Datasheet

LM4871
SNAS002F –FEBRUARY 2000–REVISED MAY 2013 www.ti.com
AUDIO POWER AMPLIFIER DESIGN
Design a 1W/8Ω Audio Amplifier
Given:
Power Output 1 Wrms
Load Impedance 8Ω
Input Level 1 Vrms
Input Impedance 20 kΩ
Bandwidth 100 Hz–20 kHz ± 0.25 dB
A designer must first determine the minimum supply rail to obtain the specified output power. By extrapolating
from the Output Power vs Supply Voltage graphs in the Typical Performance Characteristics section, the supply
rail can be easily found. A second way to determine the minimum supply rail is to calculate the required V
opeak
using Equation 4 and add the output voltage. Using this method, the minimum supply voltage would be (V
opeak
+
(V
OD
TOP
+ V
OD
BOT
)), where V
OD
BOT
and V
OD
TOP
are extrapolated from the Dropout Voltage vs Supply Voltage curve in
the Typical Performance Characteristics section.
(4)
Using the Output Power vs Supply Voltage graph for an 8Ω load, the minimum supply rail is 4.6V. But since 5V is
a standard voltage in most applications, it is chosen for the supply rail. Extra supply voltage creates headroom
that allows the LM4871 to reproduce peaks in excess of 1W without producing audible distortion. At this time, the
designer must make sure that the power supply choice along with the output impedance does not violate the
conditions explained in the POWER DISSIPATION section.
Once the power dissipation equations have been addressed, the required differential gain can be determined
from Equation 5.
(5)
R
f
/R
i
= A
VD
/2 (6)
From Equation 5, the minimum A
VD
is 2.83; use A
VD
= 3.
Since the desired input impedance was 20kΩ, and with a A
VD
impedance of 2, a ratio of 1.5:1 of R
f
to R
i
results
in an allocation of R
i
= 20kΩ and R
f
= 30kΩ. The final design step is to address the bandwidth requirements
which must be stated as a pair of −3dB frequency points. Five times away from a −3dB point is 0.17dB down
from passband response which is better than the required ±0.25dB specified.
f
L
= 100Hz/5 = 20Hz
f
H
= 20kHz * 5 = 100kHz
As stated in the External Components Description section, R
i
in conjunction with C
i
create a highpass filter.
C
i
≥ 1/(2π*20kΩ*20Hz) = 0.397µF; use 0.39µF
The high frequency pole is determined by the product of the desired frequency pole, f
H
, and the differential gain,
A
VD
. With a A
VD
= 3 and f
H
= 100kHz, the resulting GBWP = 150kHz which is much smaller than the LM4871
GBWP of 4MHz. This figure displays that if a designer has a need to design an amplifier with a higher differential
gain, the LM4871 can still be used without running into bandwidth limitations.
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