STEVAL-CCA003V1 STMicroelectronics, STEVAL-CCA003V1 Datasheet - Page 21
![BOARD EVAL FOR LV TS4984IQT](/photos/9/12/91255/steval-cca003v1_sml.jpg)
STEVAL-CCA003V1
Manufacturer Part Number
STEVAL-CCA003V1
Description
BOARD EVAL FOR LV TS4984IQT
Manufacturer
STMicroelectronics
Specifications of STEVAL-CCA003V1
Design Resources
STEVAL-CCA003V1 Gerber Files
Amplifier Type
Class AB
Output Type
2-Channel (Stereo)
Max Output Power X Channels @ Load
1W x 2 @ 8 Ohm
Voltage - Supply
2.2 V ~ 5.5 V
Operating Temperature
-40°C ~ 85°C
Board Type
Fully Populated
Utilized Ic / Part
TS4984
Description/function
Audio Amplifiers
Operating Supply Voltage
2.2 V to 5.5 V
Output Power
1 W
Product
Audio Modules
Silicon Manufacturer
ST Micro
Silicon Core Number
TS4984
Kit Application Type
Amplifier
Application Sub Type
Stereo Audio Power Amplifier
Kit Contents
Board
Rohs Compliant
Yes
Lead Free Status / RoHS Status
Lead free / RoHS Compliant
For Use With/related Products
TS4984
Other names
497-6396
STEVAL-CCA003V1
STEVAL-CCA003V1
Available stocks
Company
Part Number
Manufacturer
Quantity
Price
Application Information
Then, the power dissipated by each amplifier is:
and the maximum value is obtained when:
and its value is:
Note:
The efficiency, η, is the ratio between the output power and the power supply:
The maximum theoretical value is reached when V
The TS4984 has two independent power amplifiers, and each amplifier produces heat due to its power
dissipation. Therefore, the maximum die temperature is the sum of the each amplifier’s maximum power
dissipation. It is calculated as follows:
In most cases, P
or, stated differently:
This maximum value is only depending on power supply voltage and load values.
diss L
P
P
diss R
= P
diss L
diss R
Total P
= Power dissipation due to the right channel power amplifier.
= Power dissipation due to the left channel power amplifier.
, giving:
P d iss
P
diss
diss
Total P
P
dissmax
=
Total P
=
η =
=
2 2 V
----------------------- -
4 2 V
----------------------- - P
π R
d iss
P
π
sup ply
∂P d iss
--------------------- = 0
-------------------- - =
P
∂P
diss
R
CC
L
P
supply
=
π
---- - = 78.5%
4
CC
L
=
o ut
out
= P
PEAK
------------ -
π
2V
⋅
2P
2
–
R
diss L
P
cc
2
dissL
out
P
L
out
= V
out
------------------------ -
π
–
4V
V
+ P
–
CC
2P
PEAK
P
CC
, so that:
out
diss R
(
out
W
(W)
(
)
W
)
(W)
(
(
W
W
)
)
TS4984
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