TS489IQT STMicroelectronics, TS489IQT Datasheet - Page 23

• Manufacturer Part Number: TS489IQT
• Description: IC AMP AUDIO PWR .13W STER 8DFN
• Manufacturer: STMicroelectronics
• Type: Class ABr

Specifications of TS489IQT

• Output Type: Headphones, 2-Channel (Stereo)
• Max Output Power X Channels @ Load: 130mW x 2 @ 16 Ohm
• Voltage - Supply: 2.2 V ~ 5.5 V
• Features: Depop, Short-Circuit Protection, Standby
• Mounting Type: Surface Mount
• Package / Case: 8-DFN
• For Use With: 497-6363 - BOARD DEMO FOR TS489
• Lead Free Status / RoHS Status: Lead free / RoHS Compliant
• Other names: 497-5991-2

TS488-TS489
4.2
4.3
and its value is:
Note:
The efficiency is the ratio between the output power and the power supply:
The maximum theoretical value is reached when V
Total power dissipation
The TS488/9 is stereo (dual channel) amplifier. It has two independent power amplifiers.
Each amplifier produces heat due to its power dissipation. Therefore the maximum die
temperature is the sum of each amplifier’s maximum power dissipation. It is calculated as
follows:
■
■
■
Typically, P
Lower cut-off frequency
The lower cut-off frequency F
capacitors C
The input capacitor C
resistor R
the lowest frequency to be amplified (with a 3dB attenuation), the minimum value of the C
(C
out
P
P
Total P
diss R
diss L
) is:
This maximum value depends only on power supply voltage and load values.
L
= Power dissipation due to the left channel power amplifier.
= Power dissipation due to the right channel power amplifier.
diss
) of the amplifier is equivalent to a first order high pass filter. Assuming that F
diss R
out
= P
.
is equal to P
diss R
in
+ P
(output capacitor C
diss L
TotalP
diss L
CL
TotalP
of the amplifier depends on input capacitors C
(W)
, giving:
diss
C
η
C
P
out
=
diss
in
diss
η
=
------------------ -
P
=
P
=
=
2 2V
---------------------- P
MAX
supply
=
OUT
--------------------------------------- -
2π
π R
-- -
4
π
out
------------------------------------- -
2π
2P
=
=
⋅
) in serial with the input resistor R
⋅
CC
dissR
L
F
------------ - W
π
78.5%
V
=
F
2
CL
1
2
CC
1
CL
R
πV
------------------
2V
L
⋅
⋅
(
=
OUT
peak
R
CC
peak
R
2P
in
)
L
–
dissL
2P
= V
OUT
CC
/2, so
Application information
in
in
and output
(load
CL
23/32
is
in