FUNCTIONAL DESCRIPTION
(continued)
VIII.2 - How to choose the value of the time
constant
The time response of the PLL is given by its char-
acteristicequationwhich is :
(
x
±
1
)
2
+ (α + β)
(
x
±
1
) + β =
0.
Where :
α =
3
LD[5:0]
2
A
±
11
and
β =
3
LD[5:0]
2
B
±
19
.
(LD[5:0] = value of the LINE DURATION register,
A = value of the 1st time constant, AF or AS and
B = value of the 2
d
timeconstant, BF or BS).
As you can see, the solution depend only on the
LINE DURATION and the TIME CONSTANTS
given by the I
2
C registers.
If
(α + β)
2
±
4
β
≥
0 and 2
α ± β <
4, thePLL is sta-
ble
and its response is like this presented on
Figure11.
t
PLL
Frequency
f
0
f
1
t
Input
Frequency
f
0
f
1
9
Figure 11 :
Time Response of the PLL/Charac-
teristic EquationSolutions (with
Real Solutions)
If
(α + β)
2
±
4
β
≤
0, the responseof the PLL is like
this presented on Figure12.
In this case the PLL is stable if
τ
> 0.7 damping
coefficient).
t
PLL
Frequency
f
0
f
1
t
Input
Frequency
f
0
f
1
9
Figure 12 :
Time Response of the PLL/Charac-
teristic Equation Solutions(with
ComplexSolutions)
The Table 2 gives some good values for A and B
constants for different values of the LINE DURA-
TION.
Summary
For a good working of the PLL:
- Aand B time constants must be chosen among
values for which the PLL is stable,
- B mustbe equalor greater thanA and the differ-
ence betweenthem must beless than 3,
- The greater (A, B) are, the faster the captureis.
An optimalchoiceforthemostofapplicationsmight
be :
- For locking condition: AS = 0 and BS = 1,
- For capture process : AS = 2 and BS = 4.
But for eachapplicationthe time constants can be
calculated by solving the characteristic equation
and choosingthe best response.
Table 2 :
Valid Time ConstantsExamples
B \ A
0
1
2
3
4
5
6
7
0
1
2
3
4
5
6
YYYY
YYYY
NYYY
NNNY
NNNN
NNNN
NNNN
NNNN
YYYY
YYYY
YYYY
YYYY
NYYY
(1)
NNNY
NNNN
NNNN
YYYY
YYYY
YYYY
YYYY
YYYY
YYYY
NYYY
NNNY
YYYN
YYYN
YYYN
YYYN
YYYN
YYYN
YYYN
YYYN
YNNN
YNNN
YNNN
YNNN
YNNN
YNNN
YNNN
YNNN
NNNN
NNNN
NNNN
NNNN
NNNN
NNNN
NNNN
NNNN
NNNN
NNNN
NNNN
NNNN
NNNN
NNNN
NNNN
NNNN
9
Note :
1. Case of A[2:0] =1 (001) and B[2:0]= 4(100) :
LD
Valid TimeConstants
16
N
32
Y
48
Y
63
Y
Value of LINE DURATION Register (@ 3FF0) :
LD = 16 : LD[5:0] = 010000
LD = 32 : LD[5:0] = 100000
LD = 48 : LD[5:0] = 110000
LD = 63 : LD[5:0] = 111111
Table meaning :
N = No possible capture
Y = PLLcan lock
STV9422 - STV9424
13/15