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參數(shù)資料
| 型號： | TZA1047HL/M1B |
| 廠商： | NXP SEMICONDUCTORS |
| 元件分類： | 消費家電 |
| 英文描述： | SPECIALTY CONSUMER CIRCUIT, PQFP64 |
| 封裝： | 10 X 10 MM, 1.4 MM HEIGHT, PLASTIC, MS-026, SOT314-2, LQFP-64 |
| 文件頁數(shù)： | 15/69頁 |
| 文件大?。?/td> | 324K |
| 代理商： | TZA1047HL/M1B |
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9397 750 14664
Koninklijke Philips Electronics N.V. 2005. All rights reserved.
Product data sheet
Rev. 02 — 1 June 2005
22 of 69
Philips Semiconductors
TZA1047
Preprocessor IC for CD and DVD rewritable
The left-hand side of this diagram shows the Balancer and Normalizer sections. The input
signals L and R are low-pass ltered before processing. When balancing is switched on
(bit BAL = 1), it outputs a normalized, current mode, push-pull signal that can be used for
wobble read-out of a pre-groove. The gain balancer is used to balance the low-pass
ltered left and right central spot signals in such a way that the resultant normalized output
signal (ppn) is free of DC. The integrator time constant of the Balancer,
τ
PP, is controlled
by parameter TBAL[1:0].
The relationship between the Gain Balancer and Normalizer is given in the following
equations:
if BAL = 1
if BAL = 0
On the right-hand side of the diagram, the balanced and normalized signal ppn is scaled
by current IPPN, converted to a voltage by RPPOUT and nally amplied at a gain selected
by GPP[3:0]. The output signal is low-pass ltered at an adjustable bandwidth of between
2 MHz and 20 MHz to improve signal-to-noise ratio. The various ampliers and lters are
controlled by IPPN[3:0], GPP[3:0] and BWPP[2:0]. The components of the output signal
PPN are:
PPN = ppn
× I
PPN × RPPOUT × GPP
where IPPN is the decimal value of control word IPPN[3:0], and kPPN is the scaling current
factor.
If not used, the P-P Normalizer/Balancer can be switched off to save power by setting bit
PPBAL to logic 0.
The output signal at pin PPN is a bipolar voltage around 0.5VDD.
7.9.1 Test mode
The control bit PPTST1 is intended for test and evaluation use only and should be set to
logic 0 for normal operation.
L
B
1V
c
–
()L
=
R
B
1V
c
+
()R
=
ppn
R
B
L
B
–
()
L
B
R
B
+
()
------------------------
=
V
c
1
τ
PP
--------
ppn
() t
d
∫
=
V
c
0
=
I
PPN
I
0PPN
,
10
0.05
k
PPN
IPPN
×
=
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