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2003 Dec 16
57
Philips Semiconductors
Preliminary specification
30 Mbit/s to 3.2 Gbit/s A-rate
4-bit fibre optic transceiver
TZA3015HW
APPLICATION INFORMATION
Calculations on the clean-up PLL
The important specifications of the clean-up PLL are the
bandwidth(f
3dB
)andthejitterpeaking.Iftheseareknown,
the component parameters can be calculated. First
assume that the bandwidth of the VCXO control input
(f
3dB(vcxo)
)ismuchhigherthanf
3dB
andC2isleftout.This
simplifies the loop into a second order, type II PLL. In a
second order PLL, the damping factor
ζ
determines the
amount of peaking. To obtain peaking of less than 0.1 dB,
ζ
must be higher than 4.3. For peaking of less than
0.05 dB,
ζ
must be higher than 6. See Fig.30 for an
example.
Now R and C1 may be calculated with the following
formulas:
Where:
RDIV = reference divider ratio (1, 2, 4, 8, 16 or 32)
f
3dB
= clean-up PLL bandwidth in Hz
K
VCXO
= VCXO gain in Hz/V
I
CP
= charge pump current in A (100
μ
A or 1 mA,
depending on I
2
C-bus bit CLUPPLLHG)
ζ
= damping factor.
These formulas are valid if:
ζ
>> 1 and
f
3dB(VCXO)
> 2
×
f
3dB
and
C2
+
π
×
R
×
C1
×
.
Thetransferhasafirstorderroll-off(i.e.20 dB/decade),up
to the bandwidth of the VCXO control input. If a second
order roll-off is required C2 may be added, as long as
C2
+
2
π
×
C1
×
C2
×
Example: The clean-up PLL uses a VCXO with a
frequency of 20 MHz and has a gain K
VCXO
= 2000 Hz/V.
The bandwidth of the control input is f
3dB(VCXO)
= 10 kHz.
Since the reference frequency is 20 MHz, the reference
divider ratio RDIV = 1. According to the specification, the
maximum allowed jitter peaking is 0.1 dB. To add some
margin the design is for less than 0.05 dB peaking, so
ζ
= 6. Also according to the specification, f
3dB
should be
less than 100 kHz. To satisfy the conditions as previously
described, f
3 dB
< 0.5
×
f
3 dB(VCXO)
< 5 kHz. To cope with
component tolerances, f
3dB(VCXO)
= 2.5 kHz is chosen.
Choosing I
CP
= 1 mA yields R = 7854
and
C1 = 1.167
μ
F.
To calculate f
-3dB
and
ζ
, if R and C1 are known, use the
following formulas:
R
RDIV
---------K
2
×
π
×
f
CP
×
VCXO
=
C
1
K
RDIV
I
×
2
π
2
×
f
3dB
–
(
)
2
------------------------------------------------------
=
C2
×
2
2
f
3dB
–
×
>
--------C1
2
f
3dB
–
×
>
R
2
×
π
×
2500
10
×
×
2000
100
6
–
×
-1
78540
=
=
C
1
1
100
μ
F
π
2
×
×
6
2
×
×
2500
2
2000
116.7nF
=
=
f
3dB
–
K
2
CP
RDIV
×
R
×
π
×
-----------------------I
=
ζ
R
2
K
--------------RDIV
I
×
C1
×
×
=