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Quartz Crystal Modelling [复制链接]

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发表于 2009-12-30 23:23 |显示全部楼层

                Quartz crystals used in oscillators can be modelled as a series RLC circuit with a parasitic packaging capacitance (C2) in parallel, as shown in figure 1.


This equivalent circuit exhibits both series and parallel resonance. Series resonance occurs at the frequency where the reactances of L1 and C1 are equal. At a slightly higher frequency, parallel resonance occurs when the combination of L1 and C1 exhibit an inductive susceptance which resonates with C2. Assume all frequencies (f) are in radians/sec:
Series resonance: |XL1| = |XC1|
fsL1 = 1/(fsC1)
fs= (L1*C1)-1/2
Parallel Resonance: |XC2| = |XL1 + XC1| = |XL1| - |XC1|
1/(fpC2) = (fpL1)-1/(fpC1)
1/(fpC2) + 1/(fpC1) = fpL1
(1/C1)(1+C1/C2) = (fp)2L1
(1+C1/C2)/(C1*L1) = (1+C1/C2)(fs)2=(fp)2
or,
fp= fs(1+C1/C2)1/2
since C11/2= 1+(1/2)(C1/C2), thus:
fp= fs + df
where df = (1/2)(C1/C2)fs
If we assume that the Pierce oscillator circuit operates approximately midway between series and parallel resonances, it is possible to determine values which can make the model of figure 1 useful. Also assume the series resistance of the crystal is about 600 ohms, the packaging capacitance is 8 pF,and the Q is approximately 10,000.
If we accurately measure the frequency of operation of the Pierce oscillator as, say, 75398223 r/s, then
XL= f*L1 = Q*R1
thus
L1 = Q*R1/f = 10,000*600/75398223 = .07957747 Henry
If we say L1 = .08 Henry, then Q will not exactly equal 104 but that's OK. We must find C1 in two steps. First, find an approximate value by recognizing that f is approximately equal to fs. Then C1 can be computed approximately by
C1 = 1/(L1*f2) = 1/((0.08)(75398223)2) = 2.2 fF (approximately)
We can now approximate df,
df = (1/2)(C1/C2)f = (0.5)(2.2E-15/8E-12/)(75.4E6) = 10368 r/s
If the operating frequency of the circuit is midway between series and parallel resonance, then the series resonance frequency must be equal to the operating frequency minus df/2, or 75393039 r/s. We can now use this value of fs to compute a precise value to use for C1:
C1 = 1/(L1*(fs)2) = 1/((0.08)(75393039)2) = 2.199113 fF
Our model thus contains:
L1 = 0.08 H
C1 = 2.199113 fF
R1 = 600 ohms
C2 = 8 pF
fs = 75393039 r/s
df = 10362 r/s
fp = fs + df = 75393039 r/s + 10362 r/s = 75403401 r/s


In Millimetric Applications, an oscillator is a common module in many analogue circuits. The following article offers some suggestions of how to model the piezoelectric resonator that is its principal component.
The RLC circuit representing the impedance of a crystal resonator is shown by a two-terminal equivalent in figure 1.

The circuit above shows, the static capacitance C0, representing the effect of an insulator separated by two electrodes, and the motional branch L1-C1-R1 is an equivalent to an isolated mode of piezoelectric vibration, simulating inertia, restoring force and friction respectively. Devices in a metal can may need to be modelled by a three terminal equivalent that includes the capacitance of the holder, see figure 2. If necessary, other unwanted oscillation modes such as spurs may be simulated as additional motional branches in parallel with L1-C1-R1.
Accurate calculation of the component values in the equivalent circuit is vital to give a reliable simulation. Loading the standard model, usually for a 10MHz crystal, will produce the wrong frequency for parallel or series resonant modes: also, in a pulling circuit, incorrect values of motional capacitance will give an erroneous sensitivity to a varying load capacitance.
Taking, as example, a 27MHz fundamental crystal with a load capacitance of 12pF and pulling sensitivity of 20ppm/pF. The equation of motional resonance is:
Fs = 1/ [2??(L1 * C1)]
For parallel resonance :
Fp = Fs[1 + C1/(2*C0)] = 27.000MHz
For pulling sensitivity :
S = 1000000 * C1/ 2(C0 + Cl)² = 20ppm/pF
Where CL, the load capacitance, is 12pF. If we specify C0, as a shunt capacitance of 3pF, then the motional capacitance C1 is 9pF, Fs becomes 26.95956MHz and the motional inductance L1 is 3.87mH. A reasonable range of values for motional resistance would be 20 ? for a 30MHz crystal and 60 ? for a 10MHz. For other frequencies:
Frequency
R1
L1
C1
C0
MHz
Ohms
mH
pF
pF





2.00
200
520
0.012
4
5.00
50
115
0.010
3
15.00
30
12.5
0.009
5
30.00
20
4.7
0.006
3
Simulation of a ceramic resonator is more difficult, as these three terminal parts contain internal load capacitors, and the ratio of reactance to impedance is lower than the value for crystals. The resonance frequency is also lower than that for a crystal, so the size of the reactive components are further reduced. This reduction in the Quality Factor means a faster start-up but the drawback is that load capacitor values have a greater influence on oscillator frequency than with a crystal.
Typical values for two terminal resonators are:
Frequency
R1
L1
C1
C0
MHz
Ohms
mH
pF
pF
3.58
7
0.113
19.6
140
6.0
8
0.094
8.3
60
8.0
7
0.092
4.6
40
11.0
10
0.057
3.9
30
Modelling the start-up of any oscillator is also a difficult exercise. The Barkhausen criteria are: loop gain in the oscillator has to be equal to or greater than unity, and phase shift must be a multiple of 2 ?, for the circuit to start. In practice, the addition of a large resistor in parallel with crystals or resonators improves the starting, so a resistor value of 1Megohm across resonators, 10Megohm across crystals running above 1MHz, and 22Megohm across watch crystals is likely to get some positive results in a simulation too.
               
               

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