STA540 STMicroelectronics, STA540 Datasheet - Page 13

• Manufacturer Part Number: STA540
• Description: IC AMP AUD DUAL/QUAD MULTIWATT15
• Manufacturer: STMicroelectronics
• Type: Class ABr

Specifications of STA540

• Output Type: 2-Channel (Stereo) or 4-Channel (Quad)
• Max Output Power X Channels @ Load: 38W x 2 @ 4 Ohm; 16W x 4 @ 4 Ohm
• Voltage - Supply: 8 V ~ 22 V
• Features: Short-Circuit and Thermal Protection, Standby
• Mounting Type: Through Hole
• Package / Case: Multiwatt-15 (Vertical, Bent and Staggered Leads)
• Amplifier Class: AB
• No. Of Channels: 4
• Output Power: 24W
• Supply Voltage Range: 8V To 22V
• Thd + N: 0.02% @ 4W, 4ohm, VS=14.4V
• Load Impedance: 8ohm
• Operating Temperature Range: -40°C To +150°C
• Rohs Compliant: Yes
• Lead Free Status / RoHS Status: Lead free / RoHS Compliant
• Other names: 497-8880-5

STA540
5
5.1
5.1.1
5.1.2
Thermal information
In order to avoid the intervention of the thermal protection, placed at T
muting and T
resistance, R
The parameters that influence the calculation are:
"
"
"
There is also an additional term that depends on the Iq (quiescent current).
Heatsink specification examples
R
Given V
P
the maximum power dissipated in the device is:
and the required thermal resistance of the heatsink is:
R
Given V
the maximum power dissipated in the device is:
and the required thermal resistance of the heatsink is:
out
th_HS
th_HS
= 4 x 7 W then
maximum dissipated power for the device (P
maximum thermal resistance junction to case (R
maximum ambient temperature T
S
S
calculation for 4 single-ended channels
calculation for 2 single-ended channels plus 1 BTL channel
= 14.4 V, R
= 14.4 V, R
R
R
j
th_HS
=160° C for thermal shutdown, it is important to calculate the heatsink thermal
th_HS
th_HS
, correctly.
P
dmax
=
L
P
L
=
= 2x 2 Ω (SE) + 1x 4 Ω (BTL), P
= 4 Ω x 4 channels, R
dmax
150 T
----------------------------------------- - R
150 T
----------------------------------------- - R
=
–
P
–
P
2
=
dmax
dmax
amb_max
⋅
NChannel
amb_max
------------------ -
2Π
V
CC
2
R
2
L
amb_max
+
–
–
2V
----------------- -
Π
⋅
th_j-case
2
------------------ -
2Π
th_j-case
th_j-case
CC
V
R
CC
L
2
2
R
dmax
2
=
L
2 5.25
= 1.8° C/W, T
=
=
=
⋅
th_j-case
)
150 50
--------------------- - 1.8
4 2.62
out
150 50
--------------------- - 1.8
⋅
10.5
= 2 x 12 W + 1 x 26 W then
21
–
–
+
)
10.5
=
–
–
10.5W
amb_max
=
21W
=
=
Thermal information
j
7.7°C/W
=150° C for thermal
3°C/W
= 50° C and
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