參數(shù)資料
型號(hào): TPA0233DGQR
廠商: TEXAS INSTRUMENTS INC
元件分類(lèi): 音頻/視頻放大
英文描述: 2 W, 2 CHANNEL, AUDIO AMPLIFIER, PDSO10
封裝: GREEN, PLASTIC, MSOP-10
文件頁(yè)數(shù): 8/26頁(yè)
文件大?。?/td> 711K
代理商: TPA0233DGQR
TPA0233
2W MONO AUDIO POWER AMPLIFIER
WITH HEADPHONE DRIVE
SLOS278D JANUARY 2000 REVISED NOVEMBER 2002
16
POST OFFICE BOX 655303
DALLAS, TEXAS 75265
APPLICATION INFORMATION
BTL amplifier efficiency
Class-AB amplifiers are inefficient. The primary cause of inefficiencies is the voltage drop across the output
stage transistors. There are two components of the internal voltage drop. One is the headroom or dc voltage
drop that varies inversely to output power. The second component is due to 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 RMS value of the supply current, IDDrms, determines the internal power
dissipation of the amplifier.
An easy-to-use equation to calculate efficiency starts 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 power in the
load and in the amplifier, the current and voltage waveform shapes must first be understood. See Figure 26.
V(LRMS)
VO
IDD
IDD(avg)
Figure 26. Voltage and Current Waveforms for BTL Amplifiers
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
P
L +
V
LRMS
2
R
L
, and V
LRMS +
V
P
2
, therefore, P
L +
V
P
2
2R
L
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
where
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