石英晶体振荡器设计指南

1、 Design Guidelines for Quartz Crystal OscillatorsIntroductionA CMOS Pierce oscillator circuit is well known and is widely used for its excellent frequency stability and the wide range of frequencies over which they can be used. They are ideal for small, low current and low voltage battery operated p

2、ortable products especially for low frequency applications. 1,2 When designing with miniaturized quartz crystals, careful consideration must be given to the frequency, gain and crystal drive level.In this paper, the design equations used in a typical crystal controlled pierce oscillator circuit desi

3、gn are derived from a closed loop and phase analysis. The frequency, gain and crystal drive current equations are derived from this method.Basic Crystal OscillatorThe basic quartz crystal CMOS Pierce oscillator circuit configuration is shown on Figure 1. The crystal oscillator circuit consists of an

4、 amplifying section and a feedback network. For oscillation to occur, the Barkhausen criteria must be meta The loop gain must be equal to orgreater than one, andb The phase shift around the loop mustbe equal to an integral multiple of 2.The CMOS inverter provides the amplification and the two capaci

5、tors, CD and CG , and the crystal work as the feedback network. RA stabilizes the output voltage of the amplifier and is used to reduce the crystal drive level.Figure 1 - Basic Pierce Oscillator CircuitCrystal CharacteristicsIn order to analyze the quartz crystal oscillator, we must first understand

6、 the crystal itself. Figure 2 shows the electrical equivalent circuit of a quartz crystal. The L1, C1 and R1 are generally referred to as the electrical equivalent of the mechanical parameters; inertia, restoring force and friction, respectively. These parameters can be measured using a crystal impe

7、dance meter or a network analyzer. C0 is the shunt capacitance between terminals and the sum of the electrode capacitance of the crystal and package capacitance.R 1 Motional Resistance L 1 Motional Inductance C 1 Motional Capacitance C 0 Shunt CapacitanceFigure 2 Crystal Electrical Equivalent Circui

8、tTN-30Page 2 of 8This equivalent circuit can effectively be simplified as a resistance (Re in series with a reactance (Xe at a frequency f as shown in Figure 3. Figure 3 - Effective Electrical Circuit of a QuartzCrystal R e (f and Xe (f as a function of frequency are as follows:R 1X X mO(-1R X 1O (

9、+R (f=e 22 (1From equation (1 and (2, an example of themagnitude of Re and Xe as a function of frequency are shown in Figures 4 and 5 respectively for f s =32.768kHz, C1=2.4fF, and R1=28k. Thefrequency is expressed in terms of part per million (ppm above the series resonant frequency (fs of REV ACry

10、stal Oscillator DesignThe AC equivalent circuit of the amplifier andfeedback network of a pierce oscillator is shown in Figure 6. For the following analysis, RA is omitted and will be reintroduced later. Figure 6 - Pierce Oscillator AC Equivalent CircuitFrom Figure 6 ; ,andThe phase and amplitude re

11、lationship of the oscillator voltage, current and impedance are shown in Figures 7 and 8. Assume that the oscillator is oscillating at a frequency f and the amplifier output current ID is 180° out of phase with the oscillator input voltage V1. Figure 7 - Current and Voltage Phase Diagram X =X -

12、Xe e G R +X e e 2 2sin =R e, R +X e e 2 2cos =X eFigure 8 - Impedance Phase DiagramFrequency Equation From the imaginary part of the current phase diagram (y-axis(6and from the equations derived from theequivalent circuit, the voltage and impedance phasor diagram equation (6 becomesI =I +I X X XO X

13、X e DfromX = X -X e e G .Then. (7Assuming- 1X X m O (2R X 1O ( 2<<and - 1X X mO R X X 1m O <<9;HREV AEquation (2 becomesX m X X mO X (f=e 1 -(7awhere XO = 1C O and CO = CO + Cs .C s is the circuit stray capacitance across the crystal .Let and .From eq. 7a and 7b one can obtainC LX X m O

14、= X (X - X O m X=m X XX +X O O C L C L Then (8From eq. (3 and (4 From equation (8 . (9Then (10where C L = CS + CL .Equation 10 is the oscillating frequency of thecrystal oscillator. CL is called the load capacitance of the oscillator. With a specified CL , the crystal manufacturer can then match the

15、 crystal to the customers circuit to obtain the desired oscillation frequency. From the CL equation, the relationship between the other circuit parameters can beestablished (i.e. CD , CG , RO and CS as it relates to the oscillation frequency of the crystal oscillator.In a typical CMOS oscillator RO

16、generallydecreases as the supply voltage increases. This causes a decrease in load capacitance and an increase in the oscillation frequency.Figure 9 shows the effective load capacitance (CL changes as the output resistance (RO changes. Figure 9 -Effective Load Capacitance (CL vs. Output Resistance (

17、RO ssREV AGain EquationFrom the real part of the current phase diagram (x-axis;I D = IO cos + I1 sin (11and from the equation derived from the voltage, and impedance phase diagram equation (11 becomes and from Xe = Xe XG and eq. (7 whereEquation (12 gives the minimum gm required for the oscillator t

18、o maintain oscillation. In practice, 5 to 10 times the calculated value is required to insure fast start of oscillation. This equation also aids the designer in selecting the component values for CD and CG to match the CMOS amplifier and the crystal.It is important to note here that in most analyses

19、, only the first term of equation (12 is used. The second term must be taken into accountespecially for low frequency application were the second term becomes larger than the first term as shown in Figure 10, when RO is less than 1.2 M. Figure 10 - Comparison of minimum gmrequirements vs. Amplifiers

20、 output resistance (RO .Whereg m 1 = first term and gm 2 = 2ndterm of equation(12.For CD =20pF, CG =30pF, CS =1.1pF, CO =1.4pF, R 1=28K, fO =32.768kHz and CL =13pF.Using equation (12, Figures 11 and 12 show the change in the minimum gm requirements due to change in either CD or CG , while maintainin

21、g the other capacitor constant. For a 32.768kHz oscillator, as shown in Figure 11, trimming the output capacitor (CG will produce more change in g m than the input capacitor (CD . As shown in Figure 12, a decrease in the amplifiers output resistance (RO increases the minimum gm requirement.R R 1 + e

22、 1(2C C O L TN-30 Gain Equation From the real part of the current phase diagram (x-axis; ID = IO cos + I1 sin (11 Page 5 of 8 It is important to note here that in most analyses, only the first term of equation (12 is used. The second term must be taken into account especially for low frequency appli

23、cation were the second term becomes larger than the first term as shown in Figure 10, when RO is less than 1.2 M. 6 5 and from the equation derived from the voltage, and impedance phase diagram equation (11 becomes g m (u mhos 4 3 2 1 0 0 gm2 gm1 1 2 3 4 RO (M , eq. (7 and from Xe = Xe XG and Figure

24、 10 - Comparison of minimum gm requirements vs. Amplifiers output resistance (RO. Where nd gm1 = first term and gm2 = 2 term of equation(12. For CD=20pF, CG=30pF, CS=1.1pF, CO=1.4pF, R1=28K, fO=32.768kHz and CL=13pF. (12 CO where Re R1 1 + CL ( 2 Equation (12 gives the minimum gm required for the os

25、cillator to maintain oscillation. In practice, 5 to 10 times the calculated value is required to insure fast start of oscillation. This equation also aids the designer in selecting the component values for CD and CG to match the CMOS amplifier and the crystal. Using equation (12, Figures 11 and 12 s

26、how the change in the minimum gm requirements due to change in either CD or CG, while maintaining the other capacitor constant. For a 32.768kHz oscillator, as shown in Figure 11, trimming the output capacitor (CG will produce more change in gm than the input capacitor (CD. As shown in Figure 12, a d

27、ecrease in the amplifiers output resistance (RO increases the minimum gm requirement. REV A STATEK CORPORATION 512 N. MAIN ST., ORANGE, CA 92868 714-639-7810 FAX: 714-997-1256 e-mail:sales TN-30 2.5 Page 6 of 8 Where; and, 2 CG = 30pF g m (u mhos 1.5 1 0.5 0 0 10 20 30 40 CD =30pF CD or CG (pF Figur

28、e 11 - For RO= 2.5M gm comparison between CD and CG, where CS=1.1pF, CO=1.4pF, R1=28K, fO=32.768kHz Figure 13 - Oscillator AC equivalent circuit with the crystal electrical equivalent circuit. The crystal voltage, current and impedance phase relationships are shown in Figure 14 and 15; VOLTAGE CURRE

29、NT 6.7 5.7 CG = 30pF g m (u mhos 4.7 3.7 2.7 1.7 0.7 0 10 20 30 40 CD = 30pF CD or CG (pF Figure 14 - Voltage and current phase relationship with the circuit equivalent CRYSTAL IMPEDANCE Figure 12 - For RO= 500k gm comparison between CD and CG, where CS=1.1pF, CO=1.4pF, R1=28K, fO=32.768kHz Crystal

30、Drive Current In order to analyze the current flowing through the crystal, the AC equivalent circuit from Figure 6 is redrawn to show the crystals electrical equivalent circuit as shown in Figure 13. The crystal drive current is ib , and ia is the current through the shunt capacitance CO. Figure 15

31、Crystal impedance phase diagram REV A STATEK CORPORATION 512 N. MAIN ST., ORANGE, CA 92868 714-639-7810 FAX: 714-997-1256 e-mail:sales TN-30 From; |ia| = e and, XO ia = IX V Page 7 of 8 Typical Effects Of RA In The Oscillator Circuit (13 In many cases, a resistor RA is introduced between the amplifi

32、er output terminal and the crystal input terminal as shown in Figure 1. The use of RA will increase the frequency stability, since it provides a stabilizing effect by reducing the total percentage change in the amplifier output resistance RO and also increases the effective output impedance by RA as

33、 shown on Figure 9. RA also stabilizes the output voltage of the oscillator and is used to reduce the drive level of the crystal. The complete AC equivalent circuit of Figure 1 is shown in Figure 16, where Xd is the total output capacitance of the amplifier. Using the same analytical approach, the f

34、requency, gain and crystal drive current equations with RA are derived. (Re2 + Xe2 XO 1 1 = . (CO (CO + C sm where X O = From the current phase diagram of Figure 14 and the relationship . and from the crystal impedance phase diagram Figure 15 sin = Re Re + Xe 2 2 ; cos = Xe Re + Xe 2 2 . Substitutin

35、g sin and cos and ia from equation (13 or ib = Figure 16 - Pierce oscillator AC equivalent circuit with RA included From the frequency equation (10; Substituting; (10 where; and, CL = CS + CL (14 where Xe = Xe XG From eq. (14 the crystal drive can be calculated from; (in Watts where R1 = crystals mo

36、tional resistance. The gain equation is; gm > 4 f CG (CD + Cd Re + 2 2 (C + R C R d d A O Re REV A STATEK CORPORATION 512 N. MAIN ST., ORANGE, CA 92868 714-639-7810 FAX: 714-997-1256 e-mail:sales TN-30 Page 8 of 8 References: 1 S.S. Chuang and E. Burnett, “Analysis of th CMOS Quartz Oscillator”,

37、Proc. 9 Int. Congress Chronometry (Stuttgart, W. Germany, Sept. 1974 paper C2.2 2 E. Vittoz, “High-Performance Crystal Oscillator circuits: Theory and Application” IEEE Journal of Solid state circuits, vol. 23, No. June 1988 pp.774783. A version of this paper was presented at the 18 Piezoelectric Devices Conf. in Aug. 1996 by Jim Varsovia t