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參數(shù)資料
| 型號(hào)： | TPA0211DGNRG4 |
| 廠商： | TEXAS INSTRUMENTS INC |
| 元件分類： | 音頻／視頻放大 |
| 英文描述： | 2 W, 1 CHANNEL, AUDIO AMPLIFIER, PDSO8 |
| 封裝： | GREEN, PLASTIC, MSOP-8 |
| 文件頁數(shù)： | 8/27頁 |
| 文件大?。?/td> | 770K |
| 代理商： | TPA0211DGNRG4 |
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TPA0211
2-W MONO AUDIO POWER AMPLIFIER
SLOS275D – JANUARY 2000 – REVISED NOVEMBER 2002
16
POST OFFICE BOX 655303
DALLAS, TEXAS 75265
APPLICATION INFORMATION
BTL amplifier efficiency (continued)
Although the voltages and currents for SE and BTL are sinusoidal in the load, currents from the supply are very
different between SE and BTL configurations. In 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.
Efficiency of a BTL amplifier
+
P
L
P
SUP
(9)
where
(10)
P
L +
V
LRMS
2
R
L
, and V
LRMS +
V
P
2
, therefore, P
L +
V
P
2
2R
L
PL = Power devilered to load
PSUP = Power drawn from power supply
VLRMS = RMS voltage on BTL load
RL = Load resistance
and PSUP + VDD IDDavg
and
I
DDavg +
1
p
0
V
P
R
L
sin(t) dt
+ 1p
V
P
R
L
[cos(t)]
p
0 +
2V
P
p R
L
therefore,
P
SUP +
2V
DD VP
p R
L
substituting PL and PSUP into equation 9,
Efficiency of a BTL amplifier
+
V
P
2
2R
L
2V
DD VP
p R
L
+
p V
P
4V
DD
V
P +
2P
L RL
h
BTL +
p 2P
L RL
4V
DD
where
therefore,
VP = Peak voltage on BTL load
IDDavg = Average current drawn from the power supply
VDD = Power supply voltage
ηBTL = Efficiency of a BTL amplifier
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