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參數(shù)資料
| 型號： | TPA6211A1TDGNRQ1 |
| 廠商： | TEXAS INSTRUMENTS INC |
| 元件分類： | 音頻／視頻放大 |
| 英文描述： | AUDIO AMPLIFIER, PDSO8 |
| 封裝： | PLASTIC, MSOP-8 |
| 文件頁數(shù)： | 10/22頁 |
| 文件大小： | 424K |
| 代理商： | TPA6211A1TDGNRQ1 |
V(LRMS)
VO
IDD
IDD(avg)
Efficiency of a BTL amplifier +
P
L
P
SUP
Where:
P
L +
V
L
rms2
R
L
, and V
LRMS +
V
P
2
, therefore, P
L +
V
P
2
2R
L
PL = Power delivered to load
PSUP = Power drawn from power supply
VLRMS = RMS voltage on BTL load
RL = Load resistance
VP = Peak voltage on BTL load
IDDavg = Average current drawn from the power supply
VDD = Power supply voltage
ηBTL = Efficiency of a BTL amplifier
and PSUP + VDD IDDavg and IDDavg +
1
p
0
V
P
R
L
sin(t) dt
+ *
1
p
V
P
R
L
[cos(t)]
p
0 +
2V
P
p
R
L
Therefore,
P
SUP +
2 V
DD
V
P
p
R
L
substituting PL and PSUP into equation 6,
Efficiency of a BTL amplifier +
V
P
2
2 R
L
2 V
DD VP
p
R
L
+
p
V
P
4 V
DD
V
P +
2 P
L
R
L
Where:
SBOS555
– JUNE 2011
FULLY DIFFERENTIAL AMPLIFIER
EFFICIENCY AND THERMAL INFORMATION
Class-AB amplifiers are inefficient, primarily because
of voltage drop across the output-stage transistors.
The two components of this internal voltage drop are
the headroom or dc voltage drop that varies inversely
to output power, and the sinewave nature of the
output. The total voltage drop can be calculated by
subtracting the RMS value of the output voltage from
VDD. The internal voltage drop multiplied by the
average
value
of
the
supply
current,
IDD(avg),
determines the internal power dissipation of the
amplifier.
Figure 35. Voltage and Current Waveforms for
An easy-to-use equation to calculate efficiency starts
BTL Amplifiers
out as being equal to the ratio of power from the
power supply to the power delivered to the load. To
accurately calculate the RMS and average values of
Although the voltages and currents for SE and BTL
power in the load and in the amplifier, the current and
are sinusoidal in the load, currents from the supply
voltage waveform shapes must first be understood
are different between SE and BTL configurations. In
(see Figure 35).
an
SE
application
the
current
waveform
is
a
half-wave rectified shape, whereas in BTL it is a
full-wave
rectified
waveform.
This
means
RMS
conversion factors are different. Keep in mind that for
most of the waveform both the push and pull
transistors are not on at the same time, which
supports the fact that each amplifier in the BTL
device only draws current from the supply for half the
waveform. The following equations are the basis for
calculating amplifier efficiency.
(14)
18
Copyright
2011, Texas Instruments Incorporated
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