mauifan wrote:I saw that crystal equivalent before. Although physically a crystal is just a piece of quartz sandwiched between two plates, it acts like very precise components as shown in the model. Who knows what those actual values really are, but if for the sake of argument I assume that the "inductor" in a 4MHz crystal is 100uH, I would need a total capacitance of about 15pF. The manufacturer of the crystal may have constructed the crystal with a 5pF cap, so the manufacturer would say that the load capacitance is 10pF.
Sorry for the lag in replying. You've advanced to the point where the accuracy of your predictions is no longer limited by your understanding of the tools. Now it's limited by the model you're using. A crystal does things you just can't describe with a straight LC filter.
That means I need to explain the equivalent circuit model, and to do that I had to get my own understanding solid enough that I could break it down for someone else.
So.. to get started let's ignore the parallel cap and the load cap, and just concentrate on the branch that contains the resistor, inductor, and series capacitor. That's the part which represents the crystal itself. The inductor describes the crystal's inertia with respect to vibration, the capacitance describes the stiffness of the quartz, and the resistance describes energy losses in the lattice.
We can more or less ignore the resistor. It doesn't do anything complicated to the response curves. The inductor and capacitor do.
If you just look at just those two components, they make a bandpass filter. The inductor blocks high frequencies and the capacitor blocks low frequencies. The output ends up being a compromise between the fastest signals that can get through the inductor and the slowest signals that can get through the cap.
The transfer characteristics look like this:
The measurements are taken at the white circle on the right. The black line shows amplitude at a given frequency, the red line shows phase lag. As you can see, the phase changes abruptly at the resonant frequency.
I'm drawn the circuit with a load resistor instead of a load cap because every new capacitor changes the shape of the curve.
Speaking of capacitors, the parallel cap represents the capacitance between the contacts on either side of the crystal. It doesn't change the frequency much, but it does change the transfer function:
The amplitude response becomes flat and the phase response becomes zero for frequencies away from the RLC filter's resonant frequency. In other words, the effects of the parallel cap cancel the effects of the RLC filter everywhere except frequencies near resonance.
Near resonance, the effects of the RLC filter become too big for the parallel cap to swallow. We get an upwards spike where where frequencies pass through easily, and a downwards spike where frequencies are blocked.
Even the high point of the spike is 6db below zero, mostly because energy escapes through the resistor to GND. We plug that hole by replacing the load resistor with a load cap:
Now the pulses of current that come out of the RLC filter can't escape to GND. The high and low spikes are still there, but now the high spikes produce output that's significantly larger than the input.
The load cap doesn't have much effect on the overall shape of the curve, but it has a large effect on the gain. It controls which point on the curve crosses the 0db line, thus setting the resonant frequency for the circuit as a whole.
The exact point where the curve crosses the 0db line is determined by the load capacitance and the amount of inductive dominance not being cancelled out by anything else.
As far as a Pierce oscillator is concerned, the 0db crossing isn't the important bit though. That notch in the phase response is.
If you recall, we build a Pierce oscillator so it wants 120-160 degrees of phase shift from the crystal. The crystal's phase response is zero everywhere except in the notch, and there are two places that provide the exact amount of shift that we want within the notch.. on the left and one on the right:
The gain on the left side is about 10db, the gain on the right side is about -10db. The stronger one is the one that will create a positive fedback loop.
The numbers are all speculative, but the overall pattern of behavior is accurate. And now, since this post has been sitting in my browser for about a week, I'm going to hit 'Submit' and get the conversation moving again.
When you void a product warrany, you give up your right to sue the manufacturer if something goes wrong and accept full responsibility for whatever happens next. And then you truly own the product.
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