参数资料
型号: LM4871MDA
厂商: NATIONAL SEMICONDUCTOR CORP
元件分类: 音频/视频放大
英文描述: 1.5 W, 1 CHANNEL, AUDIO AMPLIFIER, UUC
封装: DIE
文件页数: 14/15页
文件大小: 454K
代理商: LM4871MDA
Application Information (Continued)
to turn-on clicks and pops. Thus, a value of C
B equal to
1.0F is recommended in all but the most cost sensitive
designs.
AUDIO POWER AMPLIFIER DESIGN
Design a 1W/8
Audio Amplifier
Given:
Power Output
1 Wrms
Load Impedance
8
Input Level
1 Vrms
Input Impedance
20 k
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 Per-
formance Characteristics section, the supply rail can be
easily found. A second way to determine the minimum sup-
ply rail is to calculate the required V
opeak using Equation 3
and add the output voltage. Using this method, the minimum
supply voltage would be (V
opeak +(VODTOP +VODBOT)), where
V
ODBOT and VODTOP are extrapolated from the Dropout Volt-
age vs Supply Voltage curve in the Typical Performance
Characteristics section.
(3)
Using the Output Power vs Supply Voltage graph for an 8
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 al-
lows the LM4871 to reproduce peaks in excess of 1W with-
out 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 Equa-
tion 4.
(4)
R
f/Ri =AVD/2
(5)
From Equation 4, the minimum A
VD is 2.83; use AVD =3.
Since the desired input impedance was 20k
, and with a
A
VD impedance of 2, a ratio of 1.5:1 of Rf to Ri results in an
allocation of R
i = 20k and Rf = 30k. 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 section, R
i in con-
junction with C
i create a highpass filter.
C
i ≥ 1/(2π*20k*20Hz) = 0.397F; use 0.39F
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 = 3 and fH = 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.
LM4871
www.national.com
8
相关PDF资料
PDF描述
LM4871MWA 1.5 W, 1 CHANNEL, AUDIO AMPLIFIER, UUC
LM4871N/NOPB 1.5 W, 1 CHANNEL, AUDIO AMPLIFIER, PDIP8
LM4872IBPX/NOPB 1 W, 1 CHANNEL, AUDIO AMPLIFIER, PBGA8
LM4872IBP/NOPB 1 W, 1 CHANNEL, AUDIO AMPLIFIER, PBGA8
LM4873MDC 1.1 W, 2 CHANNEL, AUDIO AMPLIFIER, UUC
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