Crystal Oscillator/External Clock - Analog Devices ADE9000 Technical Reference Manual

UG-1098

CRYSTAL OSCILLATOR/EXTERNAL CLOCK

The ADE9000 contains a crystal oscillator. Alternatively, a
digital clock signal can be applied at the CLKIN pin of the
ADE9000.
When a crystal is used as the clock source for the ADE9000,
attach the crystal and the ceramic capacitors, with capacitances
of CL1 and CL2, as shown in Figure 8. It is not recommended to
attach an external feedback resistor in parallel to the crystal.
When a digital clock signal is applied at the CLKIN pin, the
inverted output is available at the CLKOUT pin. This output is
not buffered internally and cannot be used to drive any other
external devices directly. Note that CLKOUT is available in
PSM0 operating mode only.
2.5MΩ
2.5kΩ
C
IN1
CLKIN 29
24.576MHz
C C
L1
P1
Figure 8. Crystal Application Circuit
Crystal Selection
The transconductance of the crystal oscillator circuit in the
ADE9000, gm, is provided in the data sheet. It is recommended
to have 3 to 5 times more gm than the calculated gm
the crystal.
The following equation shows how to calculate the gm
for the crystal from information given in the crystal data sheet:
× 1000 × (2 × π × f
gm = 4 × ESR
CRITICAL MAX
where:
gm is the minimum gain required to start the crystal,
CRITICAL
expressed in mA/V.
ESR is the maximum ESR, expressed in ohms.
MAX
f is 24.576 MHz, expressed in Hz as 24.576 × 10
CLK
C is the maximum shunt capacitance, expressed in farads.
0
C is the load capacitance, expressed in farads.
L
Crystals with low ESR and smaller load capacitance have a
lower gm and are easier to drive.
CRITICAL
The ADE9000 evaluation board uses a crystal manufactured by
Abracon (Part Number ABLS-24.576MHZ-8-L4Q-F-T), which
has a maximum ESR of 40 Ω, load capacitance of 8 pF, and
maximum shunt capacitance of 7 pF, which results in a gm
of 0.86 mA/V:
× 1000 × (2 × π × f
gm = 4 × ESR
CRITICAL MAX
= 4 × 40 × 1000 × (2 × π × 24.576 × 10
gm
CRITICAL
10 + 8 × 10 ) = 0.86
−12 −12 2
1.75kΩ
C
IN2
30 CLKOUT
C C
L2 P2
for
CRITICAL
CRITICAL
) 2 × (C + C ) 2
CLK(Hz) 0 L
.
6
CRITICAL
) × (C + C )
2 2
CLK(Hz) 0 L
6 ) 2 × (7 ×
Rev. 0 | Page 6 of 86
ADE9000 Technical Reference Manual
The gain of the crystal oscillator circuit in the ADE9000, gm,
provided in the data sheet is more than 5 times gm
therefore, there is sufficient margin to start up this crystal.
Load Capacitor Calculation
Crystal manufacturers specify the combined load capacitance
across the crystal, CL. The capacitances in Figure 8 can be
described as follows:
•
CP and CP : parasitic capacitances on the clock pins
1 2
formed due to printed circuit board (PCB) traces.
•
Cin and Cin : internal capacitances of the CLKIN and
1 2
CLKOUT pins respectively.
•
CL and CL : selected load capacitors to get the correct
1 2
combined CL for the crystal.
The internal pin capacitances, C
shown in the data sheet. To find the values of CP
measure the capacitance on each of the clock pins of the PCB,
CLKIN and CLKOUT, respectively, with respect to the AGND
pin. If the measurement is done after soldering the IC to the
PCB, subtract out the 4 pF internal capacitance of the clock pins
to find the actual value of parasitic capacitance on each of the
crystal pins.
To select the appropriate capacitance value for the ceramic
capacitors, calculate CL and CL
1
CL = [(CL + CP + C
1 1 IN1
CP + C + CL + CP + C
1 IN1 2 2
Select CL and CL such that the total capacitance on each clock
1 2
pins is
CL + CP + C = CL + CP
1 1 IN1 2
Using Equation 1 and Equation 2, the values of CL
be calculated.
Load Capacitor Calculation Example
If a crystal with load capacitance specification of 8 pF is selected
and the measured parasitic capacitances from the PCB traces
are CP1 = CP2 = 2 pF, Equation 1 implies that
CL = [(CL + CP + C
1 1 IN1
CP + C + CL + CP + C
1 IN1 2 2
8 pF = [(CL + 2 pF + 4 pF) × (CL
1
2 pF + 4 pF + CL + 2 pF + 4 pF)
2
Assuming that CL = CL , to satisfy Equation 2,
1 2
8 pF = [(CL + 6 pF) × (CL
1
8 pF = [(CL + 6 pF) × (CL
1
8 pF = (CL + 6 pF)/2
1
Therefore, CL = CL = 10 pF.
1 2
Based on this example, 10 pF ceramic capacitors are selected for
CL and CL .
1 2
CRITICAL
and C , are 4 pF each, as
in1 in2
and CP
1
from the following equation:
2
) × (CL + CP + C )]/(CL
2 2 IN2 1
)
IN2
+ C
2 IN2
and CL
1
) × (CL + CP + C )]/(CL
2 2 IN2 1
)
IN2
+ 2 pF + 4 pF)]/(CL
2
+ 6 pF)]/(CL + 6 pF + CL
1 1 1
+ 6 pF)]/[2 × (CL + 6 pF)]
1 1
;
,
2
+
(1)
(2)
can
2
+
+
1
+ 6 pF)