參數(shù)資料
型號(hào): TPA6030A4PWP
廠商: TEXAS INSTRUMENTS INC
元件分類: 音頻控制
英文描述: 2 CHANNEL(S), VOLUME CONTROL CIRCUIT, PDSO28
封裝: GREEN, PLASTIC, TSSOP-28
文件頁(yè)數(shù): 15/32頁(yè)
文件大小: 698K
代理商: TPA6030A4PWP
TPA6030A4
SLOS395B DECEMBER 2002 REVISED JANUARY 2005
www.ti.com
22
THERMAL INFORMATION
Class-AB amplifiers are inefficient. The primary cause of these inefficiencies is a 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 VCC. The internal voltage drop multiplied by the
average value of the supply current, ICCavg, 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 27).
V(LRMS)
VO
ICC
ICCavg
Figure 27. 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 waveform, 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 +
PL
PSUP
where:
PL +
VLRMS
2
RL
, and VLRMS +
VP
2
, therefore, PL +
VP
2
2RL
per channel
PSUP + VCCICCavg ) VCC ICC(q)
and ICCavg + 1p
p
0
VP
RL
sin(t) dt +*
VP
pRL
[cos(t)]
p
0 +
2VP
pRL
therefore,
PSUP +
2VCC VP
pRL
) VCC ICC(q)
PL = Power delivered to load (per channel)
PSUP = Power drawn from power supply
VLRMS = RMS voltage on BTL load
RL = Load resistance
VP = Peak voltage on BTL load
ICCavg = Average current drawn from the
power supply
VCC = Power supply voltage
ηBTL = Efficiency of a BTL amplifier
ηSE = Efficiency of a SE amplifier
where VP + 2PL RL
(10)
(11)
(12)
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