Datasheet
TPA6204A1
SLOS429B − MAY 2004 − REVISED SEPTEMBER 2009
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 V
DD
. The internal voltage
drop multiplied by the average value of the supply current,
I
DD
(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)
V
O
I
DD
I
DD(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
rms
2
R
L
, and V
LRMS
+
V
P
2
Ǹ
, therefore, P
L
+
V
P
2
2R
L
P
L
= Power delivered to load
P
SUP
= Power drawn from power supply
V
LRMS
= RMS voltage on BTL load
R
L
= Load resistance
V
P
= Peak voltage on BTL load
I
DD
avg = Average current drawn from the
power supply
V
DD
= Power supply voltage
η
BTL
= Efficiency of a BTL amplifier
and
P
SUP
+ V
DD
I
DD
avg
and
I
DD
avg +
1
p
ŕ
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
+
2V
DD
V
P
p R
L
substituting P
L
and P
SUP
into equation 6,
Efficiency of a BTL amplifier +
V
P
2
2R
L
2V
DD
V
P
p R
L
+
p V
P
4V
DD
V
P
+ 2P
L
R
L
Ǹ
Where:
h
BTL
+
p 2P
L
R
L
Ǹ
4V
DD
Therefore,
(4)
(5)