world limitations of the transistors used. eg they have internal capacitance which means the phase shift won't be exactly 180degrees in each of Q1 and Q2.

author: mrkev
author_rank: Regular Contributor
timestamp: January 27, 2014, 02:24:44 am
content: As has been said. When you have resistors in collector and emiter, the gain is about Rc/Re. It's not that hard to figure this out since the collector current is about the same as current through emiter and as that current correspond to the voltage at Re (which is about the same as change at input).So to get higher gain you would need lower impedance in q1 emiter. And that's exactly what that crystal does, it has the lowest impedance at certain frequency.Crystal itself works as any other resonator. Just imagine that you have a coil, some armature and spring. If you apply the voltage to the coil, the armature would be pulled down. When you disconect the coil, the returning armature would generate voltage with opposite direction.The important thing happends when you apply alternate voltage. Impedance of that coil "rezonator" would change with frequency, having extrem exactly at the point of natural oscilation of the spring-armature system.

author: IanB
author_rank: Super Contributor
timestamp: January 27, 2014, 02:52:54 am
content: Quote from: tree on January 27, 2014, 01:10:21 amThe key is in the crystal. It is not a simple first order device like a capacitor, but rather it has "momentum", giving it second order properties.For an analogy, consider a swing. Maybe you remember pushing your baby brother or sister on a swing? If you push the swing all the way up until you run out of push, you will notice you can't hold it there, it pushes back at you. You more or less have to give way and let the swing fall back. Similarly with the crystal. You can charge it up with voltage, but unlike a capacitor it won't simply sit there at the charged voltage. It has some "springiness" and it will attempt to "push back" to release the stored energy. When placed in a suitable circuit the crystal will therefore try to oscillate at a resonant frequency just like a swing or a weighted spring.

author: tree
author_rank: Regular Contributor
timestamp: January 27, 2014, 03:49:19 am
content: Quote from: IanB on January 27, 2014, 02:52:54 amSo in other words, the crystal is NOT a passive component. The larger the E field I apply to it, the harder it tries to push back?

author: IanB
author_rank: Super Contributor
timestamp: January 27, 2014, 06:44:24 am
content: Quote from: tree on January 27, 2014, 03:49:19 amNo, it is passive, just more complex in its properties than a single component like a capacitor alone. Hence the model of a crystal as an RLC circuit.A weight attached to a spring is a very loose analogy, but it is a reasonable one. Imagine a heavy iron bar attached to a spring such that it can move freely without friction. If you pull (or push) the iron bar and let go it will oscillate back and forth with a fixed period of oscillation determined by the weight of the bar and the spring stiffness. Next suppose the bar is a magnet and a solenoid coil is placed around it. Now when you vibrate the bar magnet an AC signal will be generated in the coil at the vibration frequency. Lastly, suppose you pick up the output signal, amplify it, and feed it back in to the coil. You will end up with an oscillator running at the mechanical vibration period of the magnet on the spring.The crystal in an oscillator is not exactly a magnet on a spring, but it behaves in a similar way and the excitation circuit does the same job. It is a feedback circuit that takes the output of the crystal, amplifies it and feeds it back in. Since the crystal will only produce an output voltage corresponding to its natural vibration frequency this is the electrical frequency that will appear in the feedback loop. Any other frequencies will get filtered out.