參數(shù)資料
型號(hào): LM4898
廠商: National Semiconductor Corporation
英文描述: 1 Watt Fully Differential Audio Power Amplifier With Shutdown Select
中文描述: 1瓦特全差分音頻功率放大器,具有關(guān)斷選擇
文件頁數(shù): 14/17頁
文件大?。?/td> 596K
代理商: LM4898
Application Information
(Continued)
AUDIO POWER AMPLIFIER DESIGN
Design a 1W/8
Audio Amplifier
Given:
Power Output
Load Impedance
Input Level
Input Impedance
Bandwidth
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
+(V
)),where V
BOT
and V
are extrapolated from the Dropout Voltage
vs. Supply Voltage curve in the Typical Performance Char-
acteristics section.
1W
8
1Vrms
20k
100Hz–20kHz
±
0.25dB
(7)
Using the Output Power vs. Supply Voltage graph for an 8
load, the minimum supply rail just about 5V. Extra supply
voltage creates headroom that allows the LM4898 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 Dissipation
section. Once the power dissipation equations have been
addressed, the required differential gain can be determined
from Equation 8.
(8)
R
f
/ R
i
= A
VD
From Equation 8, the minimum A
VD
is 2.83. Since the de-
sired input impedance was 20k
, a ratio of 2.83:1 of R
f
to R
i
results in an allocation of R
i
= 20k
for both input resistors
and R
= 60k
for both feedback resistors. The final design
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
±
0.25dB specified.
f
H
= 20kHz * 5 =100kHz
The high frequency pole is determined by the product of the
desired frequency pole, f
H
, and the differential gain, A
.With a A
= 2.83 and f
= 100kHz, the resulting GBWP =
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.
L
www.national.com
14
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