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LM4898LD bảng dữ liệu(PDF) 14 Page - National Semiconductor (TI)

[Old version datasheet] Texas Instruments acquired National semiconductor.
tên linh kiện LM4898LD
Giải thích chi tiết về linh kiện  1 Watt Fully Differential Audio Power Amplifier With Shutdown Select
PDF  17 Pages
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nhà sản xuất  NSC [National Semiconductor (TI)]
Trang chủ  http://www.national.com
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LM4898LD bảng dữ liệu(HTML) 14 Page - National Semiconductor (TI)

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Application Information (Continued)
AUDIO POWER AMPLIFIER DESIGN
Design a 1W/8
Ω Audio Amplifier
Given:
•
Power Output
1W
•
Load Impedance
8
Ω
•
Input Level
1Vrms
•
Input Impedance
20k
Ω
•
Bandwidth
100Hz–20kHz ± 0.25dB
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
DO TOP+(VDO BOT )),where VDO
BOT
and V
DO TOP are extrapolated from the Dropout Voltage
vs. Supply Voltage curve in the Typical Performance Char-
acteristics section.
(7)
Using the Output Power vs. Supply Voltage graph for an 8
Ω
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 /Ri =AVD
From Equation 8, the minimum A
VD is 2.83. Since the de-
sired input impedance was 20k
Ω, a ratio of 2.83:1 of R
f to Ri
results in an allocation of R
i = 20k
Ω for both input resistors
and R
f= 60k
Ω 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, AVD
.With a A
VD = 2.83 and fH = 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.
www.national.com
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