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TS419 数据表(PDF) 27 Page - STMicroelectronics

部件名 TS419
功能描述  360mW MONO AMPLIFIER WITH STANDBY MODE
PDF  32 Pages
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制造商  STMICROELECTRONICS [STMicroelectronics]
网页  http://www.st.com
标志 STMICROELECTRONICS - STMicroelectronics

TS419 数据表(HTML) 27 Page - STMicroelectronics

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TS419-TS421
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APPLICATION INFORMATION
s BTL Configuration Principle
The TS419 & TS420 are monolithic power
amplifiers with a BTL output type. BTL (Bridge
Tied Load) 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
the
output
power
in
single
ended
configuration.
s Gain In Typical Application Schematic
(cf. page 3 of TS419-TS421 datasheet)
In the flat region (no CIN effect), the output voltage
of the first stage is:
For the second stage : Vout2 = -Vout1 (V)
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
phased 180
° with Vin. This means that the positive
terminal of the loudspeaker should be connected
to Vout2 and the negative to Vout1.
s Low and high frequency response
In the low frequency region, CIN starts to have an
effect. CIN forms with RIN a high-pass filter with a
-3dB cut off frequency .
In the high frequency region, you can limit the
bandwidth by adding a capacitor (Cfeed) in
parallel with Rfeed. It forms a low-pass filter with a
-3dB cut off frequency .
s Power dissipation and efficiency
Hypothesis:
• Load voltage and current are sinusoidal (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 IccAVG (W)
Then, the power dissipated by the amplifier is:
Pdiss = Psupply - Pout (W)
and the maximum value is obtained when:
and its value is:
Remark : This maximum value is only dependent
upon power supply voltage and load values.
)
W
(
R
)
Vout
2
(
Pout
L
2
RMS
=
)
V
(
Rin
Rfeed
Vin
1
Vout
−
=
)
V
(
Rin
Rfeed
Vin
2
1
Vout
2
Vout
=
−
Rin
Rfeed
2
Vin
1
Vout
2
Vout
Gv
=
−
=
(Hz)
RinCin
2
1
FCL
π
=
)
Hz
(
Cfeed
Rfeed
2
1
FCH
π
=
)
V
(
t
sin
V
V
PEAK
OUT
ω
=
)
A
(
R
V
I
L
OUT
OUT =
)
W
(
R
2
V
P
L
2
PEAK
OUT =
)
A
(
R
V
2
Icc
L
PEAK
AVG
π
=
)
W
(
P
P
R
Vcc
2
2
Pdiss
OUT
OUT
L
−
π
=
0
P
Pdiss
OUT
=
∂
∂
)
W
(
R
Vcc
2
max
Pdiss
L
2
2
π
=



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