TDF8599 NXP Semiconductors, TDF8599 Datasheet - Page 36

• Manufacturer Part Number: TDF8599
• Description: I2C-bus controlled dual channel 43 W/2 W single channel 85 W/1 W class-D power amplifier
• Manufacturer: NXP Semiconductors

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NXP Semiconductors
TDF8599_1
Product data sheet
14.4 Speaker configuration and impedance
14.5 Heat sink requirements
A flat-frequency response (due to a 2
the low-pass filter components (L
impedance.
Table 22.
Remark: When using a 1
after the low-pass filter switches two 2
In some applications, it may be necessary to connect an external heat sink to the
TDF8599. Thermal foldback activates at T
relationship between the maximum power dissipation before activation of thermal foldback
and the total thermal resistance from junction to ambient;
P
function of output power is given in
function of output power (see
Example 1:
The total thermal resistance R
Where:
If an audio signal has a crest factor of 10 (the ratio between peak power and average
power = 10 dB) then T
R
Load impedance ( )
1
2
4
max
•
•
•
•
•
•
•
•
•
•
th j a
V
P
T
T
P
The required R
R
Thermal resistance from junction to case (R
Thermal resistance from case to heat sink (R
mounting)
Thermal resistance from heat sink to ambient (R
19
–
is determined by the efficiency ( ) of the TDF8599. The efficiency measured as a
j(max)
amb
p
o
max
th(j-c)
= 14.4 V
= 2
=
(1.1 + 1) = 17 K/W.
= 25 C
= 5.8 W (from
+ R
Filter components values
= 140 C
T
----------------------------------- -
Table 22
25 W into 4
j max
th(c-h)
P
max
–
th(j-a)
+ R
T
Rev. 01 — 13 November 2008
shows the required values.
amb
j
th(h-a)
will be much lower.
Figure
= 115 C/5.8 W = 19 K/W
(THD = 10 % continuous)
load impedance in Parallel mode, the outputs are shorted
Figure
L
2.5
5
10
th(j-a)
39)
LC
LC
( H)
, C
Figure
consists of:
32).
nd
LC
order Butterworth filter) is obtained by changing
Class-D power amplifier with load diagnostics
) based on the speaker configuration and
filters in parallel.
j
39. The power dissipation can be derived as a
= 145 C. The expression below shows the
th(j-c)
th(c-h)
) = 1.1 K/W
th(h-a)
) = 0.5
) would then be
C
4.4
2.2
1
LC
1 K/W (depending on
( F)
TDF8599
© NXP B.V. 2008. All rights reserved.
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