你撒下一粒善良的种_子, 会在暗中蔓延出一片灿烂的花海
分类: LINUX
2009-12-30 23:23:33
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 C1<
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
The RLC circuit representing the impedance of a crystal resonator is shown by a two-terminal equivalent in figure 1.
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.