TS4902IST STMicroelectronics, TS4902IST Datasheet - Page 15

• Manufacturer Part Number: TS4902IST
• Description: IC AMP AUDIO PWR .7W MONO 8MSOP
• Manufacturer: STMicroelectronics
• Type: Class ABr

Specifications of TS4902IST

• Output Type: 1-Channel (Mono)
• Max Output Power X Channels @ Load: 700mW x 1 @ 8 Ohm
• Voltage - Supply: 2.2 V ~ 5.5 V
• Features: Depop, Standby, Thermal Protection
• Mounting Type: Surface Mount
• Package / Case: 8-MSOP, Micro8™, 8-uMAX, 8-uSOP,
• Lead Free Status / RoHS Status: Lead free / RoHS Compliant
• Other names: 497-8116-2; TS4902IST

The TS4902 is a monolithic power amplifier with a
BTL (Bridge Tied Load) output configuration. BTL
means that each end of the load is connected to
two single ended output amplifiers. Thus, we have:
Single ended output 1 = Vout1 = Vout (V)
Single ended output 2 = Vout2 = -Vout (V)
And Vout1 - Vout2 = 2Vout (V)
The output power is :
For the same power supply voltage, the output
power in BTL configuration is four times higher
than
configuration.
In flat region (no effect of Cin), the output voltage
of the first stage is :
The differential output voltage is
The differential gain named gain (Gv) for more
convenient usage is :
Remark : Vout2 is in phase with Vin and Vout1 is
180 phased with Vin. It means that the positive
terminal of the loudspeaker should be connected
to Vout2 and the negative to Vout1.
In low frequency region, the effect of Cin starts.
Cin with Rin forms a high pass filter with a -3dB cut
off frequency
For the second stage : Vout2 = -Vout1 (V)
BTL Configuration Principle
Gain In Typical Application Schematic
(cf. page 1)
Low and high frequency response
the
Vout2
Gv =
output
Pout
Vout1 = Vin
F
C L
Vout2 Vou t1
-------------------------------------- - = 2
–
Vo ut1 = 2Vin
=
------------------------------- -
2 (
Vin
2
–
power
–
Vout
R in Cin
R
1
L
RMS
R fe ed
------------------- - (V)
Rin
in
)
2
Rfee d
------------------- - (V)
(
Rin
single
W
Hz
Rfee d
------------------- -
)
Rin
ended
In high frequency region, you can limit the
bandwidth by adding a capacitor (Cfeed) in
parallel with Rfeed. Its form a low pass filter with a
-3dB cut off frequency
Hypothesis :
(Vout and Iout)
• Supply voltage is a pure DC source (Vcc)
Regarding the load we have :
and
and
Then, the average current delivered by the supply
voltage is:
The power delivered by the supply voltage is
Psupply = Vcc Icc
Then, the power dissipated by the amplifier is
Pdiss = Psupply - Pout (W)
and the maximum value is obtained when
Remark : This maximum value is only depending
on power supply voltage and load values.
• Voltage and current in the load are sinusoidal
and its value is:
Power dissipation and efficiency
P
di ss
F
=
Pdiss
C H
2 2 Vcc
---------------------- P
V
I
=
CC
P
O UT
R
max
O U T
---------------------------------------------- -
I
AVG
2
OU T
AVG
L
= V
--------------------- - = 0
Rfe ed Cfeed
Pdiss
=
(W)
P
= 2
=
V
---------------------- (W)
PEAK
OU T
2
OU T
V
---------------- - (A)
1
Vcc
P E A K
2
2 R
OU T
R
V
------------------- - (A)
R
P E A K
L
L
sin t (V)
L
2
–
R
2
(
P
L
W
O UT
)
Hz
(W)
TS4902
15/19