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TPA6204A1 Folha de dados(PDF) 11 Page - Texas Instruments

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Nome de Peças TPA6204A1
Descrição Electrónicos  1.7W MONO FULLY DIFFERENTIAL AUDIO POWER AMPLIFIER
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Fabricante Electrônico  TI [Texas Instruments]
Página de início  http://www.ti.com
Logo TI - Texas Instruments

TPA6204A1 Folha de dados(HTML) 11 Page - Texas Instruments

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TPA6204A1
SLOS429 − MAY 2004
www.ti.com
11
FULLY DIFFERENTIAL AMPLIFIER
EFFICIENCY AND THERMAL INFORMATION
Class-AB amplifiers are inefficient. The primary cause of
these inefficiencies is 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 average value of the supply current,
IDD(avg), 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 18).
V(LRMS)
VO
IDD
IDD(avg)
Figure 18. 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
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
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
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
V
P
p R
L
substituting PL and PSUP into equation 6,
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
R
L
Where:
h
BTL +
p 2P
L
R
L
4V
DD
Therefore,
(4)
(5)


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