lm4894itlx National Semiconductor Corporation, lm4894itlx Datasheet - Page 14

• Manufacturer Part Number: lm4894itlx
• Description: 1 Watt Fully Differential Audio Power Amplifier With Shutdown Select
• Manufacturer: National Semiconductor Corporation

www.national.com
Application Information
Similar results would occur if the input resistors were not
carefully matched. Adding input coupling capacitors in be-
tween the signal source and the input resistors will eliminate
this problem, however, to achieve best performance with
minimum component count it is highly recommended that
both the feedback and input resistors matched to 1% toler-
ance or better.
AUDIO POWER AMPLIFIER DESIGN
Design a 1W/8Ω Audio Amplifier
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 Characteris-
tics 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 mini-
mum supply voltage is (Vopeak +(V
where V
Dropout Voltage vs Supply Voltage curve in the Typical
Performance Characteristics section.
Tolerance R
20%
10%
5%
1%
0%
Given:
Power Output
Load Impedance
Input Level
Input Impedance
Bandwidth
DO BOT
0.8R
0.9R
0.95R 1.05R -0.125V
0.99R 1.01R -0.025V
R
F1
and V
R
1.2R
1.1R
R
F2
DO TOP
V
-0.500V
-0.250V
0
02
- V
are extrapolated from the
01
DO TOP
–20kHz
(Continued)
I
62.5mA
31.25mA
15.63mA
3.125mA
0
+(V
LOAD
±
DO BOT
1Wrms
0.25dB
1Vrms
20kΩ
8Ω
)),
14
Using the Output Power vs Supply Voltage graph for an 8W
load, the minimum supply rail just about 5V. Extra supply
voltage creates headroom that allows the LM4894 to repro-
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 Dissipa-
tion section. Once the power dissipation equations have
been addressed, the required differential gain can be deter-
mined from Equation 7.
From Equation 7, 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
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 LM4894 GBWP of
10MHz. This figure displays that if a designer has a need to
design an amplifier with a higher differential gain, the
LM4894 can still be used without running into bandwidth
limitations.
R
f
H
f
f
= 60kΩ for both feedback resistors. The final design
= 20kHz * 5 =100kHz
/ R
VD
i
= A
= 2.83 and f
VD
H
H
= 100kHz, the resulting GBWP =
, and the differential gain, A
i
= 20kΩ for both input resistors
±
VD
0.25dB specified.
is 2.83. Since the de-
f
to R
VD
(7)
(8)
.
i