sta540san STMicroelectronics, sta540san Datasheet - Page 16

• Manufacturer Part Number: sta540san
• Description: 4 X 10-watt Dual/quad Power Amplifier
• Manufacturer: STMicroelectronics

Thermal information
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Thermal information
In order to avoid the thermal protection intervention that is placed at T
muting) or T
value correctly.
The parameters that influence the design are:
There is also an additional term that depends on the quiescent current, Iq, but this is
negligible in this case.
Example 1: 4-channel single-ended amplifier
V
The required thermal resistance for the heatsink is
Example 2: 2-channel single-ended plus 1-channel (BTL) amplifier
V
The required thermal resistance for the heatsink is
Design notes on examples 1 and 2
The values found give a heatsink that is designed to sustain the maximum dissipated power.
But, as explained in the applications note AN1965, the heatsink can be smaller when a
realistic application is considered where a musical program is used.
When the average listening power concept is considered, the dissipated power is about 40%
less than the P
is reduced as follows:
Figure 26
CC
CC
"" R
"" R
=14.4 V, R
= 14.4 V, R
Maximum dissipated power for the device (P
Maximum thermal resistance junction to case (R
Maximum ambient temperature T
–
–
P
th_c-amb
th_c-amb
dmax
Example 1: 10.5 W - 40% = 6.3 W giving R
Example 2: 21 W - 40% = 12.6 W giving R
below shows the power derating curve for the device.
j
=160 °C (thermal shutdown), it is important to design the heatsink R
dmax
L
P
L
=
= 4 Ω x 4 channels, R
dmax
= 2 x 2 Ω (SE) + 1 x 4 Ω (BTL), P
=
=
2
. Therefore, in examples 1 and 2, the resulting average dissipated power
150 T
-------------------------------------- - R
⋅
150 T
-------------------------------------- - R
----------------- -
2Π
V
=
CC
–
P
2
–
P
NChannel
dmax
R
dmax
2
amb_max
L
amb_max
Doc ID 18306 Rev 1
+
2V
---------------- -
Π
2
CC
amb_max
R
th_j-case
–
–
L
2
⋅
----------------- -
2Π
=
th_j-case
V
th_j-case
2 5.25
CC
= 2.5 °C/W, T
2
R
⋅
2
dmax
L
out
=
th_c-amb
=
th_c-amb
th_j-case
=
)
= 2 x 12 W + 1 x 26 W
4 2.62
+
150 50
--------------------- - 2.5
150 50
--------------------- - 2.5
⋅
10.5
10.5
21
amb_max
–
)
= 5.4 °C/W
= 13.4 °C/W
–
=
=
21W
–
10.5W
–
= 50 °C, P
j
=150 °C (thermal
=
=
2.2°C/W
7°C/W
out
STA540SAN
th
= 4 x 7 W
(°C/W)