TPA0232 Datasheet, PDF (23/30 Pages) Texas Instruments – STEREO 2-W AUDIO POWER AMPLIFIER WITH DC VOLUME CONTROL AND MUX CONTROL

TPA0232
STEREO 2-W AUDIO POWER AMPLIFIER
WITH DC VOLUME CONTROL AND MUX CONTROL
SLOS286 – NOVEMBER 1999
APPLICATION INFORMATION
BTL amplifier efficiency
Class-AB amplifiers are notoriously 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 RMS value of the supply current, I_DDrms, 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 35).
VO
IDD
V(LRMS)
IDD(avg)
Figure 35. 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
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
(7)
Where:
and
+ + _Ǹ + P_L
+ + ŕ + P_SUP
V_Lrms²
R_L
,
and V_LRMS
V_DD I_DDavg _and
V_{P ,}
2
therefore,
P_L
p
I_DDavg
1
p
0
V_P2
2 R_L
V_P
R_L
sin(t)
dt
1
p
Therefore,
+ ^VP
R_L
p
[cos(t)] ₀
2V_P
p R_L
+ P_SUP
2 V_DD V_P
p R_L
substituting P_L and P_SUP into equation 7,
V_P²
+ ₊ Efficiency of a BTL amplifier
2 R_L
p V_P
+ Ǹ Where:
V_P
2 P_L R_L
2 V_DD V_P
p R_L
4 V_DD
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_DDavg = Average current drawn from
the power supply
V_DD = Power supply voltage
η_BTL = Efficiency of a BTL amplifier
+ Ǹ Therefore,
h_BTL
p 2 P_L R_L
4 V_DD
(8)
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