Datasheet
TPA0202
2-W STEREO AUDIO POWER AMPLIFIER
SLOS205B – FEBRUARY 1998 – REVISED DECEMBER 2000
26
POST OFFICE BOX 655303 • DALLAS, TEXAS 75265
APPLICATION INFORMATION
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.
I
DD
rms
2V
P
R
L
P
SUP
V
DD
I
DD
rms
V
DD
2V
P
R
L
Efficiency
P
L
P
SUP
Efficiency of a BTL Configuration
V
P
2V
DD
P
L
R
L
2
1 2
2V
DD
(3)
Where:
(4)
P
L
V
L
rms
2
R
L
V
p
2
2R
L
V
L
rms
V
P
2
Table 1 employs equation 4 to calculate efficiencies for four different output power levels. Note that the efficiency
of the amplifier is quite low for lower power levels and rises sharply as power to the load is increased resulting
in a nearly flat internal power dissipation over the normal operating range. Note that the internal dissipation at
full output power is less than in the half power range. Calculating the efficiency for a specific system is the key
to proper power supply design. For a stereo 1-W audio system with 8-Ω loads and a 5-V supply, the maximum
draw on the power supply is almost 3.25 W.
Table 1. Efficiency Vs Output Power in 5-V 8-Ω BTL Systems
OUTPUT POWER
(W)
EFFICIENCY
(%)
PEAK-TO-PEAK
VOLTAGE
(V)
INTERNAL
DISSIPATION
(W)
0.25 31.4 2.00 0.55
0.50 44.4 2.83 0.62
1.00 62.8 4.00 0.59
1.25 70.2 4.47
†
0.53
†
High peak voltages cause the THD to increase.
A final point to remember about linear amplifiers (either SE or BTL) is how to manipulate the terms in the
efficiency equation to utmost advantage when possible. Note that in equation 4, V
DD
is in the denominator. This
indicates that as V
DD
goes down, efficiency goes up.