Datasheet

www.ti.com
R
L
C
C
V
O(PP)
V
O(PP)
V
DD
-3 dB
f
c
BTL AMPLIFIER EFFICIENCY
V
L(RMS)
V
O
I
DD
I
DD(RMS)
TPA321
SLOS312C – JUNE 2000 – REVISED JUNE 2004
APPLICATION INFORMATION (continued)
Figure 24. Single-Ended Configuration and Frequency Response
Increasing power to the load does carry a penalty of increased internal power dissipation. The increased
dissipation is understandable considering that the BTL configuration produces 4× the output power of a SE
configuration. Internal dissipation versus output power is discussed further in the thermal considerations section.
Linear 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 sine-wave 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 RMS value of the supply current, I
DD(RMS)
, 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 values of power in the load and in the
amplifier, the current and voltage waveform shapes must first be understood (see Figure 25 ).
Figure 25. 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.
14