lm4898mmx National Semiconductor Corporation, lm4898mmx Datasheet - Page 14

• Manufacturer Part Number: lm4898mmx
• Description: 1 Watt Fully Differential Audio Power Amplifier With Shutdown Select
• Manufacturer: National Semiconductor Corporation

www.national.com
Application Information
AUDIO POWER AMPLIFIER DESIGN
Design a 1W/8Ω Audio Amplifier
Given:
A designer must first determine the minimum supply rail to
obtain the specified output power. The supply rail can easily
be found by extrapolating from the Output Power vs. Supply
Voltage graphs in the Typical Performance Characteristics
section. A second way to determine the minimum supply rail
is to calculate the required Vopeak using Equation 7 and add
the dropout voltages. Using this method, the minimum sup-
ply voltage is (Vopeak +(V
BOT
vs. Supply Voltage curve in the Typical Performance Char-
acteristics section.
Using the Output Power vs. Supply Voltage graph for an 8Ω
load, the minimum supply rail just about 5V. Extra supply
voltage creates headroom that allows the LM4898 to repro-
• Power Output
• Load Impedance
• Input Level
• Input Impedance
• Bandwidth
and V
DO TOP
are extrapolated from the Dropout Voltage
DO TOP
+(V
100Hz–20kHz
DO BOT
(Continued)
)),where V
±
0.25dB
1Vrms
20kΩ
1W
8Ω
(7)
DO
14
duce peaks in excess of 1W without producing audible dis-
tortion. 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 8.
R
From Equation 8, the minimum A
sired input impedance was 20kΩ, a ratio of 2.83:1 of R
results in an allocation of R
and R
step is to address the bandwidth requirement which must be
stated as a single -3dB frequency point. Five times away
from a -3dB point is 0.17dB down from passband response
which is better than the required
f
The high frequency pole is determined by the product of the
desired frequency pole, f
.With a A
150kHz which is much smaller than the LM4898 GBWP of
10MHz. This figure displays that if a designer has a need to
design an amplifier with a higher differential gain, the
LM4898 can still be used without running into bandwidth
limitations.
H
f
= 20kHz * 5 =100kHz
/ R
i
f
= 60kΩ for both feedback resistors. The final design
= A
VD
VD
= 2.83 and f
H
H
= 100kHz, the resulting GBWP =
i
, and the differential gain, A
= 20kΩ for both input resistors
±
VD
0.25dB specified.
is 2.83. Since the de-
f
to R
(8)
VD
i